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My professor presented the following proof and I am wondering if it makes sense. There is one piece of it that I do not quite grasp.

If $d|a$ and $d|b$ and $\displaystyle\gcd(\frac{a}{d}, \frac{b}{d})=1$ then $\gcd(a,b)=d$

Proof: We know that there are $x$ and $y$ such that $a=dx$ and $b=dy$. But since we are given $\displaystyle\gcd(\frac{a}{d}, \frac{b}{d})=1$ we have that $\gcd(x,y)=1$. Any common divisor of $a$ and $b$ that is greater than $d$ would have to be a common divisor of $x$ and $y$. But $x$ and $y$ have no common divisors greater than 1. So $\gcd(a,b)=d$.

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It is not clear to me that the bolded statement above is true. If it is true then could someone verify that it is true and possibly elaborate on why it must be true.

Jac Frall
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    The bolded statement doesn't make sense. Perhaps the intended statement was "any common divisor of $a$ and $b$ that is greater than $d$ must be of the form $de$ where $e$ is a common divisor of $x$ and $y$". But that (while true) requires proof, essentially equivalent to the statement to be proved in the first place. – Greg Martin Jan 23 '20 at 21:24
  • @GregMartin I agree the bolded statement doesn't make sense. However, your intended statement of "any common divisor of $a$ and $b$ that is greater than $d$ must be of the form $de$ where $e$ is a common divisor of $x$ and $y$" is not always true. For example, have $d = 2$, $a = 6$ and $b = 18$. In this case, $3$ is greater than $d = 2$ and is a common divisor of $a$ and $b$, but you don't have $3 = de$ where $e$ is an integer. I believe you perhaps meant something more like what I suggest in my answer below. – John Omielan Jan 23 '20 at 22:35
  • The argument is either incorrect or incomplete without any explicit justification of the highlighted claim. It is impossible to know which is the case without further elaboration from the author. The theorem is a special case of the fundamental GCD Distributive Law for which there are various possible proofs (follow the link). – Bill Dubuque Jan 23 '20 at 23:00
  • @JohnOmielan Sorry, I misread the OP and was thinking that $d=\gcd(a,b)$ was known; now that I see that's not the case, I completely agree with your comment. – Greg Martin Jan 23 '20 at 23:37

1 Answers1

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I agree with the part of Greg Martin's comment which says

The bolded statement doesn't make sense.

I obviously can't be exactly sure what your professor was trying to convey, but I believe it was something along the line of

Any common divisor of $a$ and $b$ that is greater than $d$ and is of the form $de$ would have $e$ be a common divisor of $x$ and $y$.

The rest of the stated proof would then follow. Using this same basic approach, here is how I would write a formal proof. Let

$$f = \gcd(a,b), \; a = fg, \; b = fh, \; \gcd(g,h) = 1 \tag{1}\label{eq1A}$$

Since $d \mid a$ and $d \mid b$, then $d \mid f$. Then for some integer $e \ge 1$, you have

$$f = ed \tag{2}\label{eq2A}$$

This gives

$$\begin{equation}\begin{aligned} \gcd\left(\frac{a}{d},\frac{b}{d}\right) & = \gcd\left(\frac{(ed)g}{d},\frac{(ed)h}{d}\right) \\ & = \gcd(eg,eh) \\ & = e(\gcd(g,h)) \\ & = e \\ & = 1 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Note going from the second to third lines used, as Bill Dubuque's comment states, the GCD distributive law, with multiple proofs for this such as in his answer. You thus have $e = 1$, so using \eqref{eq1A}, \eqref{eq2A} and \eqref{eq3A} gives

$$\gcd(a,b) = f = ed = d \tag{4}\label{eq4A}$$

John Omielan
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  • It seems to beg the principle by using $,\gcd(eg,eg) = e\gcd(g,h), $ in $(3)$. In any case, we already have plenty of proofs of the gcd distributive law. – Bill Dubuque Jan 23 '20 at 22:40
  • @BillDubuque Thanks for the feedback. I've added an explicit mention of using the GCD distributive law, along with links to your comment & your answer you provided. – John Omielan Jan 23 '20 at 22:55