Download Mathematical Induction

Mathematical Induction
Example: Consider the sum of
n=1
n=2
n=3
n=4
the first n natural numbers:
sum = 1
sum = 1 + 2 = 3
sum = 1 + 2 + 3 = 6 = (3)(4)/2
sum = 10
= (4)(5)/2
Based on the above, one might guess that for n = k, 1 + 2 + 3 + · · · + k = k(k + 1)/2, and
this turns out to be correct. The question is how to prove it. We can’t just prove it for a few
values of n and then claim it’s true for all n. What we really have to do is prove an infinite
number of statements {Pk : k = 1, 2, . . .}, where
Pk is the statement
k
X
n = k(k + 1)/2.
n=1
The way to do this is by using mathematical induction, which we now explain. First, we
need an axiom:
Axiom: Any nonempty subset of N has a smallest element
(This might seem “obvious”, but remember that we haven’t really defined N rigorously. This
is normally done by assuming the existence of 1, and then defining a “successor” operation
which effectively adds 1 to anything. So from 1 we get 2, and from 2 we get 3, etc. You
may see this in some other math class like algebra, but we’re not going to do it. Therefore
the property stated in the axiom is not provable with the tools at hand, so we’re going to
assume it.
Theorem: (Mathematical induction) If {Pn : n ∈ N} is a set of propositions
indexed by n such that
1. P1 is true.
2. Pk ⇒ Pk+1 , ∀k ∈ N,
then Pn is true for all n ∈ N.
Proof: Suppose not, and let K ⊆ N be the set of all natural numbers for which the
proposition is false. If K 6= ∅, then K has a smallest number, say m. Now m 6= 1, since
P1 is true. Thus m − 1 ∈ N \ K. So Pm−1 is true, and therefore, by the second part of the
hypothesis above, so is Pm . So such an m doesn’t exist, which means, by the axiom that
K = ∅, and the theorem is true.
Example: We’ve used the following identity to prove the convergence of the geometric
series for |x| < 1:
1 − xn+1
.
1 + x + x2 + · · · + xn =
1−x
1
But we’ve never proven it! Let’s do so, using mathematical induction:
Step 1: Show that P1 is true: If n = 1, then the statement reads
1+x=
1 − x2
,
1−x
which is true as can be verified by multiplying both sides by (1 − x).
Step 2: Show that Pk ⇒ Pk+1 : We assume that Pk holds:
1 + x + · · · + xk =
1 − xk+1
.
1−x
(This is called the induction hypothesis). We now must show that Pk+1 holds. But
k
1 + x + ··· + x + x
k+1
1 − xk+1
=
+ xk+1 (induction hypothesis)
1−x
1 − xk+1 + xk+1 − xk+2
=
1−x
k+2
1−x
=
1−x
So we’ve shown that Pk ⇒ Pk+1 . Therefore, Pn is true for all n ∈ N.
Remark: Sometimes, Pn is not true for all n ∈ N, but is true for n sufficiently large, say
n ≥ j. To use induction here, show that Pj holds and that Pk ⇒ Pk+1 for any k ≥ j.
♣ Exercise: (1): Show that
n
X
m = n(n + 1)/2.
m=1
(2): Find an expression for the sum of the first n squares
mathematical induction.
2
Pn
m=1
m2 and prove it using