Prove that, for any natural number n,
.
Proof: First we check P(1):
Okay! Now the induction: suppose that P(k) is true, and consider the case of k+1:
but
Now
so P(k+1) is also true. Q.E.D., by mathematical induction.
Prove: that 
for 
.
Proof:
. Check.
),
,
and consider 
:
Hence P(k+1), and the result is proven by mathematical induction. Q.E.D.
Exercise 60/64b, p. 110/109
.
First of all, we check the base case: since the sum of the interior angles of a
triangle is 180, we have P(3). Suppose that P(r) is true for all 
,
and consider the case of a k+1-sided polygon ( 
): the idea is to ``divide
and conquer''.
We split the k+1-sided polygon into two polygons by connecting
two of the unconnected vertices in such way that we create two new polygons, of
m and n sides each. Now m+n-2=k+1, since we double count the new side in
the two smaller polygons. Since 
and 
, they satisfy our
hypothesis, and we can write the sum of their interior angles as
The sum of the interior angles of the two sub-polygons is equal to the sum of the interior angles of the initial polygon, since we have simply subdivided two of the angles in the initial polygon and counted part of the measure in one of the new sub-polygons and the rest in the other. Hence, P(k+1); Q.E.D., by mathematical induction.
Exercise 68, p. 186 (not in Edition 5)
Proof: by contradiction. The principle of well-ordering says that every non-empty set of positive integers contains a smallest member. Given a non-empty set of positive integers T without a smallest member, we will prove that this is a contradiction if the first principle of induction holds.
Since T is non-empty, it contains some positive integer: call it
a. Consider P(n) defined by ``every member of T is greater than
n''. Clearly P(1), since 1 would be the smallest member otherwise. Suppose
P(k) to be true, and consider P(k+1): since all members of T are greater
than k, if k+1 is in T, then k+1 is the smallest member. Since there is
no smallest member, k+1 is not in T, and hence P(k+1). Thus, by the first
principle of mathematical induction, we have demonstrated that P(n) is true
for all .
But P(a) is false, since a is in the set! This is a contradiction, which means that T did, in fact, have a smallest member, which establishes the principle of well-ordering.
Backwards induction (The prisoner's last request)
A prisoner, condemned to die by the Sultan of an antique land, asked for one last request: ``Please Sultan, if you would only grant me two favors: one, that you have me executed in the month of January (next month), and two, that you don't allow me to know the day of my death until 10 a.m. of the day I am to die.'' The Sultan, being a merciful man, granted these requests, whereupon the prisoner demonstrated that it was impossible to execute him subject to these conditions. How?
