Transcript
Conjectures and proof Book page 24 - 30
copycgrahamphysicscom 2016
What is a conjecture bull A conjecture is used to describe a pattern in
mathematical terms
bull When a conjecture has been proved it becomes a theorem
bull There are many types of proofs some of which are - logical deductive reasoning - direct proof - induction
copycgrahamphysicscom 2016
Alan Turing
bull English Mathematician who broke the enigma code in WW2
bull Alan Turing - Celebrating the life of a genius
copycgrahamphysicscom 2016
Direct proof bull A direct proof can be checked by an established method
bull Using our knowledge of odd and even numbers we can proof that the sum of two odd numbers is always even
bull Odd number 2k + 1 Even number 2k
bull Let 119886 = 2119896 + 1 119886119899119889 119887 = 2119897 + 1
bull 2119896 + 1 + 2119897 + 1 = 2119896 + 2119897 + 2 = 2(119896 + 119897 + 1)
bull Let 119896 + 119897 + 1 = 119904 119904 isin 119937 119886 + 119887 = 2119904 which is an even number
copycgrahamphysicscom 2016
Example 2 bull Consider the pattern 1 = 12 1 + 3 = 22 1 + 3 + 5 = 32 1 + 3 + 5 + 7 = 42
bull The pattern shows that the sum of the first two positive odd integers is a perfect square the sum of the first three positive odd integers is a perfect square and the sum of the first four positive odd integers is a perfect square
bull Can we then say based on the first few lines of this pattern that
bull 1 + 3 + 5 + 7 + hellip + (2n ndash 1) = 1198992 bull ie the sum of the first n odd positive integers is a perfect
square
copycgrahamphysicscom 2016
Solution bull This can be checked as the positive odd integers form an
arithmetic progression with a = 1 and d = 2 The sum of the first n terms is given by
bull In this example we had a method to verify our guess
bull This will not always be the case
copycgrahamphysicscom 2016
Exercise 1G page 24 all
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
What is a conjecture bull A conjecture is used to describe a pattern in
mathematical terms
bull When a conjecture has been proved it becomes a theorem
bull There are many types of proofs some of which are - logical deductive reasoning - direct proof - induction
copycgrahamphysicscom 2016
Alan Turing
bull English Mathematician who broke the enigma code in WW2
bull Alan Turing - Celebrating the life of a genius
copycgrahamphysicscom 2016
Direct proof bull A direct proof can be checked by an established method
bull Using our knowledge of odd and even numbers we can proof that the sum of two odd numbers is always even
bull Odd number 2k + 1 Even number 2k
bull Let 119886 = 2119896 + 1 119886119899119889 119887 = 2119897 + 1
bull 2119896 + 1 + 2119897 + 1 = 2119896 + 2119897 + 2 = 2(119896 + 119897 + 1)
bull Let 119896 + 119897 + 1 = 119904 119904 isin 119937 119886 + 119887 = 2119904 which is an even number
copycgrahamphysicscom 2016
Example 2 bull Consider the pattern 1 = 12 1 + 3 = 22 1 + 3 + 5 = 32 1 + 3 + 5 + 7 = 42
bull The pattern shows that the sum of the first two positive odd integers is a perfect square the sum of the first three positive odd integers is a perfect square and the sum of the first four positive odd integers is a perfect square
bull Can we then say based on the first few lines of this pattern that
bull 1 + 3 + 5 + 7 + hellip + (2n ndash 1) = 1198992 bull ie the sum of the first n odd positive integers is a perfect
square
copycgrahamphysicscom 2016
Solution bull This can be checked as the positive odd integers form an
arithmetic progression with a = 1 and d = 2 The sum of the first n terms is given by
bull In this example we had a method to verify our guess
bull This will not always be the case
copycgrahamphysicscom 2016
Exercise 1G page 24 all
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Alan Turing
bull English Mathematician who broke the enigma code in WW2
bull Alan Turing - Celebrating the life of a genius
copycgrahamphysicscom 2016
Direct proof bull A direct proof can be checked by an established method
bull Using our knowledge of odd and even numbers we can proof that the sum of two odd numbers is always even
bull Odd number 2k + 1 Even number 2k
bull Let 119886 = 2119896 + 1 119886119899119889 119887 = 2119897 + 1
bull 2119896 + 1 + 2119897 + 1 = 2119896 + 2119897 + 2 = 2(119896 + 119897 + 1)
bull Let 119896 + 119897 + 1 = 119904 119904 isin 119937 119886 + 119887 = 2119904 which is an even number
copycgrahamphysicscom 2016
Example 2 bull Consider the pattern 1 = 12 1 + 3 = 22 1 + 3 + 5 = 32 1 + 3 + 5 + 7 = 42
bull The pattern shows that the sum of the first two positive odd integers is a perfect square the sum of the first three positive odd integers is a perfect square and the sum of the first four positive odd integers is a perfect square
bull Can we then say based on the first few lines of this pattern that
bull 1 + 3 + 5 + 7 + hellip + (2n ndash 1) = 1198992 bull ie the sum of the first n odd positive integers is a perfect
square
copycgrahamphysicscom 2016
Solution bull This can be checked as the positive odd integers form an
arithmetic progression with a = 1 and d = 2 The sum of the first n terms is given by
