How would one use Bézout's theorem to prove that if $d = \gcd(a,b)\ \text{then} \ \gcd(\dfrac{a}{d}, \dfrac{b}{d}) = 1$.

Question

Note: I have checked the questions with the same title and I am after something more specific.

I am doing my first course in discrete mathematics, and came across the following proposition that I was asked to prove:

Let $a,b,d \in \mathbb{Z}$. If $d = \gcd(a,b)\ \text{then} \ \gcd\Bigl(\dfrac{a}{d}, \dfrac{b}{d}\Bigr) = 1$.

My first thought was to prove it by contradiction, and I did it as follows,

Let $a,b,c,d \in \mathbb{Z}$ and suppose that $d = \gcd(a,b).$

Assume that $\ \gcd\Bigl(\dfrac{a}{d}, \dfrac{b}{d}\Bigr) = c$, where $c \neq 1 $.

Then $c\mid\frac{a}{d} \ \text{and} \ c\mid\frac{b}{d}$, that is, $a = cmd \ \text{and} \ b = cnd$ , where $m,n \in \mathbb{Z}$

This implies that there is a integer $cd$ that divides both $a$ and $b$, where $cd > d$. But, $d$ is the greatest common divisor of $a$ and $b$, which yields a contradiction. Therefore, the assumption is false, and $\ \gcd\Bigl(\dfrac{a}{d}, \dfrac{b}{d}\Bigr) = 1$.

This is my approach, but the solution presented by the TA's notes uses a different approach which, given the way it was presented, I could not understand how it would prove the proposition. The approach uses Bézout's theorem, which was presented to us in following manner:

Consider the equation $$ax+by=c,$$ where $a,b,c$ are integers, with $a$ and $b$ not both zero. if $ c=d$, where $d$ is the greatest common divisor of $a$ and $b$ then the equation has a solution in integers $x,y$.

if $d\mid c$ then the equation has a solution in integers.

if $d\nmid c$ then the equation has no solution in integers.

The proof presented went on to apply this theorem to prove the proposition:

Consider the equation$$ax+by=d,$$ where $d = \gcd(a,b)$ (with integer coefficients).

Dividing both sides by $d$ yields,

$$\frac{a}{d}x+\dfrac{b}{d}y=1,$$ where $\dfrac{a}{d}$ and $\dfrac{b}{d}$ are both integers (follows from definition of that $\gcd$)

and then it goes on to say by Bézout's theorem, we can conclude that $\ \gcd\Bigl(\dfrac{a}{d}, \dfrac{b}{d}\Bigr) = 1$

Now, I am really confused as to what role Bézout's theorem (the way it is presented to us) has played in their conclusion; the theorem does not say that if there are integer solutions then the RHS must be the gcd of the coefficients. $\textbf{And}$ if they claimed that $\dfrac{a}{d}$ and $\dfrac{b}{d}$ are relatively prime, wouldn't that mean that, by definition, their greatest common divisor must be 1? Because if so, then I really do not see the need to use the theorem in the first place.

Given that some of the proofs for this proposition (that I have seen) here and on other websites use Bézout's theorem, I am inclined to believe that there is something wrong with my way of thinking, as in there is something I am not seeing, so I would appreciate it if you could clarify this for me. I have attached my proof just to see if it is correct in case the proof presented in my notes turned out to be incomplete or incorrect.

edit: small corrections

Answer 1

You use Bézout's theorem twice. First, if $\gcd(a,b)=d$, Bézout's theorem says that the equation $ax+by=d$ has integer solutions, since $d\mid d$.

Now fix the values $x=x_1,y=y_1$ that solve it. They also solve the equation $\frac adx+\frac bd y=1$. If $\gcd(\frac ad,\frac bd)\not\mid 1$, Bézout's theorem would say that this equation has no solutions. But we know it does have a solution, so $\gcd(\frac ad,\frac bd)\mid 1$, which implies it is equal to $1$.

Answer 2

By Bézout's theorem, we can write $ax+by=d$ with $x,y\in \mathbb Z$ and so $\frac ad x+\frac bd y=1$.

Therefore, every common divisor of $\frac ad$ and $\frac bd$ divides the LHS and so divides the RHS, and so must divide $1$.

Thus $\gcd(\frac ad,\frac bd)=1$.