Prove that if $n \mid a$ and $n \mid b$ then $n \mid (a+b)$

Asked May 13 '22 at 01:16

Active May 13 '22 at 01:24

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As stated in the title, I need to prove that if $n \mid a$ and $n \mid b$ then $n \mid (a+b)$.

So far this is what I have:

Assume that $n \mid a$ and $n \mid b$. Then there exists two integers $d$ and $f$ such that $n \mid a = nd$ and $n \mid b = nf$. I am stumped at this point. I remember there was a rule for something like this but I can't remember the name of it. Any help is appreciated, thanks!

edited May 13 '22 at 01:24

Sammy Black

25,273

asked May 13 '22 at 01:16

Josh

2

If n | a then a = nf, and if n | b then b = nd for some integers f & d. Then you can add a & b to get nd + nf. Which then you can factor out the n and get that a + b= n(f+d), which implies that n | a + b – krum May 13 '22 at 01:19
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" n|a = nd " makes no sense, check your definitions – jjagmath May 13 '22 at 01:22
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And this can be generalized to $n|ax+by$, where x and y are integers. It also works for any number of variables. – wsz_fantasy May 13 '22 at 01:24
@jjagmath it makes sense to me. – MJD May 13 '22 at 01:27
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@MJD It doesn't make sense as definition of $n \mid a$ , which is the current context. – jjagmath May 13 '22 at 01:36
@jjagmath I agree – Peter May 13 '22 at 10:05

Prove that if $n \mid a$ and $n \mid b$ then $n \mid (a+b)$

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