Assume $a$ and $b$ are positive integers. Consider the following set of positive integers $$X =\{xa + yb > 0\,|\,x,y ∈ \mathbb{Z}\}$$ (where $x$ or $y$ can be negative or $0$). $X$ is non-empty since it contains both $a$ and $b$. Therefore it contains a smallest element $d$. Prove that $d = \gcd(a, b)$
I have to solve the question above, but I'm not sure how to do it. I think it has to do with Bézout’s theorem, but I have have no idea how to apply it in this particular question. Any help?
First, Bezout's identity tells us there exist $x, y \in \Bbb Z$ such that $ax+by=\gcd(a, b)$. This means $\gcd(a, b) \in X$, so since $d$ is the least element in $X$, $d \leq \gcd(a, b)$.
Then, for any $ax+by\in X$, we know that if $m=\gcd(a, b)$, then since $m$ is a factor of both $a$ and $b$, we can write $a=mt$ and $b=ms$ for $t, s \in \Bbb{Z}$. Thus:
$$ax+by=mtx+msy=m(tx+sy)$$
This clearly shows that $ax+by$ is a multiple of $m$, so since $ax+by > 0$ and $m > 0$, we know that $m \leq ax+by$ since a factor is always than less than or equal to a positive multiple of that factor. Thus $m$ is less than or equal to all of the elements in $X$, so since $d \in X$, $m \leq d\rightarrow \gcd(a, b)\leq d$.
We now have $d \leq \gcd(a, b)$ and $\gcd(a, b)\leq d$, so $d=\gcd(a, b)$.
