Prove that if $a$ and $bc$ are nonzero integers, then $(ca,cb) = |c|(a,b)$.

Question

Prove that if a and bc are nonzero integers, then $$(ca,cb) = |c|(a,b).$$

Basically, I was confused by the statement of the question. In particular, I was unsure if choosing a and bc to be nonzero was arbitrary. In other words, why choose a and bc to be nonzero instead of just saying "Prove that if a, b, and c are nonzero integers..." or, as the person who provided a solution suggested, why not just say "Prove if a, b, and c are integers with abc nonzero..."

I'm not sure if there is some advantage to wording the question such that a and bc are given to be nonzero integers rather than, say, some other equivalent form to word the question, such as the examples I added to my posted question (after editing).
However, I think that is really where my question lies. I assumed there was some premeditation behind the wording such that it was intended to be a hint or something. For example, maybe that I was supposed to view bc as a product for some particular reason.

But I guess the wording of the question was arbitrary? — GaMbiTaaaa, Jan 29 '14 at 00:20
There's no reason why it should be or has been written that way. A slightly stronger statement is that it is true if c is nonzero and a and b are not both 0 - perhaps they were trying to include that case and did it (very) wrongly. — user11977, Jan 29 '14 at 00:28

score 2 · Accepted Answer · answered Jan 28 '14 at 09:17

2

Let $d = (ca, cb)$ and $d' = |c|(a,b)$.

By the property of gcd, $\exists x,y$ s.t. $d = cax + cby$. $\frac{d}{|c|(a,b)} = \frac{c}{|c|} \frac{a}{(a,b)} x + \frac{c}{|c|} \frac{b}{(a,b)} y$. Note that each fraction on the right is an integer since $|c|$ divides c and $(a,b)$ divides $a$ and $b$. Hence $d'|d$.

Now $d' = |c|(au+bv) = |c|au + |c|bv$ for some $u,v$ by the property of gcd. $(ca,cb)$ divides $|c|a$ and $|c|b$ so $d|d'$.

Since both $d$ and $d'$ are positive, $d = d'$ which is the required result.

answered Jan 28 '14 at 09:17

Kelvin Soh

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Thanks. That is essentially the same solution I came up with, but wasn't sure because the statement of the problem confused me. I was thinking the problem would make more sense if it said let a and b be integers both not zero. Not sure why it says "a and bc are nonzero." – GaMbiTaaaa Jan 28 '14 at 09:32
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$c$ needs to be nonzero as well (unless we define the gcd of 0 to be trivially zero). $bc$ nonzero implies both $b$ and $c$ are both nonzero. So technically the question could have written the condition as the product $abc$ is nonzero. – Kelvin Soh Jan 28 '14 at 14:56

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

This is the fundamental gcd distributive law $\rm\:(ax,bx) = (a,b)x.\:$ It is instructive to prove it in various ways. here are $3$ proofs, using Bezout's identity, universal gcd laws, unique factorization.

First we show that the gcd distributive law follows immediately from the fact that, by Bezout, the gcd may be specified by linear equations. Distributivity follows because such linear equations are preserved by scalings. Namely, for naturals $\rm\:a,b,c,x \ne 0$

$\rm\qquad\qquad \phantom{ \iff }\ \ \ \:\! c = (a,b) $

$\rm\qquad\qquad \iff\ \: c\:\ |\ \:a,\:b\ \ \ \ \ \ \&\ \ \ \ c\ =\ na\: +\: kb,\ \ \ $ some $\rm\:n,k\in \mathbb Z$

$\rm\qquad\qquad \iff\ cx\ |\ ax,bx\ \ \ \&\ \ \ cx = nax + kbx,\ \,$ some $\rm\:n,k\in \mathbb Z$

$\rm\qquad\qquad { \iff }\ \ cx = (ax,bx) $

The reader familiar with ideals will note that these equivalences are captured more concisely in the distributive law for ideal multiplication $\rm\:(a,b)(x) = (ax,bx),\:$ when interpreted in a PID or Bezout domain, where the ideal $\rm\:(a,b) = (c)\iff c = gcd(a,b)$

Alternatively, more generally, in any integral domain $\rm\:D\:$ we may employ the universal definitions of GCD, LCM to generalize the above proof.

Theorem $\rm\ \ (a,b)\ =\ (ax,bx)/x\ \ $ if $\rm\ (ax,bx)\ $ exists in $\rm\:D.$

Proof $\rm\quad\: c\ |\ a,b \iff cx\ |\ ax,bx \iff cx\ |\ (ax,bx) \iff c\ |\ (ax,bx)/x\ \ \ $ QED

Such universal definitions often serve to simplify proofs, e.g. see this proof of the GCD * LCM law.

Alternatively, comparing powers of primes in unique factorizations, it reduces to the following $$ \min(a+c,\,b+c)\ =\ \min(a,b) + c$$

The proof is precisely the same as the prior proof, replacing gcd by min, and divides by $\le$, and

$$\begin{eqnarray} {\rm employing}\quad\ c\le a,b&\iff& c\le \min(a,b)\quad&&\rm[universal\ definition\ of\ \ min]\\ \rm the\ analog\ of\quad\ c\ \, |\, \ a,b&\iff&\rm c\ \ |\ \ gcd(a,b)\quad&&\rm[universal\ definition\ of\ \ gcd] \end{eqnarray}$$

Prove that if $a$ and $bc$ are nonzero integers, then $(ca,cb) = |c|(a,b)$.

2 Answers2