EE1J2 - Slide 1 EE1J2 – Discrete Maths Lecture 12 Number theory Mathematical induction Proof by induction Examples.

EE1J2 - Slide 1

EE1J2 – Discrete Maths Lecture 12

Number theory Mathematical induction Proof by induction Examples

EE1J2 - Slide 2

Integers and Number Theory

Recall that the ℤ denotes the set of integers {…-2, -1, 0, 1, 2…}

We take for granted the fact that it x and y are integers, then x + y and x y are also integers

Mathematicians say that the set of integers is closed under addition and multiplication

Number Theory is concerned with properties of the integers with the operations ‘+’ and ‘’

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Properties of + and ℤ together with + and has several properties

which we take for granted: If x=a and y=b then x+a=y+b and xy=ab a+b= b+a and a b=b a (Commutative Laws) (a+b)+c=a+(b+c) and (a+b)+c=a+(b+c)

(Associative Laws) a (b+c)=a b + a c (Distributive Law)

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Properties of + and (cont.) There exist unique integers 0 and 1 such that

a , ℤ a+0=0+a=a and a 1=1a=a 0 is called the additive identity 1 is called the multiplicative identity

For every a there is an integer –a such that a+(-a)=(-a)+a=0 (-a is the additive inverse of a)

If a b=a c and a0, then b=c

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{ ,+, ℤ }

Sometimes use the notation { ,+, ℤ } as a reminder that number theory is concerned not just with the integers, but also with + and

{ ,+, ℤ } is a common domain of discourse for Predicate Logic

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Example

Suppose P(x) = ‘x is even’ Then,

x:(P(x) P(x2))

x:(P(x2) P(x))

are statements in Predicate Logic for which the domain of discourse is { ,+, ℤ }

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Proofs in Number Theory

Consider the statement x:(P(x) P(x2)) A formal proof of the validity of this statement

requires many steps (see, for example, Anderson, p 106)

In practice many of these steps are missed out and a typical ‘acceptable’ proof might look as follows:

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Proof of x:(P(x) P(x2))

Let x be even Then there exists yℤ such that x=2y (this is

the definition of an even number) Therefore x2=(2y)2=(2y) (2y)

=2(2y2) So, if z= 2y2 then x2=2z Therefore x2 is even.

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Proof of x:(P(x2) P(x))

In this case we’ll use ‘Equivalence of Contrapositive’

I.e. a b b a So, proving P(x2) P(x) is equivalent to

proving P(x) P(x2)

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Proof of P(x) P(x2) Assume P(x). In other words x is odd Then there exists yℤ such that x=2y+1 So, x2=(2y+1)2=4y2 + 4y + 1 Let z=2y2 + 2y Then x2=2z + 1, so x2 is odd. I.e. P(x2) is

true So P(x) P(x2), hence P(x2) P(x)

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Mathematical Induction

Let P(n) be the predicate

This is true for particular values of n, e.g:

2

1321

nnnnP

2

)14(4104321

2

)13(36321

2

)12(2321

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Mathematical Induction

Want to show that P(n) is true for all integers n I.e.: n:P(n) The easiest way to prove such a statement is to

use the method of proof by induction

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Mathematical Induction

If P(1) is true

…and for all k, P(k) P(k+1)

Then P(n) is true for all values of n This is the principle of mathematical

induction

nPnkPkPkP :1:1

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Proof by Induction Intuitively…

1. Show that P(k) P(k+1) for any k

2. Show that P(1) is true

3. Then from 1 and 2, P(2) is true

4. …and from 1 and 3, P(3) is true

5. …and from 1 and 4, P(4) is true

6. etc

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Example 1

Case n=1: , so P(1) is true Now assume P(k) is true

Need to show P(k+1) is true I.e. need to show that

1,

2

1321:

n

nnnnSnP

2

11111

S

2

211

kkkS

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Example 1 (continued)

2

212

121

12

1

1

1211

kk

kkk

kkk

kkS

kkkS

Step 1: Write S(k+1) in terms of S(k)

Step 2: Use the fact that P(k) is true

Step 3: Manipulate to get the right formula

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Example 1 (concluded)

So, Therefore P(k+1) is true Therefore, by the principle of mathematical

induction, n:P(n)

2

211

kkkS

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Example 2 Let P(n) be the predicate defined by:

P(n): ‘n3-n is divisible by 3’

Show that n:P(n) Case n=1: n3-n=1-1=0, which is divisible

by 3. Hence P(1) is true Now assume that P(k) holds Need to prove that P(k+1) holds

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Example 2 (continued)

Case n=k+1: (k+1)3 - (k+1) = (k3+3k2+3k+1) - (k+1)

= (k3-k)+(3k2+3k)

= (k3-k) + 3(k2+k)

Divisible by 3 since P(k) is true Clearly

divisible by 3

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Example 2 (concluded)

Hence if k3-k is divisible by 3, then (k+1)3-(k+1) is also divisible by 3

In other words, k:(P(k) P(k+1)) Also, P(1) is true Therefore, by mathematical induction,

n:P(n)

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Example 3

Let P(n) be the predicate ‘if S is a finite set such that |S| = n, then |P(S)|=2n’

Claim: n:P(n) This is another way of saying that for any

finite set S, |P(S)| = 2|S| (See lecture 7, slide18)

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Example 3 (continued) Case n=1: If |S|=1, then S has just one

element, x say. Then S = {x}

In this case P(S)={,S}, so |P(S)|=2=21=2|S|

Hence P(1) is true Now assume P(k) is true for some k In other words, if S is a set such that |S| = k,

then |P(S)|=2k

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Example 3 (continued) Let Sk+1 be a set such that |Sk+1|=k+1

Write Sk+1 = {x1, x2,…,xk+1}

= Sk {xk+1}, where Sk= {x1, x2,…,xk}

We can write P(Sk+1) = Pk Pk+1, where:

Pk is the set of subsets of S which don’t include xk+1

Pk+1 is the set of subsets of S which do include xk+1

But Pk=P(Sk), so | Pk|=|P(Sk)|=2k by assumption

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Example 3 (continued) Now consider Pk+1, the set of subsets which

include xk+1

Every subset in Pk+1 must arise by taking a subset of Sk and adding xk+1 to it

…so | Pk+1| = |P(Sk)| = 2k

Hence | P(Sk+1)| = |Pk| |Pk+1|

= 2k + 2k = 2 2k = 2k+1

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Example 3 (concluded)

Hence P(1) is true, and if P(k) is true then P(k+1) is also true

Hence, by mathematical induction, n:P(n)

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Notes Mathematical induction can only be used to prove arguments

for which the domain of discourse is the positive, whole numbers.

For example, if P(b) is the predicate: the roots of the equation x2+bx+1 are given by

Then b:P(b) is true, but cannot be proved by induction 2

42

bbx

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Summary

Properties of the integers Proof by induction