30 Slides--Induction II - CS103 HO#30 Slides-Induction II...

CS103 HO#30 Slides--Induction II April 26, 2010
1
The Principle of Mathematical Induction
A proof by mathematical induction that a proposition P
n
is true
for every positive integer n consists of two steps:
BASE CASE: Show that P
1
is true.
INDUCTIVE STEP: Assume that P
k
is true for an arbitrarily
chosen positive integer k and show that under that assumption,
P
k+1
must be true.
From these two steps we conclude (by the principle of mathematical
induction) that for all positive integers n, P
n
is true.
Note that in the inductive step,
you are proving a conditional:
IF P
k
for some arbitrary k,
THEN P
k+1
.
P
1
k P
k
. . .
. . .
P
k+1
∀ n P
n
Here is
a Formal
version
Peano's Axioms (1889)
• The equality relation is reflexive, symmetric, and transitive,
and the natural numbers are closed under equality.
• There is a natural number 0.
• Every natural number has a successor, denoted by S(a).
• There is no natural number whose successor is 0, i.e.,
∀ x (S(x) ≠ 0).
• Two natural numbers with the same successor are
themselves equal, i.e., ∀ x ∀ y(S(x) = S(y) → x = y)
• If a property is possessed by 0 and if the successor of
every natural number possessing the property also
possesses it, then it is possessed by every number, i.e.,
[Q(0) ∧ ∀ x(Q(x) → Q(S(x))] → ∀ xQ(x)
Bad Induction
Show that for any positive n, n
3
– n + 1 is a multiple of 3.
P
n
: n
3
– n + 1 is a multiple of 3
We will show that if we assume P
k
is true, then P
k+1
is true
(k + 1)
3
– (k + 1) + 1 = k
3
+ 3k
2
+ 3k + 1 – k = k
3
– k + 1 + 3(k
2
+ k)
Divisible by 3
by inductive
hypothesis
Obviously
divisible by 3
So P
k+1
is true and P
n
is true for all n.
What's wrong?
Bad Induction
Show that for any positive n, n
3
– n + 1 is a multiple of 3.
P
n
: n
3
– n + 1 is a multiple of 3
We will show that if we assume P
k
is true, then P
k+1
is true
(k + 1)
3
– (k + 1) + 1 = k
3
+ 3k
2
+ 3k + 1 – k = k
3
– k + 1 + 3(k
2
+ k)
Divisible by 3
by inductive
hypothesis
Obviously
divisible by 3
So P
k+1
is true and P
n
is true for all n.
What's wrong?
There is no base case. Not only is P
1
not true, P
n
is false for all n .
The Well-Ordering Property
Every non-empty set of positive integers has a least element.
From this axiom, we can prove that the
Principle of Mathematical Induction is correct.
The Well-Ordering Property
Every non-empty set of positive integers has a least element.
From this axiom, we can prove that the
Principle of Mathematical Induction is correct.
Suppose that for some P
n
we show the base case P
1
and the
inductive step ∀ k ≥ 1 (P
k
→ P
k+1
). In order to derive a contradiction,
suppose that it is not the case that ∀ nP
n
where n ≥ 1. Then ∃ n(¬P
n
).