$gcd(ka,kb) = k \times gcd(a,b)$

Question

I know this this a duplicate, but I would like to know if my solution is fine.

$gcd(ka,kb) = (ka)f + (kb)j$ for some $k,a,b,j,f \in Z$

since $gcd(f,j) = 1$ then we can write;

$(ka)f + (kb)j = k(af + bj) = gcd(a,b) \times k$.

Is this correct? if not why?

The idea is there but the formulation is not complete, you have to explain where $f,j$ come from and the final assertion is not sufficient to conclude — Elaqqad, Apr 05 '15 at 18:37
As advocated below by @ktoi and me, a better approach might be a proof in any GCD domain (not necessarily a Bézout domain), using only the natural definition of "the" GCD. Such a proof may be found for instance here, in user2345215's answer. — Anne Bauval, Sep 22 '23 at 15:34

score 1 · Answer 1 · answered Apr 05 '15 at 21:39

A few things,

Firstly on the presentation side, you have used Bézout's identity but have not explicitly stated that you are using it. I'm assuming this was proved beforehand at some point, though otherwise you should prove it.

Also, you are proving that for any integers $a,b,k,$ the equality holds. So the proof should start by letting these variables be arbitrary. Then by Bézout's identity, there exists $f,j \in \Bbb Z$ s.t. $(ka)f + (kb)j = \gcd(ka,kb).$

Note that the values of $f,j$ are dependent of $a,b,k,$ so you have to define them separately, making it clear what they are.

Secondly, there is a flaw in your proof. Unless you have additional restrictions (in which case you must clearly state and possibly prove if you haven't done so), the converse of Bézout's identity does not hold in general, even if the coefficients are relatively prime.

As a hint, working from the definition of the greatest common divisor may be a better approach. Your lecturer may have defined it differently, but I'm thinking of this definition:

Given two integers $a,b,$ we define $\gcd(a,b)$ to satisfy the following properties,

$$ \gcd(a,b) \mid a \ \ \mathrm{and} \ \ \gcd(a,b) \mid b, $$ $$ \forall d \in \Bbb Z, \ d \mid a \ \ \mathrm{and} \ \ d \mid b \implies \gcd(a,b) \mid d.$$

Try to show that $k\gcd(a,b)$ satisfies these properties. If your lecturer used a different definition, you should either prove the equivalence of the definitions first, or try a different approach.

+1 but misprint: replace $\gcd(a,b)\mid d$ by $d\mid\gcd(a,b).$ Working from the definition of the GCD is indeed a better approach because it proves the result in any GCD domain (not necessarily a Bézout domain). Your two lines definition may be reformulated by a one line equivalence: $$\forall d\in\Bbb Z\quad[(d\mid a\text{ and }d\mid b)\iff d\mid\gcd(a,b)].$$ — Anne Bauval, Sep 22 '23 at 15:10

$gcd(ka,kb) = k \times gcd(a,b)$

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