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Consider the following:

$\forall c\in\mathbb{Z}^{+}, a \equiv b \pmod n \Leftrightarrow a \equiv b,b+n,b+2n,..,b+(c-1)n \pmod {cn}$.

It seems false, so I suspect there is a typo in the statement, could anyone tell me where is it?

Thanks a lot.

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The commas denote "or," not "and." – Qiaochu Yuan Sep 28 '11 at 15:47
@QiaochuYuan explains a lot. – newbie Sep 28 '11 at 15:52

1 Answer 1

up vote 2 down vote accepted

It's correct: $\rm\: a\equiv b\pmod{n}\ $ iff $\rm\ \exists\ i:\ a = b + i\ n\:.\:$ By the Division Algorithm $\rm\: i = j + k\ c\:,\ \ 0\le j < c\:,\:$ so $\rm\: \exists\ i:\ a = b + i\ n\ $ iff $\rm\ \exists\ k,\:j,\ 0\le j < c:\ a = b + j\ n + k\ (c\:n)\:\equiv\: b + j\ n\pmod{c\:n}\:.$

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