Dec 2012
The Netherlands
11010001001_{2} Posts

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 ab, 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 welldefined (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.
