I have been trying to prove this via divisibility, assuming that $a=\frac{n}{m}$ and $b=\frac{r}{q}$ for some $n,m,r,q$ in
Ints($m$,$q$ not $0$), but I'm completely stuck here. Any help?
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If $x$ is divisibly by $p$ and $y$ is divisible by $q$ then $xy$ is divisible by $pq$. And so is $qx$, $py$, and $qx+py$. – mniip Jul 19 '15 at 02:35
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Note that conversely, $a+b,ab$ both integers means either that $a,b$ are both integers or both non-rational. E.g., $a=1+\sqrt 2,b=1-\sqrt 2$. – abiessu Jul 19 '15 at 02:42
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is there a way to prove this by disproving the contradiction? Specifically, we can case this by when a,b are both non-integers, and when one of a or b is an integer. – user3467433 Jul 19 '15 at 03:45
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Let $n:=a+b$ and $m:=ab.$ Then $a^2-an+m=0$ and hence $$a=\dfrac{n\pm\sqrt{n^2-4m}}{2}.$$ note that since $a\in\Bbb Q,$ then $n^2-4m$ must be a perfect square.
Then necessarily $2\mid n\pm\sqrt{n^2-4m}$ (if $n$ is odd, then $n^2-4m$ is odd $\Longrightarrow$ $n\pm\sqrt{n^2-4m}$ is even; if $n$ is even, then $n^2-4m$ is even $\Longrightarrow$ $n\pm\sqrt{n^2-4m}$ is even).
CIJ
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