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I don't really have an idea on how to prove this.

My thinking:

$f(x,y) = ax+by = \gcd(a,b) = 1$ by Bézout's identity.

To prove a function is surjective: Let $f(x,y)=m, m\in \mathbb{Z}$

Then prove $\exists x,y \in \mathbb{Z} \times\mathbb{Z} $ such that $f(x,y) = ax+by=m$

Though I don't quite understand how $\gcd(a,b)=1$ is relevant to all of this. I've looked at other answers on this fourm but none make sense at all really. I know $ax+by=1$ has something to do with this but I don't quite know what. Any help would be appriciated!

user26857
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someman112
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    Let $n\in\mathbb Z$. Since $\gcd(a,b)=1$, there are integers $x$ and $y$ such that $ax+by=1$, so $f(xn,yn)=a(xn)+b(yn)=n$. Since $n$ is arbitrary, this proves that $f$ is surjective. – Joe Apr 09 '22 at 20:35
  • @JoséCarlosSantos I've looked at the answers but it doesn't make sense to me. – someman112 Apr 09 '22 at 20:35
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    This answer completely answers your question. – José Carlos Santos Apr 09 '22 at 20:41
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    @Joe I think I got it. So since $ax+by=1$ all you do is multiply both sides of the equation by any number in $\mathbb{Z}$, say $n$, and then your new $x = xn$ and your new $y = yn$. – someman112 Apr 09 '22 at 20:44
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    @Yaya123: Yes. It perhaps help to use different variable names, in an effort to avoid a notational conflict: you want to show that $(\exists s)(\exists t)(as+bt=n)$. We know that $ax+by=1$ for some $x,y\in\mathbb Z$, and multiplying both sides by $n$ we get $a(xn)+b(yn)=n$. Therefore, we can set $s=xn$, $t=yn$. – Joe Apr 09 '22 at 20:47

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