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

64 The Nuts and Bolts of Proof, Third Edition
We want to compare the two functions g and h. They are both defined for
all real numbers as they are inverses of/ To compare them, we have to
compare their outputs for the same value of the variable. While we have a
formula for g, we do not have a formula for h. So we need to use the
properties of h and g:
g{x) = g{foh{x))
= g{f{h{x)))
= (gof)(hix)) = h(x).
Therefore, g = h. So, the inverse of/is unique.
•
A shorter and less expHcit proof of the existence part of the statement in
Example 2 relies on a broader knowledge of functions and inverse function.
We will mention it for sake of completeness. The function/is one-to-one and
onto; therefore, it will have an inverse function. See proof 1 in the Exercises
for this section.
EXAMPLE 3. We proved that if n is an integer number larger than 1, then
n is either prime or a product of prime numbers. Thus, we can write:
n = pixp2X '•- xpk
where the pj are prime numbers, and p\ S Pi S • - • S Pk- This factorization of
n is unique.
Proof: Let us assume that there are at least two ways of writing n as the
product of prime factors listed in nondecreasing order. Therefore,
p\xp2X
•" xpk = n = qix qix '" X qs.
Thus, the prime factor p\ divides the product q\ x qjx -" x qs (indeed
qi X q2 X • •' X qs/pi = P2 x ps x -- - x pk). This implies that pi divides at
least one of the qj. Let us assume that p\ divides qi (we can reorder the qj).
As qi is prime, this implies that pi = qi. Therefore, after simplifying pi
and ^1, we have:
P2X'--xpk = q2X"'Xqs.
Similarly, p2 divides <?2 x • • • x ^5. So, we can assume that p2 divides q2.
This again implies that p2 = q2- Thus,
P3 X " • X pk = q3 y< "' ^ Qs-
So, if k < s, we obtain 1 = qk+i x -- • x qs.
This equaUty is impossible because all the qj are larger than 1 (they are
prime numbers).