Abstract Mathematics Lecture 14

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Consider this statement:

Conjecture

The sum of the first n odd natural numbers equals n2. How to prove it?

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n sum of the first n odd natural numbers n2 1 1 = . . . 1 2 1 + 3 = . . . 4 3 1 + 3 + 5 = . . . 9 4 1 + 3 + 5 + 7 = . . . 16 5 1 + 3 + 5 + 7 + 9 = . . . 25 .. . ... ... n 1 + 3 + 5 + 7 + 9 + 11 + · · · + (2n − 1) = . . . n2 .. . ... ...

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Note that in the first five lines of the table, the sum of the first n odd numbers really does add up to n2.

The table raises a question. Does the sum

1 + 3 + 5 + 7 + 9 + 11 + · · · + (2n − 1) really always equal n2? Mathematical induction answers just this kind of question, where we have an infinite list of statements

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Let S1 : 1 = 12 S2 : 1 + 3 = 22 S3 : 1 + 3 + 5 = 32 .. . Sn: 1 + 3 + 5 + 7 + · · · + (2n − 1) = n2 .. .

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The Simple Idea Behind Mathematical Induction

Statements are lined up like dominoes.

(1) Suppose the first statement falls (is proved true).

(2) Suppose the kth falling always causes the (k + 1)th to fall;

Then all must fall (all are proved true). S 1 S2 S3 S4 S5 S6 S 1 S2 S3 S4 S5 S6 S k Sk+₁ Sk+₂ Sk+₃ Sk Sk Sk+1 Sk+1 Sk+2 Sk+2 Sk+2 Sk+3 Sk+3 Sk+3 Sk+4 Sk+4 Sk+4 · · · · · · · · · · · · · · · · · · · · · · · · S1 S_k S_k+1 S1 S2 S3 S4 S5 S6 S2 S3 S4 S5 S6

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Proof by Induction

Proposition

The statements S1, S2, S3, S4, . . . are all true.

Proof.

(Induction).

(1) Prove that the first statement S1 is true.

(2) Given any integer k ≥ 1, prove that the statement Sk → Sk+1 is

true.

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Example

Proposition

If n ∈ N, then 1 + 3 + 5 + 7 + · · · + (2n − 1) = n2.

Proof.

We will prove this with mathematical induction.

1 _{Observe that if n = 1, this statement is 1}2_{= 1, which is obviously}

true.

2 _{That is, we must show that if 1 + 3 + 5 + 7 + · · · + (2k − 1) = k}2_,

then 1 + 3 + 5 + 7 + · · · + (2(k + 1) − 1) = (k + 1)2.

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Cont.

We use direct proof. Suppose 1 + 3 + 5 + 7 + · · · + (2k − 1) = k2. Then 1 + 3 + 5 + 7 + · · · + (2(k + 1) − 1) = 1 + 3 + 5 + 7 + · · · + (2k − 1) + (2(k + 1) − 1) = 1 + 3 + 5 + 7 + · · · + (2k − 1) + (2(k + 1) − 1) = k2 + (2(k + 1) − 1) = k2+ 2k + 1 = (k + 1)2 Thus 1 + 3 + 5 + 7 + · · · + (2(k + 1) − 1) = (k + 1)2. It follows by induction that 1 + 3 + 5 + 7 + · · · + (2n − 1) = n2 for every n ∈ N.

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Try to prove the following statement.

Proposition

If n is a non-negative integer, then 5|(n5− n).

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Sometimes in an induction proof it is hard to show that Sk → Sk+1.

It may be easier to show some lower Sm (with m < k) implies Sk+1.

For such situations there is a slight variant of induction called strong induction.

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Outline for Proof by Strong Induction

Proposition

The statements S1, S2, S3, S4, . . . are all true.

Proof.

(Strong induction).

(1) Prove the first statement S1. (Or the first several Sn, if needed.)

(2) Given any integer k ≥ 1, prove (S1∧ S2∧ · · · ∧ Sk) → Sk+1.