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.

@Xam: This question seems better written and (arguably) has better answers. I'd be tempted to close the other one instead. — Ilmari Karonen, Sep 15 '17 at 00:07
This is a special case of the linked GCD DIstributive Law, whose general proof is no more difficult than this special case — Bill Dubuque, Sep 01 '19 at 16:10

score 10 · Accepted Answer · answered Sep 14 '17 at 18:38

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.

score 2 · Answer 2 · edited Sep 14 '17 at 18:35

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 $pd$.

edited Sep 14 '17 at 18:35

Michael Hardy

1

answered Sep 14 '17 at 18:29

Apurv Anand

332

score 2 · Answer 3 · edited Sep 14 '17 at 18:41

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

DanielWainfleet · Answer 4 · 2017-09-14T19:30:21.130

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)). $

score -1 · Answer 5 · answered Sep 14 '17 at 23:37

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.

score -1 · Answer 6 · answered Mar 03 '18 at 16:53

-1

since d is a common divisor of a and b, we can have integers r and s such that a=dr and b=ds. now gcd(a/d,b/d)=1 which implies gcd(r,s)=1, multipliying d both sides leads to d.gcd(r,s)=d------(1) we know by theorem, that gcd(ka,kb)=k.gcd(a,b). thus d.gcd(r,s)=gcd(dr,ds)=gcd(a,b)-----(2)

from (1)and (2) gcd(a,b)=d

answered Mar 03 '18 at 16:53

manas pandey

1

First, please use MathJax. Second, if you are simply assuming that $d$ is some common divisor, you can't conclude that $\gcd(a/d,b/d)=1$. – rogerl Mar 03 '18 at 17:59
first, i will (if i can) do so.please guide me how to use MathJax.. Second, thanks....... . gcd(a/d,b/d)=1 is supposed to be condition/assertion which concludes gcd(r,s)=1. – manas pandey Mar 10 '18 at 07:14
a/d and b/d are integers implies d is a divisor of both, and is common too. – manas pandey Mar 10 '18 at 07:39

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

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