I'm relatively new to this, I got no idea how to proceed at all. I'm just bad at this chapter :( $$ m\mid (ab- (a \bmod m \cdot b \bmod m)). $$ How do I proceed after that? Is this even the first step?
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1What do you want to prove? – Vinyl_cape_jawa Nov 05 '19 at 22:25
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I just need to prove that ab = [ (a mod m) * (b mod m) ] (mod m) – Dark of the knight Nov 05 '19 at 22:30
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Since, by definition, $m$ divides $a-a\bmod m$ and $b-b\bmod m$, rewrite the expression as \begin{align} ab-a\bmod m\cdot b\bmod m&=(a-a\bmod m)b+a\bmod m\cdot b-a\bmod m\cdot b\bmod m\\ &=(a-a\bmod m)b+a\bmod m\cdot(b-b\bmod m) \end{align}
Bernard
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Sorry I'm really new to this... I don't understand what does this even leads to... I'm looking at my textbook and still dumbfounded by the whole proving method... Someone even put this as duplicate and I don't see any correlation with the linked question – Dark of the knight Nov 05 '19 at 23:41
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Look, in the last lie the facot $a-a\bmod m$ is a multiple of $m$ (by definition), hence its product by $b$ is too. Similarly, $b-b\bmod m $ is a multiple of $m$, hence its product by $a\bmod m$. What can you say of the sum of two multiples of $m$? – Bernard Nov 05 '19 at 23:47
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that actually means u can already divisible by M! I hope im going on the right track! – Dark of the knight Nov 06 '19 at 00:04
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the whole equation is divisible by m. Sorry for not clarifying – Dark of the knight Nov 06 '19 at 00:09
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