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Proof.

Let c = (a/d, b/d). Then c | a/d, and so cd | a. Also c | b/d, and so cd | b.

My question is, did this happen because of this simple algebra? ie. c | a/d So cx = a/d for some integer x then cx/d = a

Next step of proof.. Thus, cd is a common divisor of a and b. Therefore, cd <= d, which implies c = 1.

My question for this part: how did I conclude cd <= d?

JustinL
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1 Answers1

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If $cd$ is a divisor of $a$ and $b$, it is a divisor of the greatest common divisor, which is $d$. Hence $cd|d$.

Remark:

If you are familiar with Bezout's identity.

We have $$ax+by=d, x, y \in \mathbb{Z}$$

$$x\left( \frac{a}d\right)+ y\left( \frac{b}d\right)=1$$

and we can conclude.

Siong Thye Goh
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  • Sorry sir, i am not familiar with Bezout’s identity. Can this be done without that knowledge? – JustinL Nov 08 '18 at 04:22
  • $cd$ is a divisor of $a$ and $b$, it can't be larger than the greatest common divisor, hence $cd \le d$. btw, in your working $cx=\frac{a}d$, the next line should be $cxd=a$ rather than dividing by $d$. – Siong Thye Goh Nov 08 '18 at 04:23