INDUCTION

byArtificial Intelligence

QUESTION 1

What do you show in the basis step and inductive step when you use

ordinary math induction to prove that a property involving an integer n is true for all integers more than or

equal to some initial integer?

Review…

Basis Step: Prove that P(n) is true. 

Induction Step: Prove that if P(k) is true (called induction hypothesis), then P(k+1) is true. 

Example : Let P(n) be "2n > n² whenever n is an integer greater than 4 ".

Basis step: P(5) is true since 25 = 32 > 25 =5². Inductive step: Assume that P(n) is true, that is, 2n> n².  Then 2n+1 = 2*2n> n² + n² > n² + 4n  n²+ 2n + 1 = (n+1)²   since n > 4.Hence P(n + 1) is true.

QUESTION 2

What is the inductive hypothesis in a proof by (ordinary)

mathematical induction?

Inductive hypothesisThe predicate P(n) is called the

induction hypothesis.

You have to proof a statement of the method:

If P(x) is true, then P(x + 1) is true.

We have to proof the inductive hypothesis is true before we can proof

that P(x + 1) is true.

QUESTION 3

Are you able to use (ordinary) mathematical induction to construct proofs involving formulas, divisibility

properties and inequalities?

3.1 involving formulaProve that;12+ 2² + 3² +…+n² = n(n + 1) (2n + 1)/ 6 for

all positive integers n.

Solution:Statement P(n) is defined by

1² + 2² + 3² + ... + n² = n (n + 1) (2n + 1)/ 6

BASIS STEP: Show that P(1) is true.Left side = 1² = 1Right side = 1 (1 + 1) (2*1 + 1)/ 6 = 1Both sides are equal hence P(1) is true.

INDUCTION STEP:

Assume P(k) is true 1² + 2² + 3² +…+k² = k(k + 1) (2k + 1)/ 6

For P(k + 1) ,add (k + 1)² to both sides above 1² + 2² + 3² +…+k² + (k + 1)² =

k(k + 1) (2k + 1)/ 6 + (k + 1)²

Set common denominator and factor k + 1 on the

right side = (k + 1) [ k(2k + 1) + 6(k + 1)] / 6

Expand k(2k + 1) + 6(k + 1) = (k + 1) [ 2k² + 7k + 6] / 6

Now factor 2k² + 7k + 6 = (k + 1) [ (k + 2) (2k + 3)] / 6

We have started from the statement P(k) and have shown that1² + 2² + 3² +…+k² + (k + 1)²

= (k + 1) [ (k + 2) (2k + 3) ]/6Which is the statement P(k + 1).

3.2 involving divisible properties

Problem: For any natural number n , n3 + 2n is divisible by 3.

Solution:BASIS STEP

If n = 0, then n3 + 2n = 03 + 2*0 = 0. So it is divisible by 3.

INDUCTION STEP

Assume that for an arbitrary natural number n, n3 + 2n is divisible by 3.

subtitute n with n+1 in n3 + 2n

( n + 1 )3 + 2( n + 1 ) = ( n3 + 3n2 + 3n + 1 ) + ( 2n + 2 ) = ( n3 + 2n ) + ( 3n2 + 3n + 3 ) = ( n3 + 2n ) + 3( n2 + n + 1 )

which is divisible by 3, because ( n3 + 2n ) is divisible by 3 by the induction hypothesis.

3.3 involving inequalities

if x ≥ 5, then 2 ≥ x² ͯ

Basis step :

Let x = 5

2^5 ≥ 5²

32 ≥ 25 → proven!

Inductive step:

Substitute x with x + 1;

2x ≥ x2

2x+1 ≥ (x+1)2

2∙2x ≥ (x+1)2 → 1

2∙2x ≥ 2x2 → 2

Relates 1 and 2

2∙2x ≥ 2x2 ≥ (x+1)2

2x2 ≥ (x+1)2

2x2 ≥ x2 + 2x + 12x2 ≥ x2 + 2x + 1(x2 ≥ 2x + 1) divide by xx ≥ 2 + (1/x) → shown!

END