I know that if $m\mid(a-b)$, then $a-b=mk$ where $k$ is an integer. And $a=b +mk.$ So does $m\mid [ (b+mk) -b] = m \mid mk$? I know $d\mid m$ iff $m= dk.$ So does $b= mk - a$? Then it would be $m \mid [( mk - a) - (b + mk)] = m | mk - a - b - mk = $? I know that if $m\mid (a-b)$ and $m\mid (b-a)$ are true, but I don’t know how to go on with my proof.
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4Suppose $m\mid (a-b)$, then $a-b=mk$ for some integer $k$. It follows then that $b-a=m(-k)$... – JMoravitz Nov 07 '18 at 00:10
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How would you show that if $x$ divides $y$ then it also divides $-y$? – Ethan Bolker Nov 07 '18 at 00:11
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take absolute value – user29418 Nov 07 '18 at 00:11
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Nov 07 '18 at 00:12
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The set of multiples of $m$ are not only closed under negation but also under subtraction and scaling, so more generally if $,a,b,$ are multiples of $m$ then so too is $,j,a + k,b,$ for all integers $,j,k.,$ It's very easy to prove and fundamental in number theory, e.g. see the LCM universal Property. – Bill Dubuque Nov 07 '18 at 00:26
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If m $\mid$ (a-b) then nm = a - b for some integer n. Now multiply both sides by -1. Then we have $$-nm = -(a-b) = b - a.$$ So we have that m $\mid$ b- a.
Joel Pereira
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