Page 254 - HOW TO PROVE IT: A Structured Approach, Second Edition

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240 Functions
Commentary . It is often most efficient to write the proof of an iff statement as
a string of equivalences, if this can be done. In the case of statement 1 this is
easy, using rules of logic. For statement 2 this strategy doesn’t quite work, but
it does give us an equivalence that turns out to be useful in the proof.

Example 5.2.4. Let A = R \{−1}, and define f : A → R by the formula
2a
f (a) = .
a + 1
Prove that f is one-to-one but not onto.
Scratch work
By part 1 of Theorem 5.2.3, we can prove that f is one-to-one by proving the
equivalent statement ∀a 1 ∈ A∀a 2 ∈ A( f (a 1 ) = f (a 2 ) → a 1 = a 2 ). Thus, we
let a 1 and a 2 be arbitrary elements of A, assume f (a 1 ) = f (a 2 ), and then prove
a 1 = a 2 . This is the strategy that is almost always used when proving that a
function is one-to-one. The remaining details of the proof involve only simple
algebra and are given later.
To show that f is not onto we must prove ¬∀x ∈ R∃a ∈ A( f (a) = x). Re-
expressing this as a positive statement, we see that we must prove ∃x ∈ R∀a ∈
A( f (a) = x), so we should try to find a particular real number x such that
∀a ∈ A( f (a) = x). Unfortunately, it is not at all clear what value we should
use for x. We’ll use a somewhat unusual procedure to overcome this difficulty.
Instead of trying to prove that f is not onto, let’s try to prove that it is onto! Of
course, we’re expecting that this proof won’t work, but maybe seeing why it
won’t work will help us figure out what value of x to use in the proof that f is
not onto.
To prove that f is onto we would have to prove ∀x ∈ R∃a ∈ A( f (a) = x),
so we should let x be an arbitrary real number and try to find some a ∈ A such
that f (a) = x. Filling in the definition of f, we see that we must find a ∈ A
such that
2a
= x.
a + 1
To find this value of a, we simply solve the equation for a:
2a x
= x ⇒ 2a = ax + x ⇒ a(2 − x) = x ⇒ a = .
a + 1 2 − x
Aha! The last step in this derivation wouldn’t work if x = 2, because then
we would be dividing by 0. This is the only value of x that seems to cause
trouble when we try to find a value of a for which f (a) = x. Perhaps x = 2is
the value to use in the proof that f is not onto.