ordinary induction
DESCRIPTION
Questions-answersTRANSCRIPT
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!
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