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Statement: If $a$ divides both $b$ and $c$, then $a$ divides $b+c$ Proof: Assume that $a$ does not divide $b+c$. Then there is no integer $k$ such that $ak=b+c$. However, $a$ divides $b$, so $am=b$ for some integer $m$; and $a$ divides $c$, so $an=c$ for some integer $n$. Thus $am+an=a(m+n)=b+c$. Therefore $k=m+n$ is an integer satisfying $ak=b+c$. Thus the assumption that $a$ does not divide $b+c$ is false, and $a$ does divide $b+c$.

Is my proof valid?

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It is valid, but a bit too complex.

You could have skipped the "assume the opposite part" and just do take $b = am$, $c = an$, and then conclude that $b+c = am + an = a(m+n)$.

So, basically what you did, but stripped of the unnecessary intro with the assumption that you do not need (nor use!).

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Yes, but it is sufficient to say

$a$ divides $b$, so $am=b$ for some integer $m$; and $a$ divides $c$, so $an=c$ for some integer $n$. Thus $am+an=a(m+n)=b+c$. Therefore $k=m+n$ is an integer satisfying $ak=b+c$.

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  • $\begingroup$ I edited your answer for spelling (dicides -> divides). I hope you don't mind! $\endgroup$ – Cameron Williams Oct 18 '13 at 17:07

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