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Sec. 1.7 – Mathematical Induction
Principle of Mathematical Induction:
• Suppose that we have a propositional
function S(n) whose domain of discourse
is the set of positive integers.
• Further suppose that S(1) is true ( basis
step ), and for all n ≥ 1, if S(n) is true, then
S(n+1) is true ( inductive step ).
• Then, S(n) is true for every positive
integer n.
2
n
n
1
n
n
n + 1
le t S = 1 + 2 +. .. + n.
n ( n + 1 ) p r o v e : S = f o r a ll n 1. 2
p r o o f :
1 2 b a s i s s t e p : S = = 1 2
n ( n + 1 ) i n d u c t i v e s t e p : S = 2
a s s u m i n g t h a t S i s t r u e , w e m u s t s h o w t h a t
( n + 1 ) ( n + 2 ) S = i s t 2
≥
×
r u e.
Example induction proof
3
by definition, S n+1 = 1+2+ ... + n + (n+1) = S + (n+1)n
n(n+1) n(n+1) 2(n+1) = + (n+1) = + =
2 2 2
n(n+1) + 2(n+1) (n+1)(n+2)
4
n-
1-1 0
the factorial function is defined as follows:
1, if n = 0 n! = n(n-1)(n-2)...1, if n 1
prove: n! 2 for all n 1.
proof:
basis step: let n=1, then 1! = 1 1 = 2 = 2.
inductive step
≥
≥ ≥
≥
n-
n
n-
n-
n
: assume n! 2 is true.
then, we need to prove that (n+1)! 2.
(n+1)! = (n+1)n! (n+1)
2 2 (since n+1 2)
= 2
≥
≥
≥
≥ × ≥
7
n+ 2 n n+
n+ 2 n n+1 n+
n+
inductive step: we want to prove:
a(r -1) a + ar + ar + ... + ar + ar =.
r-
Assuming that the formula is true for n, we have:
a(r -1) a + ar + ar + ... + ar + ar = + ar
r-
a(r -1)
r-
n+1 n+1 n+2 n+
n+2 n+
ar (r -1) ar - a + ar - ar
ar - a a(r -1)
r - 1 r - 1
