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Old 2016-03-30, 10:20   #10
Nick
 
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Dec 2012
The Netherlands

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May I suggest that the modern algebraic way of looking at this offers some advantages here?

If we are working with integers modulo 5, for example, then we regard integers a and b as equivalent if (and only if) 5 divides a-b, so we have 5 equivalence classes:
{...,-15,-10,-5, 0,5,10,15,...}
{...,-14,-9,-4, 1,6,11,16,...}
{...,-13,-8,-3, 2,7,12,17,...}
{...,-12,-7,-2, 3,8,13,18,...}
{...,-11,-6,-1, 4,9,14,19,...}
For any integer a, we write \(\bar{a}\) to denote the equivalence class of a, i.e. the entire set of all integers equivalent to a modulo 5 (this is the set in the above list which a appears in).
We then define addition and multiplication on these sets by:
\[
\begin{eqnarray*}
\bar{a}+\bar{b} & = & \overline{a+b} \\
\bar{a}\cdot\bar{b} & = & \overline{a\cdot b}
\end{eqnarray*}
\]
for any integers a & b, and show that this is well-defined (i.e. the definition does not depend on the representative elements chosen). Thus, for example, we can write \(\bar{4}^2=\overline{-1}^2=\bar{1}\).

Viewed this way, we are not tempted to write something like "30 mod 5 mod 7" because 30 mod 5 is then the set \(\overline{30}=\bar{0}\) and not all elements of the set have the same remainder on division by 7, so the final "mod 7" is not a valid operation.
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