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My original assignment was to prove that the set $\{\,x+iy \mid x,y\in\mathbb{Q}\,\}$ is countable. Since I now that $\mathbb{Q}\times\mathbb{Q}$ is countable, I immediately thought of the injection $f\colon \{\,x+iy \mid x,y\in\mathbb{Q}\,\}\rightarrow\mathbb{Q}\times\mathbb{Q}$ with $f(x+iy)=(x,y)$. My teacher then asked me, whether or not this function is well defined. I'd say it's pretty obvious that $f$ is well defined, since unique $x+iy$, gives unique pairs of $x$ and $y$, but I'm not quiet sure on how I can show it more strictly?

Hagen von Eitzen
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njlieta
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    Take rational numbers $a,b,x,y$ and assume that $a+ib=x+iy$. The goal is to prove that $a=x$ and $b=y$ (actually the goal is $(a,b)=(x,y)$, but this is equivalent). Now to proving this depends a bit on what you can use. – Git Gud Oct 02 '16 at 10:26
  • *now proving this – Git Gud Oct 02 '16 at 10:31

1 Answers1

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Assume $x+iy=u+iv$ with $x,y,u,v\in\Bbb Q$. If $y=v$, we immediately get $x=u$. So assume $y\ne v$. Then $i=\frac{x-u}{v-y}\in\Bbb Q$ - but is it?

Apart from that, it suffices to have the obvious surjection $\Bbb Q\times \Bbb Q\to \{\,x+iy\mid x,y\in\Bbb Q\,\}$.

Hagen von Eitzen
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  • Well defined here mean (a priori) that $f(x+iy)\in\mathbb Q$. – Surb Oct 02 '16 at 10:35
  • @Surb Do you mean $\mathbb Q\times \mathbb Q$? Well defined means a conjunction of things, what you mention is part of it, but perhaps it is too trivial to be worthy of mention. – Git Gud Oct 02 '16 at 10:37