bull In this example we had a method to verify our guess
bull This will not always be the case
copycgrahamphysicscom 2016
Exercise 1G page 24 all
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Direct proof bull A direct proof can be checked by an established method
bull Using our knowledge of odd and even numbers we can proof that the sum of two odd numbers is always even
bull Odd number 2k + 1 Even number 2k
bull Let 119886 = 2119896 + 1 119886119899119889 119887 = 2119897 + 1
bull 2119896 + 1 + 2119897 + 1 = 2119896 + 2119897 + 2 = 2(119896 + 119897 + 1)
bull Let 119896 + 119897 + 1 = 119904 119904 isin 119937 119886 + 119887 = 2119904 which is an even number
copycgrahamphysicscom 2016
Example 2 bull Consider the pattern 1 = 12 1 + 3 = 22 1 + 3 + 5 = 32 1 + 3 + 5 + 7 = 42
bull The pattern shows that the sum of the first two positive odd integers is a perfect square the sum of the first three positive odd integers is a perfect square and the sum of the first four positive odd integers is a perfect square
bull Can we then say based on the first few lines of this pattern that
bull 1 + 3 + 5 + 7 + hellip + (2n ndash 1) = 1198992 bull ie the sum of the first n odd positive integers is a perfect
square
copycgrahamphysicscom 2016
Solution bull This can be checked as the positive odd integers form an
arithmetic progression with a = 1 and d = 2 The sum of the first n terms is given by
bull In this example we had a method to verify our guess
bull This will not always be the case
copycgrahamphysicscom 2016
Exercise 1G page 24 all
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example 2 bull Consider the pattern 1 = 12 1 + 3 = 22 1 + 3 + 5 = 32 1 + 3 + 5 + 7 = 42
bull The pattern shows that the sum of the first two positive odd integers is a perfect square the sum of the first three positive odd integers is a perfect square and the sum of the first four positive odd integers is a perfect square
bull Can we then say based on the first few lines of this pattern that
bull 1 + 3 + 5 + 7 + hellip + (2n ndash 1) = 1198992 bull ie the sum of the first n odd positive integers is a perfect
square
copycgrahamphysicscom 2016
Solution bull This can be checked as the positive odd integers form an
arithmetic progression with a = 1 and d = 2 The sum of the first n terms is given by
bull In this example we had a method to verify our guess
bull This will not always be the case
copycgrahamphysicscom 2016
Exercise 1G page 24 all
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Solution bull This can be checked as the positive odd integers form an
arithmetic progression with a = 1 and d = 2 The sum of the first n terms is given by
bull In this example we had a method to verify our guess
bull This will not always be the case
copycgrahamphysicscom 2016
Exercise 1G page 24 all
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Exercise 1G page 24 all
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Proof by induction copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Problem direct proof is not useful bull Pascalrsquos triangle is named after the French Mathematician
Blaise Pascal in 1653
bull The first rows are shown below
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Pascalrsquos Triangle
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Things to think about a) Find the sum of the numbers in
I row 0 ii row 1 iii row 2 iv row 3 v row 4
b) What do you suspect is the sum of the numbers in row n
c) Suppose your suggestion in (b) is true for the nth row of Pascalrsquos triangle i) can you then show that it is true for the (n + 1)th row as well ii) How do we know we have proven a statement to be true
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Proof by induction bull Proof by induction is a mathematical way to prove
statements about sequences
bull Mathematical Induction
bull Imagine n ndash dominos arranged in a row in the same way
If the first falls the second one will fall ndash 2nd step
N dominoes added at end will also fall ndash 3rd step
We can add as many dominos as we like they will fall if the 1st is tilted towards the 2nd
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
This process can be summarized Step 1 the first expression must be true (the first domino falls) Step 2 assuming that a general expression is true (assume that some domino in the series falls) Step 3 prove that the next expression is true (prove that the next domino in the series falls) Step 4 if all of these events happen then we know by induction that all of the expressions are true and thus the original formula is true (all the dominoes will fall)
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Proof by induction
The domino analogy can be applied to mathematics
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Strong and weak induction bull Weak induction uses if p(k) is true then p(k+1) is true while in
strong induction you use if p(i) is true for all i less than or equal to k then p(k+1) is true where p(k) is some statement depending on the positive integer k
bull They are NOT identical but they are equivalent
