Page 309 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Strong Induction 295
obvious contradictions, but they do lead to the conclusion that p and q must both
be even. Thus, in the fraction p/q we can cancel a 2 from both numerator and
denominator, getting a new fraction with smaller numerator and denominator
√
that is equal to 2.
How can we derive a contradiction from this conclusion? The key idea is to
√
note that our reasoning would apply to any fraction that is equal to 2. Thus, in
any such fraction we can cancel a factor of 2 from numerator and denominator,
and therefore there can be no smallest possible numerator or denominator for
such a fraction. But this would violate the well-ordering principle! Thus, we
have our contradiction.
This idea is spelled out more carefully in the following proof, in which we’ve
applied the well-ordering principle to the set of all possible denominators of
√
fractions equal to 2. We have chosen to put this application of the well-
ordering principle at the beginning of the proof, because this seems to give
the shortest and most direct proof. Readers of the proof might be puzzled at
first about why we’re using the well-ordering principle (unless they’ve read
this scratch work!), but after the algebraic manipulations with the equation
√
p/q = 2 are completed, the contradiction appears almost immediately. This
is a good example of how a clever, carefully planned step early in a proof can
lead to a wonderful punch line at the end of the proof.

√
+
Proof. Suppose that 2 is rational. This means that ∃q ∈ Z ∃p ∈ Z (p/q =
+
√ √
+ +
2), so the set S ={q ∈ Z |∃p ∈ Z (p/q = 2)} is nonempty. By the well–
ordering principle we can let q be the smallest element of S. Since q ∈ S,
√
2
2
we can choose some p ∈ Z + such that p/q = 2. Therefore p /q = 2,
2
2
2
so p = 2q and therefore p is even. We now apply the theorem from
2
Example 3.4.2, which says that for any integer x, x is even iff x is even. Since
2
p is even, p must be even, so we can choose some ¯ p ∈ Z such that p = 2 ¯ p.
+
2
2
2
2
Therefore p = 4 ¯ p , and substituting this into the equation p = 2q we get
2
2
2
2
2
4 ¯ p = 2q ,so2 ¯ p = q and therefore q is even. Appealing to Example 3.4.2
+
again, this means q must be even, so we can choose some ¯ q ∈ Z such that
√
q = 2¯ q. But then 2 = p/q = (2 ¯ p)/(2¯ q) = ¯ p/¯ q,so ¯ q ∈ S. Clearly ¯ q < q,
so this contradicts the fact that q was chosen to be the smallest element of S.
√
Therefore 2 is irrational.
Exercises
1. This exercise gives an alternative way to justify the method of
∗
strong induction. All variables in this exercise range over N. Suppose