Proof by Induction
Faculty Name: Ruhi Fatima
Topics Covered
• Mathematical Induction
• Strong Induction
• Smallest Counterexample
Proof by Induction
• First step (1) is called the basis step, second step (2) is called
the inductive step and assumption that Sk is true is called the
inductive hypothesis.
• Example: Prove the following proposition.
Proposition If n ϵ N, then 1+3+5+7+ ….+(2n-1) = n2.
• Proof:
Cont…
Cont…
• Example 2: Prove the following proposition.
Proposition If n is a non-negative integer, then 5l(n5 - n).
Proof:
Cont…
• Example 3: Prove the following proposition.
Proposition For each n ϵ N, it follows that 2n ≤ 2n+1- 2n -1 -1.
Proof:
Proof by Strong Induction
• Sometimes in induction proofs it is difficult to show that Sk forces Sk+1 to be
true. Instead you may find that you need to use the fact that some “lower”
statement Sm (with m < k) forces Sk+1 to be true. These situations use a
slight variant of induction called strong induction.
• Example: Prove the following proposition.
Proposition If n ϵN, then 12 l (n4 - n2).
• In regular induction, start by showing 12 l (n4 - n2) is true for n = 1. This
part is easy because it reduces to 12 l 0, which is clearly true. Next
assume that 12 l (k4 - k2) and try to show this implies 12 l ((k+1)4 – (k+1)2)
Cont…
• Strong induction can get us out of this bind.
• The idea is to show Sk-5→Sk+1 instead of Sk →Sk+1. Thus basis step
involves checking that S1,S2,S3,S4,S5,S6 are all true so that Sk-5 doesn’t
“run off the end.”
(1) If n=1, 12 divides n4-n2 = 14-12 =0; If n=2, 12 divides n4-n2 = 24-22 =12
If n=3, 12 divides n4-n2 = 34-32 =72; If n=4, 12 divides n4-n2 = 44-42 =240
If n=5, 12 divides n4-n2 = 54-52 =600; If n=6, 12 divides n4-n2 = 64-62 =1260
(2) Let k ≥6 and assume 12 l (m4-m2) for 1≤ m ≤ k.
• Show that 12 l ((k+1)4- (k+1)2). Since Sk-5 is true, set m=k-5. It is known that
12 l (m4-m2) means (m4-m2)=12a for some integer a. Observe
• Therefore, by strong induction it follows that 12 l (n4-n2 ) for every positive
integer n.
Proof by smallest Counterexample
• It is a hybrid of induction and proof by contradiction.
• Example : Prove the following proposition.
Proposition: If n ϵN, then 4 l (5n -1).
Proof:
(1) If n =1, then the statement is 4 l (51 -1), or 4 l 4, which is obviously true.
(2) For sake of contradiction, suppose it’s not true that 4 l (5n -1) for all n.
(3) Let k > 1 be the smallest integer for which
(4) Then 4 l (5k-1 -1), so there is an integer a for which (5k-1 -1)= 4a. Then:
• This means 4 l (5n -1), a contradiction of Step 3.
• Therefore 4 l (5n -1) is true for every n.