GCD of two numbers divided by their greatest common divisor is 1

Question

Im trying to prove that, given $a,b$ with at least one of $a,b \neq 0$, $$ \gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)=1 $$ I have tried to prove the identity $$ \gcd(c\cdot a, c\cdot b) = c\cdot \gcd(a,b) $$ with $c = \dfrac{1}{\gcd(a,b)}$

However I'm having trouble understanding the proof.

Thank you for your time.

Answer 1

Let $c = \gcd\left( \dfrac a {\gcd(a,b)}, \dfrac b {\gcd(a,b)} \right).$

Then $c$ divides $\dfrac a {\gcd(a,b)}$ and $c$ divides $\dfrac b {\gcd(a,b)}.$

Thus $c\cdot\gcd(a,b)$ divides $a$ and $c\cdot\gcd(a,b)$ divides $b.$

So $c\cdot\gcd(a,b)$ is a common divisor of $a$ and $b$.

If $c>1$ then $c\cdot\gcd(a,b)$ is a greater common divisor than the greatest common divisor.

Answer 2

Let $d = (a,b)$. If $\big ( \frac{a}{d}, \frac{b}{d} \big ) = p$, not 1.
Then, $pd \mid a$ and $pd \mid b$ and so the gcd of $a$ and $b$ will become $pd$.

Answer 3

If you know the fact that whenever $\gcd(a,b) = d $, we have integers $x$ and $y$ such that $ax+by = d$. This can be shown by the division algorithm. Suppose $ax+by =1$ for some integers $x$ and $y$, then $d \mid (ax+by)$ as $d\mid a$ and $d\mid b$, therefore $d\mid 1$. As $d$ is integer $d=1$. So we proved that $ax +by=1$ iff $a$ and $b$ are co-prime. Then as $1=(a/d)x+(b/d)y$ where $a/d, b/d$ are integers as $d$ is $\gcd$, $1 = \gcd(a/d, b/d).$

Answer 4

All variable-symbols stand for non-zero integers here.

We have $a=a'd$ and $b=b'd$ where $d=\gcd (a,b).$

Suppose $n$ divides both $a'$ and $b'.$ Then $a'=na''$ and $b'=nb'',$ so $a=a''(nd)$ and $b=b''(nd).$ So $nd$ is a common divisor of $a$ and $b,$ and by the def'n of $\gcd$ we have $d\geq |nd|.$ And we also have $nd\ne 0.$ So $|n|=1.$

Therefore $1=\gcd (a',b')=\gcd (a/d,b/d)=\gcd (a/\gcd (a,b), b/\gcd (a,b)). $

Answer 5

Let g be a common divisor of both a and b, and let g' be a common divisor of both a/g and b/g. Then (a / g) / g' = a / (gg') is an integer, and so is (b / g) / g' = b / (gg'). Therefore g*g' is a common divisor of a and b.

If g is the greatest common divisor of a and b, then g*g', which is also a common divisor, is not greater than g, and therefore g' = 1.