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Mathematical Induction

Proving a divisibility property by mathematical

induction

- Proposition For any integer n1,
- 7n - 2n is divisible by 5. (P(n))
- Proof (by induction)
- 1) Basis step
- The statement is true for n1 (P(1))
- 71 21 7 - 2 5 is divisible by 5.
- 2) Inductive step
- Assume the statement is true for some k1

(P(k)) - (inductive hypothesis)
- show that it is true for k1 . (P(k1))

Proving a divisibility property by mathematical

induction

- Proof (cont.) We are given that
- P(k) 7k - 2k is divisible by 5.

(1) - Then 7k - 2k 5a for some a?Z . (by

definition) (2) - We need to show
- P(k1) 7k1 - 2k1 is divisible by 5.

(3) - 7k1 - 2k1 77k - 22k 57k 27k - 22k
- 57k 2(7k - 2k) 57k 25a (by (2))
- 5(7k 2a) which is divisible by 5. (by

def.) - Thus, P(n) is true by induction.

Proving inequalities by mathematical induction

- Theorem For all integers n5,
- n2 lt 2n. (P(n))
- Proof (by induction)
- 1) Basis step
- The statement is true for n5 (P(5))
- 52 25 lt 32 25.
- 2) Inductive step
- Assume the statement is true for some k5

(P(k)) - show that it is true for k1 .

(P(k1))

Proving inequalities by mathematical induction

- Proof (cont.) We are given that
- P(k) k2 lt 2k. (1)
- We need to show
- P(k1) (k1)2 lt 2k1.

(2) - (k1)2 k22k1 lt k2 2k (since k5)
- lt 2k 2k (based on (1))
- 22k 2k1.
- Thus, P(n) is true by induction.

Proving inequalities by mathematical induction

- Theorem For all integers n4,
- 2n lt n! . (P(n))
- Proof (by induction)
- 1) Basis step
- The statement is true for n4 (P(4))
- 24 16 lt 24 4! .
- 2) Inductive step
- Assume the statement is true for some k4

(P(k)) - show that it is true for k1 .

(P(k1))

Proving inequalities by mathematical induction

- Proof (cont.) We are given that
- P(k) 2k lt k! (1)
- We need to show
- P(k1) 2k1 lt (k1)! (2)
- 2k1 22k
- lt 2k! (based on (1))
- lt (k1)k! (since k4)
- (k1)!
- Thus, P(n) is true by induction.