0

Lemma. Suppose two integers, m and n, are relatively prime. Then, every integer d that divides the product mn can be written in exactly one way, d = bc, with b dividing m and c dividing n.

This is the proof given in the book :

Proof. We will show the contrapositive. That is, “P implies Q” is logically the same as “not Q implies not P.” Suppose that d divides mn. We can write d = bc and also d = b'c', with both b and b' dividing m and both c and c' dividing n. We need to show that m and n are not relatively prime. We canwrite the integer (mn)/d in two ways:

$\frac{(mn)}{d} = \frac{m}{b} \frac{n}{c} = \frac{m}{b'} \frac{n}{c'} $ .

So, cross multiplying,

$\frac{bc}{b'c'} = 1$.

Since, b is not equal to b', some prime p dividing b does not divide b' (or vice versa); it must divide c' or it won’t cancel out to give 1 on the right side. So, p divides b, which divides m, and p divides c', which divides n; so, m and n are both divisible by p. divisible by p.

My problem with the proof : Say 4 is not equal to 2 but then 2|4 (See the highlighted line in bold and italics), so shouldn't the proof explicitly say that b and b' are relatively prime but then the proof wouldn't be valid.

Am I saying something wrong here ?

Bill Dubuque
  • 272,048
Samyak Parashar
  • 1
  • 1
    The argument has many errors, e.g. it only (attempts to) prove uniqueness, not existence, and the claim is false as written since $,d = bc = (-b)(-c).,$ If that is characteristic of proofs in this textbook then you should consider finding a better book. – Bill Dubuque Jul 20 '23 at 15:00
  • For existence, show that $d=\gcd(d,m)\gcd(d,n)$. Also, one should restrict to positive integers to avoid having factorizations with two negative factors as in Bill Dubuque's comment. – Geoffrey Trang Jul 20 '23 at 15:26
  • You can fix it using multiplicity, i.e. some prime $p$ occurs to higher power in $b$ or $b'$ (wlog $b$) so $p$ must also divide $c'$ (by Euclid's Lemma). Alterntively we can use gcds as here when (equivalently) proving uniqueness of reduced fractions – Bill Dubuque Jul 20 '23 at 15:29
  • Existence is easy by gcd laws: $,de = mn \Rightarrow (d,m)(d,n) = d(d,\color{#c00}{m,n},e) = d,$ by $\color{#c00}{(m,n)=1},\ d > 0\ \ $ – Bill Dubuque Jul 20 '23 at 15:35

0 Answers0