Proof involving gcd.

Question

Here is the question that I am working on:

Suppose $a,b \in \mathbb{Z^+}$ with $g = \gcd(a,b)$. Show that $\gcd(\frac{a}{g},\frac{b}{g}) = 1.$

I would like to solve this using Bézout's Identity (I believe that there is a nicer way to show this result holds).

My alternate proof

We know that $g|a$ and $g|b$. Thus, $\exists k_1,k_2 \in \mathbb{Z}$ such that

$$gk_1 = a$$ and $$gk_2 = b.$$

We want to show that $gcd(k_1,k_2) = 1.$ Suppose that $d|k_1$ and $d|k_2$ where $d \in \mathbb{N}.$ We want to show that $d = 1.$ Thus, $\exists l,m \in \mathbb{Z}$ such that

$$a = gdl$$ and $$b = gdm.$$

Thus, $gd|a$ and $gd|b.$ So, $gd$ is a common divisor. So, $gd \leq g,$ or $d \leq 1,$ but $d \in \mathbb{N}$. Thus, $d \geq1$. Thus, $d = 1.$ so, the only common divisor of $k_1$ and $k_2$ is $1$. Thus, $gcd(k_1, k_2) = gcd(\frac{a}{g}, \frac{b}{g}) = 1.$

QED

Answer 1

HINT: You know there are integers $m$ and $n$ so that $ma+nb = g$. Divide by $g$. What does this tell you?

Answer 2

Notice that $\gcd(\frac a{\gcd(a,b)}, \frac b{\gcd(a,b)})$ divides both $\frac a{\gcd(a,b)}$ and $ \frac b{\gcd(a,b)}$. So $\gcd(\frac a{\gcd(a,b)}, \frac b{\gcd(a,b)})*\gcd(a,b)$ divides both $\frac a{\gcd(a,b)}*\gcd(a,b) = a$ and $ \frac b{\gcd(a,b)}*\gcd (a,b) = b$.

So $\gcd(\frac a{\gcd(a,b)}, \frac b{\gcd(a,b)})*\gcd(a,b)$ is a common factor $a$ and $b$.

But $\gcd(a,b)$ was the greatest common factor so......

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Bezout's identity: Let $ak + bj = \gcd(a,b)$ then $\frac a{\gcd(a,b)} *k + \frac b{\gcd(a,b)}*j = 1$. So .....