LECTURE 14: INDUCTION GENERALIZED (
§
6.2)
There are two kinds of people, those who finish what they start and so on. Robert Byrne
1.
Generalization
Induction works if we start our first step anywhere in
Z
.
Result.
For every nonnegative integer
n
we have
2
n
> n
.
Proof.
We work by induction on
n
≥
0.
Base case: If
n
= 0 then the inequality is 1
>
0 which is true.
Inductive step: Assume 2
k
> k
for some
k
≥
0. We want 2
k
+1
> k
+ 1. We compute
LHS = 2
k
+1
= 2
·
2
k
>
2
·
k
=
k
+
k.
If
k
≥
1 then
k
+
k
≥
k
+ 1 = RHS and we are done.
But we could have
k
= 0, in which case
2
k
+1
= 2
>
1 =
k
+ 1. So the inductive step works in either case.
Result.
For every integer
n
≥
5
, we have
2
n
> n
2
.
Let’s sketch how the inductive step might go. We assume 2
k
> k
2
(for some
k
≥
5) and we want
2
k
+1
>
(
k
+ 1)
2
=
k
2
+ 2
k
+ 1. We compute
LHS = 2
k
+1
= 2
·
2
k
>
2
k
2
.
So we want 2
k
2
> k
2
+ 2
k
+ 1. Is this true? We might be able to use the fact that
k
≥
5 to get this.
In fact
2
k
2
=
k
2
+
k
2
≥
k
2
+ 5
k
=
k
2
+ 2
k
+ 3
k
≥
k
2
+ 2
k
+ 15
> k
2
+ 2
k
+ 1
.
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 Spring '11
 Smith
 Logic, Mathematical Induction, Inductive Reasoning, Natural number, inductive step

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