Download Greatest common divisor as a product of primes

Greatest common divisor as a product of primes
Clive Newstead, 18th March 2014
The Fundamental Theorem of Arithmetic (FTA) is all about products of primes: it tells you that
every natural number greater than 1 has a representation as a product of primes, and moreover
that expression is unique up to reordering. We can use this to find an expression of the greatest
common divisor of two natural numbers by looking at their prime factorisation.
Let a, b ∈ N and let p1 , p2 , . . . , pn be a list of prime numbers exhausting all the primes that appear
in the factorisations of a and b. (Note that if a = 1 or b = 1 then there are no primes in their
prime factorisations, but that’s okay because raising a natural number to the power 0 gives 1.)
We can then write
a = pα1 1 pα2 2 · · · pαnn
b = pβ1 1 pβ2 2 · · · pβnn
where αi , βi > 0 for all 1 6 i 6 n. (We need to allow them to be zero in case one of the primes
appears in the factorisation of a but not b, for example.)
For 1 6 i 6 n write δi = min{αi , βi }.
Theorem. Given natural numbers a, b as above,
gcd(a, b) = pδ11 pδ22 · · · pδnn
That is, the gcd can be calculated by raising all the prime factors that appear in the factorisations
of a and b to the smallest power that appears.
Proof. Let G = gcd(a, b) and P = p1δ1 pδ22 · · · pδnn . We’ll prove P 6 G and G 6 P separately;
together these will imply that P = G, and then we’ll be done.
First we prove P 6 G. Since δi 6 αi and δi 6 βi , we can write αi = δi + ki and βi = δi + `i for
some ki , `i > 0. Hence
a = pα1 1 pα2 2 · · · pαnn
=
=
=
=
(prime factorisation)
p1δ1 +k1 pδ22 +k2 · · · pδnn +kn
(pδ11 pk11 ) · (pδ22 pk22 ) · · · (pδnn pknn )
(pδ11 pδ22 · · · pδnn ) · (pk11 pk22 · · · pknn )
P · (pk11 pk22 · · · pknn )
(since αi = δi + ki )
(properties of indices)
(re-ordering the factors)
(definition of P )
Since pk11 pk22 · · · pknn ∈ Z, we can deduce that P | a. Similarly we have
b = P · p`11 p`22 · · · p`nn
and so P | b. Thus P is a common divisor of a and b. By definition of the gcd, P 6 G.
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It remains to show G 6 P . We’ll do this by first showing that G | P . Since G, P ∈ N, the fact
that G 6 P will follow from Proposition 1(iv) in the lecture notes.
We can write G = pλ1 1 pλ2 2 · · · pλnn k, where λi > 0 for all 1 6 i 6 n and k ∈ N, such that none of
the pi divide k. (That is, we’ve just taken the prime factorisation of G, and clustered together
all the primes not amongst the pi s and called this number k.) We’ll prove:
(1) k = 1. Then we’ll just have G = pλ1 1 pλ2 2 · · · pλnn .
(2) λi 6 δi for all 1 6 i 6 n. Then we’ll have G | P (for the same reason as P | a above).
We prove (1) and (2) by contradiction.
For (1), suppose k 6= 1. Then k > 1, so it has a prime factor by FTA, say q. By definition of
k, q 6= pi for any i, because we specified that none of the pi s divide k. But q | G, and we know
that G | a since G = gcd(a, b), so by Proposition 1(i), q | a. By FTA, q appears in the prime
factorisation of a, so q = pi for some i. But this contradicts our assumption that q 6= pi for any
i. We conclude that k = 1.
For (2), suppose λi > δi for some i. Without loss of generality, i = 1 (otherwise just re-label
the prime factors). Since δ1 = min{α1 , β1 }, we must have δ1 = α1 or δ1 = β1 . Without loss of
generality, suppose δ1 = α1 . (The case δ1 = β1 is identical.) Then λ1 > α1 . Now G | a, so there
exists s ∈ N such that sG = a. Looking at the prime factorisations, we have:
s · pλ1 1 pλ2 2 · · · pλnn = pα1 1 pα2 2 · · · pαnn
Dividing by pα1 1 gives, using laws of indices, that
s · p1λ1 −α1 pλ2 2 · · · pλnn = pα1 1 −α1 pα2 2 · · · pαnn
which, noticing that pα1 1 −α1 = p01 = 1, can be rewritten as
s · p1λ1 −α1 pλ2 2 · · · pλnn = pα2 2 · · · pαnn
Since λ1 > α1 , we have λ1 − α1 > 0, so p1 divides the left-hand side of this equation. Hence p1
divides the right-hand side. But by FTA, if p1 divides the right-hand side then it would appear
in the factorisation. . . so we have a contradiction! We conclude that λi 6 δi for all 1 6 i 6 n.
As mentioned above, this implies that G | P , and hence G 6 P by Proposition 1(iv).
Since G 6 P and P 6 G, we have P = G by antisymmtery of 6, and we’re done.
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