Proofing Modulus Operator Associativity

[Bohrok]

Homework Statement

We have a problem that basically defines an operator [·] (it's a dot inside a box) as a[·]b = (a·b) mod 6 where a, b ε ℕ. Is this operator associative?

Homework Equations

a[·]b = (a·b) mod 6

I tried looking for any modulus identities online but didn't have any luck besides this one:
(a·b) mod c = [(a mod c) · (b mod c)] mod c

but I couldn't find a proof for it and I wasn't sure how I'd prove it either.
While doing my proof, I also found that (a mod b) mod b = a mod b seems to be true but I couldn't prove that either.

The Attempt at a Solution

Is (a[·]b)[·]c = a[·](b[·]c) true?


(a[·]b)[·]c

= {[(a·b) mod 6]·c} mod 6

= ({[(a·b) mod 6] mod 6}·(c mod 6)) mod 6 (here using (a·b) mod c = [(a mod c) · (b mod c)] mod c)

= {[(a·b) mod 6]·(c mod 6)} mod 6 (here using (a mod b) mod b = a mod b)

= [(a·b)·c] mod 6

= [a·(b·c)] mod 6

= {(a mod 6)·[(b·c) mod 6]} mod 6

= ((a mod 6)·{[(b·c) mod 6] mod 6} mod 6 (here using (a mod b) mod b = a mod b)

= {a·[(b·c) mod 6]} mod 6 (here using (a·b) mod c = [(a mod c) · (b mod c)] mod c)

= a[·](b[·]c)


If I could prove those two equations I used in the proof, would that be okay? Is there a better way of proving this?

Might be an error or two; I've got to go now.

 

[sutupidmath]

It is not difficult to show that

$$a\cdot b( mod c)=a(mod c)\cdot b(mod c)$$

Can you show this?

If yes, then the associativity of the operation [^.] follows immediately.

That is:
$$a[\cdot](b[\cdot]c)=a[\cdot](bcmod6)=a(bcmod6)mod6=(amod6)(bcmod6)=(amod6)(bmod6)(cmod6)$$

all you need to do now is use the fact that the last expression is associative and work your way back.