epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

80 The Nuts and Bolts of Proof, Third Edition
EXAMPLE 2. Let f{x) = ^^, and let y and z be two real numbers larger
than l.If/();)=/(z), then ); = z.
(This proves that the function / is one-to-one on the interval (1, -hoc).
See the front material of the book for the definition of one-to-one.)
Proof: Because/(y) =f{z), it follows that
);2 + 1 z^-\-l'
We can now multiply both sides of the equation by (y^ + l)(z^ + 1). This
is a nonzero expression because y^ + 1 / 0 and z^ + 1 7^ 0. Therefore, we
obtain:
which can be simplified as:
zy^ + z = z^y + y
(z-y)(l-yz) = 0
Thus, either z — y = 0 or 1 — yz = 0.
The first equahty implies that y = z.
The second equality implies that yz=l. This is not possible because
y and z are two real numbers larger than 1. Therefore, the only possible
conclusion is y = z. •
There are at least two special ways to prove that a number is equal to
zero, and both of them use the absolute value function and its properties.
(See front material of the book for the definition of absolute value.)
Method 1. To prove that a = 0, we can prove that |fl| — 0.
(This is true because, by definition of absolute value of a number, |a| = 0 if
and only if a = 0.)
Method 2. Let a be a real number. Then a=:0 if and only if \a\ < s for
every real number s > 0.
The second method is often used in calculus and analysis. We can prove
that the two methods are equivalent.
EXAMPLE 3. Let a be a real number. Then |fl| = 0 if and only if |a| < sfor
every real number ^ > 0.
Proof: Because this is an equivalence statement, the proof has two parts.
Part 1. If |a| =0, then \a\ < ^ for every real number s > 0.