bull It is easy to see that if weak induction is true then strong induction is true if you know that statement p(i) is true for all i less than or equal to k then you know that it is true in particular for i=k and can use simple induction
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
bull It is harder to prove but still true that if strong induction is true then weak induction is true That is what we mean by equivalent
bull The reason this distinction remains is that it serves a pedagogical purpose The first proofs by induction that we teach are usually sequences
bull The proofs of these naturally suggest weak induction which students learn as a pattern to mimic
bull Later we teach more difficult proofs where that pattern no longer works To give a name to the difference we call the new pattern strong induction so that we can distinguish between the methods when presenting a proof in lecture Then we can tell a student try using strong induction which is more helpful than just try using induction
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example 1 bull Proof by induction that the sum of the first n odd numbers
is 1198992
bull That is 1 + 3 + 5 + 7 + ⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
Solution
bull First we need to state what our proposition is We do this as follows Let P(n) be the proposition that 1 + 3 + 5 + 7 +⋯ + 2119899 minus 1 = 1198992 forall119899 ge 1
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Using the 3 steps
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example 2
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example 3
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Answer
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Plan A didnrsquot work what is YOUR plan B
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
What do you know
Three steps show true for base case Assume true for n = k RTS true for n = k+1
Two different types 1 Use sequences and add the next term to show n = k + 1 is true copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Type 1 proof - easy
bull Prove that 1 + 5 + 9 + hellip+ (4n ndash 3) = n(2n ndash 1)
bull Show true for n = 1
bull Assume true for n = k 4k ndash 3 = k (2k ndash 1)
bull RTS using the assumption true for n = k + 1
bull 4 (k + 1) ndash 3 = (k + 1) [2(k +1)-1] = (k+ 1)(2k + 1)
bull Rewrite left side adding the last two terms of the sequence
bull (4k ndash 3) + (4(k + 1) ndash 3) and replace (4k ndash 3) with assumption copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Type 2 proof ndash more difficult
bull Manipulate the expression
bull Rewrite sums to get the sum of the assumption
bull Insert bracket to get common factor
bull Regroup to factor
bull Add and subtract the same number
bull Keep in mind the expression you want to prove
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Yesterdayrsquos example
bull Prove 1198992 gt 7119899 + 1 forall119899 ge 8
bull Easier solution Rewrite the equation
bull 1198992 minus 7119899 minus 1 gt 0 bull Flight Of The Phoenix
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example
bull Prove by induction that3119899 gt 1 + 2119899 forall119899 gt 1
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Another way to prove ndash read at home
bull You chose which method you prefer
bull 3119899 ge 1 + 2119899 forall119899 isin 119885 119899 ge 0
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Try this on your own
bull Prove by induction that 2119899 ge 1 + 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Another example
bull Prove by induction that 2119899 ge 119899 forall119899 ge 1
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Can you do this bull 8119899 ge 1198993 forall119899 ge 1
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
An easy example
bull 5 + 10 + 15 + ⋯ 5119899 =5
2119899(119899 + 1)
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example 4
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
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Example 5
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copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
More on sequence proofs
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example 5
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
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copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Practice
Book page 29 exercise 1H 2
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example 6 bull Prove by induction that 9119899 minus 1 is divisible by 8 for all
n ge 1 where n is an integer
bull Solution
bull Let P(n) be the proposition that 9119899 minus 1 is divisible by 8 for all n ge 1 where n is an integer
bull Step 1 This is true when n = 1 since 91 minus 1 = 8 which is divisible by 8
bull Step 2 Assume P(n) is true for n = k Assume true for 9119896 minus 1 = 8119898 where m is an integer
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Example 7
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
Practice
Book page 30 exercise 1I all
copycgrahamphysicscom 2016
hellip continued
copycgrahamphysicscom 2016
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