UCI ICS/Math 6D, Fall 2007
4Induction1
Induction
Let P(n) be a predicate (“propositional function”) regarding
integers n≥n
0
for some n
0
(typically n
0
=0 or 1).
“
“
Normal” Induction
Normal” Induction
:
If
you prove two things:
1) Basis step
: P(n
0
) is true
and
2) Inductive step
: For all k≥n
0
: If
P(k) true then
P(k+1) true
then
it follows that P(n) is true for all n≥n
0
.
Example: Show that P(n) = “1+2+.
..+n=n(n+1)/2” is true for all n≥1.
1)
Basis (for n
0
=1): “1=1·(1+1)/2” is true.
1)
Inductive step:
Assuming
P(k) [that is, assuming 1+2+.
..+k=k(k+1) ]
prove
P(k+1) [that is, prove 1+2+.
..+k+(k+1)=(k+1)(k+2)/2 ]
Ref:
http://en.wikipedia.org/wiki/Mathematical_induction
The inductive step is a
proof of an implication
P(k) => P(k+1)
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4Induction2
Induction Examples
For n≥0, n
2
<4
n
.
P(n)=“n
2
<4
n
”
Basis step: P(0)=“0<4
0
=1” is true
Inductive step: Assuming
k
2
<4
k
, prove (k+1)
2
<4
k+1
.
The sum of the first n odd positive integers is n
2
Basis step: “1=1
2
” is true
Inductive step: Assuming
1+3+5+.
..+(2k1)=k
2
,
prove that 1+3+5+.
..+(2k1)+(2k+1)= (k+1)
2
A set with n elements has 2
n
subsets (P(S)=2
s
)
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 Fall '07
 Jarecki
 Natural number, inductive step, UCI ICS/Math 6D

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