If $c \mid a, c \mid b$ and $d = \gcd(a, b)$, then $\gcd(\frac{a}{c},\frac{b}{c}) = \frac{d}{c}$.

Question

what I've already done is I showed

$d \mid ck_{1}$ and $d \mid ck_{2}$ where $ck_1$ and $ck_2$ are $a$ and $b$

since $d$ is the gcd, then $d\geq c$ and if $d = c$, then $\gcd(\frac{a}{c},\frac{b}{c}) = \gcd(k_1,k_2)= 1 = \frac{d}{c}$.

but now I don't know how to prove when $c<d$

Maybe I am in wrong direction? Could anyone show me how to do this step?

Answer 1

It's very good that you looked at a special case.

Have you learned about Bezout's Theorem?

This theorem tells us that since $\gcd (a, b)=d$, there exist integers $x, y$ such that $ax+by=d$. Now dividing both sides by $c$ we get $\dfrac ac x + \dfrac bc y = \dfrac dc$. All of the fractions are integers (why?). Applying Bezout's Theorem again we conclude that $\gcd \left(\dfrac ac, \dfrac bc \right)$ divides $\dfrac dc$. Now if we can show the other way around, namely that $\dfrac dc$ divides $\gcd \left(\dfrac ac, \dfrac bc\right)$, we would be done. How can we show this? By showing that $\dfrac dc$ divides both $\dfrac ac$ and $\dfrac bc$. And indeed, since $d|a$, $a=dk$ for some $k \in \mathbb{Z}$ so $\dfrac ac = \dfrac dck$; and again the fractions are integers. The exact same argument can be used for $\dfrac bc$, so we are done.

Answer 2

$\exists r \text{ and } s\in\mathbb Z: ra+sb=d \implies r\frac ac +s \frac bc=\frac dc\implies gcd(\frac ac, \frac bc) | \frac dc$...
But, conversely, $\frac dc | \frac ac$ since the ratio is $\frac ad$...

Similarly, $\frac dc | \frac bc$...

Now using that $t \frac ac+u\frac bc=gcd(\frac ac, \frac bc)$, for some $t,u\in \mathbb Z$, we get $\frac dc| gcd(\frac ac, \frac bc)$...

Answer 3

Let $a =\prod_{p \in P} p^{a_p} $, $b =\prod_{p \in P} p^{b_p} $, $c =\prod_{p \in P} p^{c_p} $, $d =\prod_{p \in P} p^{d_p} $.

Since $c|a, c|b$, $c_p \le a_p$ and $c_p \le b_p$ so $c_p \le \min(a_p, b_p)$.

Since $d = \gcd(a, b)$, $d_p = \min(a_p, b_p)$.

$a/c =\prod_{p \in P} p^{a_p-c_p} $, $b/c =\prod_{p \in P} p^{b_p-c_p} $, so $\gcd(a/c, b/c) =\prod_{p \in P} p^{\min(a_p-c_p, b_p-c_p)} $.

$d/c =\prod_{p \in P} p^{d_p-c_p} $, so if $d_p-c_p =\min(a_p-c_p, b_p-c_p) $ then we are done.

But $\min(a_p-c_p, b_p-c_p) =\min(a_p, b_p)-c_p =d_p-c_p $ and we are done.