Principle of Strong Mathematical Induction

1
Principle of Strong Mathematical Induction

• Let P(n) be a statement defined for integers n
• a and b be fixed integers with ab.
• Suppose the following statements are true
• 1. P(a), P(a1), , P(b) are all true
• (basis step)
• 2. For any integer kgtb,
• if P(i) is true for all integers i with
  ailtk,
• then P(k) is true. (inductive step)
• Then P(n) is true for all integers na.

2
Example Divisibility by a Prime

• Theorem For any integer n2,
• n is divisible by a prime. P(n)
• Proof (by strong mathematical induction)
• 1) Basis step
• The statement is true for n2 P(2)
• because 2 2 and 2 is a prime number.
• 2) Inductive step
• Assume the statement is true for all i with
  2iltk P(i)
• (inductive hypothesis)
• show that it is true for k . P(k)

3
Example Divisibility by a Prime

• Proof (cont.)
• We have that for all i?Z with 2iltk,
• P(i) i is divisible by a prime number. (1)
• We must show
• P(k) k is also divisible by a prime. (2)
• Consider 2 cases
• a) k is prime. Then k is divisible by itself.
• b) k is composite.
• Then kab where 2altk and 2bltk.
• Based on (1), pa for some prime p.
• pa and ak imply that pk (by
  transitivity).
• Thus, P(n) is true by strong induction.

4
Proving a Property of a Sequence

• Proposition
• Suppose a0, a1, a2, is defined as follows
• a01, a12, a23,
• ak ak-1ak-2ak-3 for all integers k3.
• Then an 2n for all integers n0. P(n)
• Proof (by strong induction)
• 1) Basis step
• The statement is true for n0 a01 120 P(0)
• for n1 a12 221 P(1)
• for n2 a23 422 P(2)

5
Proving a Property of a Sequence

• Proof (cont.)
• 2) Inductive step For any kgt2,
• Assume P(i) is true for all i with 0iltk
• ai 2i for all 0iltk . (1)
• Show that P(k) is true ak 2k (2)
  ak ak-1ak-2ak-3
• 2k-12k-22k-3 (based on (1))
• 20212k-32k-22k-1
• 2k-1 (as a sum of geometric sequence)
• 2k
• Thus, P(n) is true by strong induction.