172 METHODS OF MATHEMATICAL PROOF, PART I Chapter 5Note that the proof of Example 3 employed a straightforward substi-
tution, whereas that of Example 4 was somewhat more intricate, involving
a fairly "clever" choice of x', in terms of the arbitrary x that we chose at
the outset of the proof. Some of the most ingenious proofs in mathematics
involve specialization, especially in the latter form. You will often encounter
proofs in future studies, and perhaps already have, that left you wondering;
"How did anyone ever think of that?' The next time this occurs, try to
notice whether the key step is, in fact, a clever application of the
skcialization technique. We now apply this technique to Example 1.
Solution to Example 1 We must prove that C has y-axis symmetry, given
that it has x-axis and origin symmetry. To do this, we begin by letting
(x, y) be an arbitrary element of C. We must prove that (-x, y) E C.
The question is how to take advantage of our two hypotheses. Well,
first, note that since C has x-axis symmetry, by hypothesis, and since
(x, y) E C, by assumption, then (x, -y) E C. Now comes the tricky part.
We have shown that (x, -y) E C and have not yet used the hypothesis
that C is symmetric with respect to the origin. This hypothesis says that
(-x, - y) E C whenever (x, y) E C; in particular then, since (x, - y) E C,
we must have that (-x, -(-y))~ C. But (-x, -(-y)) = (-x, y), so
that ( - x, y) E C, as desired.The proof in Example 1 illustrates the power and generality of universal
quantification {recall the theorem of the predicate calculus [(Vx)(p(x))] -,
p(a), Theorem 2(d), Article 3.3). If a statement, say, of the form p(x) -+ q(x)
is known to be true for all x, then p(a) -+ q(a) is true for any specific value
of a, so that, if p(a) can be proved, we may conclude q(a) by the principle of
modus ponens [Example 5 and Theorem 2(e), Article 2.31. This description
applies to the preceding proof, letting Ax, y) represent "(x, y) E C" and
q(x, y) stand for "(-x, -y) E C," while a is taken to be the ordered pair
(x, - y). Other examples of applications of this technique occur in Exercises
5(c) and 8._------
DIVISION INTO CASES
Often, in the course of setting up a proof, we arrive at a point where there
is a natural division of the argument into a finite number of cases. As we
will see, these cases must always be "exhaustive" and are often "mutually
exclusive." As one example, it may be that, in trying to prove something
about an integer m, there is an advantage in considering the two possibilities
(i.e., dichotomy) m even and m odd. Or else, in dealing with a real number
x, we might be able to use the three cases (i.e., trichotomy), x c 0, x = 0,
and x > 0, to advantage. If we know that an object x is in the union of
two sets A and B and are trying to derive some conclusion from this, the
two cases x E A and x E B (which may fail to be mutually exclusive in many