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MATHEMATICALINDUCTION


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HASSAM BAIG

ABDUL SATTAR

MUHAMMAD REHAN

FAHAD BAJWA


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1-INTRODUCTION

2-HISTORY

3-DESCRIPTION

4-EXAMPLES


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Mathematical induction is a method of mathematical prooftypically used to

establish that a given statement is true of all natural numbers. Itis done by

proving that the first statement in the infinite sequence of

statements is true and then proving that if any one statement in the infinite

sequence of

statements is true, then so is the next one.

The method can be extended to prove statements about moregeneral

well-founded structures, such as trees; this generalization,known as

structural induction, is used in mathematical logic and computer

science. Mathematical induction in this extended sense is closelyrelated

to recursion.


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The earliest implicit proof by mathematical inductionfor arithmetic sequences was introduced in the al-Fakhri written by al-Karaji around 1000 AD, who usedit to prove the binomial theorem, Pascal's triangle,and the sum formula for integral cubes. The sum

formula for integral cubes is the (true) propositionthat every integer can be expressed by the sum ofcubed natural numbers. It is a particular case of whatis referred to as Waring's Problem. His proof was thefirst to make use of the two basic components of aninductive proof. First, he notes the truth of the

statement for n = 1. That is, 1 is the sum of a singlecube because 1 = 13. Secondly, he derives the truthfor n = k from that of n = k 1. For example, when n= 2, it is true that 2 = 13 + 13. When n = 3, it is truethat 3 = 13 + 13 + 13.


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The truth of the statement can be

extrapolated in this way without limit. Of

course, as n grows larger, some of the sums

of 13 can be rewritten as the cubes of othernatural numbers: for example when n=8 then

8 = 23 = [13 8]. Of course, this second

component is not explicit since, in some

sense, al-K

araji's argument is in reverse; thisis, he starts from n = 10 and goes down to 1

rather than proceeding upward."


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The simplest and most common form ofmathematical induction proves that a statementinvolving a natural number n holds for all valuesof n. The proof consists of two steps:

Basic step:-

The basis (base case): showing that the

statement holds when n is equal to the lowestvalue that n is given in the question. Usually, n =0 or n = 1.


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Inductive step:-

The inductive step: showing that if the statement holds for somen, then the statement also holds when n + 1 is substituted for n.

The assumption in the inductive step that the statement holds

for some n is called the induction hypothesis.To perform theinductive step, one assumes the induction hypothesis and thenuses this assumption to prove the statement for n + 1.

The choice between n = 0 and n = 1 in the base case is specific tothe context of the proof: If 0 is considered a natural number, asis common in the fields of combinatorics and mathematical logic,

then n = 0. If, on the other hand, 1 is taken as the first naturalnumber, then the base case is given by n = 1.


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Show that for each positive integer n, 1 + 2 + 3 + ... + n = n(n + 1) / 2

Proof: let S(n) be the statement 1 + 2 + 3 + ... + n = n(n + 1) / 2

Basis step: S(1) is the statement 1 = 1(1 + 1) / 2. Thus S(1) is true.

Inductive step: We suppose that S(k) is true and prove that S(k + 1) istrue. Thus, we assume that

1+ 2 + 3 + ... + k = k(k +

1) / 2

Put S(k+1)

1 + 2 + 3 + ... + k + k + 1 = (k + 1)( k + 1 + 1) / 2.

1 + 2 + 3 + ... + k + k + 1

= k(k + 1) / 2 + (k + 1)

= (k + 1)(k / 2 + 1)

= (k + 1)(k / 2 + 2 / 2)

= (k + 1)(k + 2) / 2

= (k + 1)(k + 1 + 1) / 2

Hence, if S(k) is true, then S(k + 1) is true.

Therefore, 1 + 2 + 3 + ... + n = n(n + 1) / 2 for each positive integer n.


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There is another proof technique that isvery similar to the principle ofmathematical induction.

It is called the second method ofmathematical induction.

It can be used to prove that a

propositional function P(n) is true for anynatural number n.


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The second principle of mathematicalinduction:

show that p(0) is true.

(Base step).

Show that if p(0) and p(1) and p(n),then p(n+1)is true for any natural number.

(Inductive step).


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Then p(n) must be true for any natural

numbers.

Conclusion

Thus by concluding that by the help of

mathematical induction we can solve many

conplex problems .

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