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

Special Kinds of Theorems 79
(For definitions of the least common multiple and the greatest common
divisor and their properties, see the front material of the book.)
By definition of GCD(a, b), we can write:
a = dn and b = dp
with n and p positive, relatively prime integers. Therefore,
ab
(dn)(dp)
Let M = dpn. We want to prove that M = L.
Part 1. We will prove that L< M.
Clearly, M is a multiple of both a and b. Indeed, M = pa and M = nb, with
n and p positive integers. As M is a common multiple, it will be larger than
(or equal to) the least common multiple, L.
Thus, L<M.
Part 2. We will prove that M <L,
By definition. Lis a multiple of both a and b. Thus,
L = at
L = bs
and
where t and s are positive integers. Therefore, at = bs. Substituting a and b,
we obtain:
Thus,
L = dnt = dps.
nt = ps.
This implies that n divides ps. As p and n are relatively prime, it follows
that n divides s. Thus, s = nk for some integer fc > 1. So,
nt = pnk,
that is,
for some integer /c> 1. Then,
t = pk
L = at = (dn)(pk) = dnpk.
This implies that L = Mk, with fc> 1.
Thus, L>M.
Because L<M and L> M, we can conclude that L = M. •