COMP250: Induction proofs.

Jérôme WaldispühlSchool of Computer Science

McGill UniversityBased on slides from (Langer,2012)

for any n ≥ 2, n2 ≥ 2n ?

g(x) = 2x

f(x) = x2

for any n ≥ 5, n2 ≤ 2n ?

g(x) = 2x

f(x) = x2

Motivation

for anyn ≥1, 1+ 2+3+ 4+!+ n = n ⋅ (n+1)2

for anyn ≥1, 1+3+ 5+ 7+!+ (2 ⋅n−1) = n2

for anyn ≥ 5, n2 ≤ 2n

How to prove these?

And in general, any statement of the form:“for all n≥n0 , P(n)” where P(n) is some proposition.

Mathematical induction

Many statement of the form “for all n≥n0 , P(n)” can be proven with a logical argument call mathematical induction.

The proof has two components:• Base case: P(n0) • Induction step: for any n≥n0 , if P(n) then P(n+1)

0 1 2 3 4 5 ... k k+1 …

n0

Principle

0 1 2 3 4 5 ... k k+1 …

n0

Implies

0 1 2 3 4 5 ... k k+1 …

etc.

Example 1

for anyn ≥1, 1+ 2+3+ 4+!+ n = n ⋅ (n+1)2

Claim:

Proof:

• Base case:

• Induction step:

n =1 1= 1⋅22

✔

for anyk ≥1, if 1+ 2+3+ 4+!+ k = k ⋅ (k +1)2

then 1+ 2+3+ 4+!+ k + (k +1) = (k +1) ⋅ (k + 2)2

Example 1

Assume 1+ 2+3+ 4+!+ k = k ⋅ (k +1)2

then 1+ 2+3+ 4+!+ k + (k +1)

=k ⋅ (k +1)2

+ (k +1)

=k ⋅ (k +1)+ 2 ⋅ (k +1)

2

=(k + 2) ⋅ (k +1)

2

Induction hypothesis

Summary

Base case: P(1) ✔

Induction step: for any k≥1, if P(k) then P(k+1) ✔

Thus for all n≥1, P(n)

Example 2

for anyn ≥1, 1+3+ 5+ 7+!+ (2 ⋅n−1) = n2Claim:

Proof:

• Base case:

• Induction step:

n =1 1=12 ✔

for anyk ≥1, if 1+3+ 5+ 7+!+ (2 ⋅ k −1) = k2

then 1+3+ 5+ 7+!+ (2 ⋅ (k +1)−1) = (k +1)2

Example 2

Assume 1+3+ 5+ 7+!+ (2 ⋅ k −1) = k2

then 1+3+ 5+ 7+!+ (2 ⋅ k −1)+ (2 ⋅ (k +1)−1)= k2 + 2 ⋅ (k +1)−1= k2 + 2 ⋅ k +1= (k +1)2

Induction hypothesis

Example 3

f(x) = 2x+1

g(x) = 2x

Example 3

for anyn ≥ 3, 2 ⋅n+1< 2nClaim:

Proof:

• Base case:

• Induction step:

n = 3 2 ⋅3+1= 7 < 23 = 8 ✔

for anyk ≥ 3, if 2 ⋅ k +1< 2k

then 2 ⋅ (k +1)+1< 2k+1

Example 3

Assume 2 ⋅ k +1< 2k

then 2 ⋅ (k +1)+1= 2 ⋅ k + 2+1< 2k + 2≤ 2k + 2k for k ≥1= 2k+1

Induction hypothesis

Stronger than we need, but that works!

Example 4

g(x) = 2x

f(x) = x2

Example 4

for anyn ≥ 5, n2 ≤ 2nClaim:

Proof:

• Base case:

• Induction step:

n = 5 25≤ 32 ✔

for anyk ≥ 5, if k2 ≤ 2n

then (k +1)2 ≤ 2k+1

Example 4

Assume k2 ≤ 2k

then (k +1)2

= k2 + 2 ⋅ k +1≤ 2k + 2 ⋅ k +1≤ 2k + 2k

= 2k+1

Induction hypothesis

From previous example

Example 5Fibonacci sequence:Fib0 = 0 base caseFib1 = 1 base caseFibn = Fibn-1 + Fibn-2 for n > 1 recursive case

Claim: For all n≥0, Fibn

Example 5

Assume that for all i ≤ k, Fibi < 2i (Note variation of induction hypothesis)

Then Fibk+1 = Fibk + Fibk-1< 2k + 2k-1

≤ 2k + 2k for k≥1= 2k+1

Induction hypothesis (x2)