I am working with Paul Halmos's *Linear Algebra Problem Book* and the seventh problem asks you to show that multiplication modulo 6 is commutative and associative. As examples of multiplication modulo 6:

4 * 5 = 2

2 * 3 = 0

3 * 9 = 3

The answer in the back for this problem states that:

** If each of a and b is one of the numbers 0, 1, 2, 3, 4, 5, and if the largest multiple of 6 that doesn't exceed their ordinary product is, say, 60, so that ab = x + 60, where x is one of the numbers 0, 1, 2, 3, 4, 5, then, because ordinary multiplication is commutative, the same conclusion holds for ba. **

The reasoning to prove associativity works similarly – the language and the notation have to be chosen with care but there are no traps and no difficulties.

Unfortunately I don't totally understand this solution. I see how it works for commutativity, but I can't work it out for associativity. We want to show that

(a*b)*c = a*(b*c)

where * denotes multiplication modulo 6. According to the above,

a*b = x

because ab = x + [largest multiple of 6 not exceeding ab]. Then we can say that

b*c = y,

where bc = y + [largest multiple of 6 not exceeding bc]. Then what we want to prove is

x*c = a*y

Since xc = m + [largest multiple of 6 not exceeding xc] and ay = n + [largest multiple of 6 not exceeding ay],

x*c = m

a*y = n

After this I get confused and don't know what to do. I have no idea if I'm on the right track. What confuses me is the imprecise nature of the "largest multiple of 6 not exceeding..." term. I think I should be working with that instead, but I can't think straight. Any ideas?