How can one prove that $a*c \equiv b*c \:\: mod(n*c)$ equals $ a \equiv b \: \: mod(n)$, the only thing I know thus far is that factor cancellation is only allowed if gcd(a,n)=1 i.e when a and n are relatively prime. I thought that using a Direct Proof would be the easiest way so we assume our first congruence and than try to get to our second congruence via arithmetic manipulations.
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In order to help you prove what you say you know we need to know where you are starting. If you [edit] the question to tell us how you might start this proof we may be able to help. Use mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Oct 19 '23 at 13:56
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Quick beginner guide for asking a well-received question + please avoid "no clue" questions: show some personal work. – Anne Bauval Oct 19 '23 at 16:20
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Duplicate of this and this and many others. It's best for site health to delete this question. – Bill Dubuque Oct 19 '23 at 17:54
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Okay so after thinking a little bit more about the problem my answer to the question would be as follows:
We can write $$a*c \equiv b*c \: \: mod(c*n)$$ as $$(a*c-b*c)=k*(n*c), k \in \mathbb Z$$
So now we have a ordinary equation where we can just cancel out the factor c via division on the left and right hand side and end up with $$k*n = a-b$$ which in turn is equal to $$ a\equiv b \: \: mod(n)$$ and this basically concludes the proof because one has shown that the first congruence equation is equal to the second. Would this be somewhat acceptable as prove?
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This question goes against our quality standard policy. Instead of posting an answer, you should encourage the person who posted the question to improve it. Moreover, we should anyway not edit more answers to such a megaduplicate. See for instance https://math.stackexchange.com/questions/2641382 or https://math.stackexchange.com/questions/1883921 – Anne Bauval Oct 19 '23 at 16:27
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Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 19 '23 at 17:54