0

I understand the definition of coprime, which is $\gcd(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}) = 1$ or $x(\frac{a}{\gcd(a,b)}) + y(\frac{b}{\gcd(a,b)}) = 1$.

I am pretty sure that I have to use the logic that $a = qb +r$ for $r = \{0,1,2,3, ..., b-1\}$

with the algorithm of Euclides this would mean that $$\gcd(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}) = \gcd(\frac{qb+r}{\gcd(a,b)},\frac{b}{\gcd(a,b)}) = \gcd(\frac{r}{\gcd(a,b)},\frac{b}{\gcd(a,b)})$$

Now given for that \gcd, I am unsure how to prove that this would mean that $\gcd(\frac{r}{\gcd(a,b)},\frac{b}{\gcd(a,b)}) = 1$.

Martin Sleziak
  • 56,060
Texts
  • 47

2 Answers2

0

Since $k$ is a divisor of both $ku$ and $kv$ you have $$\gcd(ku,kv)=k\gcd(u,v)$$ so

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

Dividing by $\gcd(a,b)\neq 0$ you get $\gcd\left(\frac a{\gcd(a,b)},\frac b{\gcd(a,b)}\right)=1$

zwim
  • 29,833
0

If $r$ is a $gcd(a,b)$, then there exist $x,y$ integers such that $ax+by=r$. Dividing both sides by $r$, we get $x \dfrac{a}{r}+y\dfrac{b}{r}=1$, which proves that $gcd(\dfrac{a}{r},\dfrac{b}{r})$ is $1$.

Mihail
  • 2,341
  • 1
  • 14
  • 26