Proving that $\gcd\left(\frac a {\gcd(a,b)},\frac b {\gcd(a,b)}\right) =1$

Question

I need to show that the $$\gcd\left(\frac a {\gcd(a,b)},\frac b {\gcd(a,b)}\right) =1$$

I'm not sure exactly how to approach this .. I tried using the fact that $\gcd(a,b)=d$ so $d=ma+nb$ but I didn't get too far

Could anyone suggest where to start?

Answer 1

No need to break apart the abstraction: we can deduce this just by the algebraic properties of the $\gcd$ operator.

If we pull out the common divisor:

$$ \gcd\left( \frac{a}{\gcd(a,b)}, \frac{b}{\gcd(a,b)} \right) = \frac{1}{\gcd(a,b)} \gcd(a, b) = 1$$

I personally find algebraic calculation, when it can be used, much easier than reasoning about divisors or linear combinations.

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In general, when the quotients $a/c$ and $b/c$ are integers, we have

$$\gcd\left( \frac{a}{c}, \frac{b}{c} \right) = \frac{1}{c} \gcd(a, b) $$

If that's not obvious that this rule should be true, rewrite it as

$$c \gcd\left( \frac{a}{c}, \frac{b}{c} \right) = \gcd(a, b) $$

and then apply the more familiar rule

$$ x \gcd(y, z) = \gcd(xy, xz) $$

(If signs are relevant to you, then that should be $ |x| \gcd(y,z) = \gcd(xy, xz) $, assuming you always take the positive value for the $\gcd$)

(In case of degeneracy, I believe the correct convention is $\gcd(0,0) = 0$, so the last identity holds even when $x=0$)

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In a suitably generalized sense, the identity

$$\gcd\left( \frac{a}{c}, \frac{b}{c} \right) = \frac{1}{c} \gcd(a, b) $$

holds even when $a/c$ and $b/c$ aren't integers. e.g. we can consider divisibility for rational numbers saying that $x$ divides $y$ if $y/x$ is an integer. Then, for example, the greatest common divisor of $1/2$ and $1/3$ is $1/6$. And we can verify

$$ \gcd \left( \frac{1}{2}, \frac{1}{3} \right) = \gcd \left( \frac{3}{6}, \frac{2}{6} \right)= \frac{1}{6} \gcd(3, 2) = \frac{1}{6} $$

Answer 2

Write $g = \gcd(a,b).$ Then write $\alpha = \frac{a}{g},$ and $\beta = \frac{b}{g}.$ So far we have $a = g \alpha, \; b = g \beta.$ Now, assume that the $\gcd(\alpha, \beta) > 1,$ so that some $t \mid \alpha$ and $t \mid \beta$ with $t > 1.$ Then $gt$ is a divisor of both $a,b$ that is larger than $g,$ which is a contradiction of the definition of $g.$

Answer 3

There is a very simple proof, Let $$\gcd(a,b) = d = am+bn$$ Now let’s multiply both side with ${1}/{d}$ $$\frac{1}{d}\cdot d = \frac{1}{d} (am+bn)$$ $$1= \frac{a}{d}\cdot m+\frac{b}{d}\cdot n$$ And this is just the definition of the linear combination, So $$\gcd\left(\frac{a}{b},\frac{b}{d}\right)=1$$

Answer 4

This is a lot simpler than it seems. Let $d=\gcd(a,b)$, by definition this is the greatest factor that $a,b$ have in common. We can therefore write $a=a'd$ and $b=b'd$, where $$\gcd(a',b')=1$$ But notice that this expression is precisely $$\gcd\left(\frac{a}{d},\frac{b}{d}\right )=\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right )=1$$

Answer 5

Let $d=gcd(\frac{a}{gcd(a,b)},\frac{b}{gcd(a,b)})$. Then $d$ is a common divisor implying $d|\frac{a}{gcd(a,b)}$ and $d|\frac{b}{gcd(a,b)}$.This tells us that $d\cdot gcd(a,b)|a$ and $d\cdot gcd(a,b)|b$ implying that $d\cdot gcd(a,b)$ is a common divisor of $a$ and $b$. Therefore, by definition of common divisor, $d\cdot gcd(a,b)|gcd(a,b)$. Because of this and since $gcd(a,b)|d\cdot gcd(a,b)$.it follows that $d\cdot gcd(a,b)$ and $gcd(a,b)$ are associates. Therefore $d$ is a unit, $d=1$.

Answer 6

Just imagine any $d>1$ dividing both $\frac a{\gcd(a,b)}$ and $\frac b{\gcd(a,b)}$. Then the product $d\gcd(a,b)$ divides both $a$ and $b$. But as per the meaning of "greatest" it cannot, because $d\gcd(a,b)>\gcd(a,b)$.

Answer 7

It is simply the special case $\: c = (a,b)\:$ of the $\,$ gcd distributive law $\ (a/c,b/c) = (a,b)/c$

See here for a few proofs of distributivity (by linearity (Bezout); universality; prime factorization).