# visual proof of the chinese remainder theorem?

I have seen visual proofs for fermats little theorem, gauss sum and many other things. I find them very useful. Is there a visual proof for the chinese remainder theorem? Thanks in advance.

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Which visual proofs have you seen? Any links or references? –  lhf Nov 2 '12 at 1:09
I really don't see how Fermat's little theorem could be made "visual"... just read up the proof and it's an identity that works, but maybe someone got original. I'd be interested in a visual proof for Fermat's. –  Patrick Da Silva Nov 2 '12 at 1:12
Patrick Da Silva: youtube.com/watch?v=XPMzosLWGHo , and here is a formal proof running with that idea: math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Bishop.pdf –  ante.ceperic Mar 12 at 14:18

$$\begin{array}{cc|cccccc} & & 0 & 1 & 2 & 3 & 4 & \pmod 5 \\\hline \\ & 0 & 0 & 6 & 12 & 3 & 9 \\ \pmod 3 & 1 & 10 & 1 & 7 & 13 & 4 \\ & 2 & 5 & 11 & 2 & 8 & 14 \end{array}$$

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Fine, it's a grid... I don't see how that helps! Maybe the OP disagrees with me but anyway. –  Patrick Da Silva Nov 2 '12 at 1:11
I think, it's visual. $0,1,2,3,..$ written in the diagonal, continuously. Pairs of remainders mod 3 and 5 are corresponded to remainder mod 15. –  Berci Nov 2 '12 at 1:28
Danger often lurks behind a "visual" proof @Berci, e,g, $$\begin{array}{cc|cccccc} & & 0 & 1 & 2 & 3 & 4 & 5 &\rm (mod\ 6) \\\hline \\ & 0 & 0 & & 8 & & 4 \\ \pmod 4 & 1 & & 1 & & 9 & & 5 \\ & 2 & 6 & & 2 & & 10 \\ & 3 & & 7 & & 3 & & 11 \end{array}$$ –  Bill Dubuque Nov 2 '12 at 2:33