Multiplication Properties of Equivalence Classes

[jcfaul01]

Homework Statement

Prove or disprove and salvage if possible: for [a], ∈ Zn for a positive integer n, if [a]·=[0], then either [a]=[0] or =[0].

The Attempt at a Solution

I've managed to disprove the statement:
Let n=6,[a]=3,and=[4]. The[a]·=[ab]=[3·4]=[12]. Since12≡0(modn), 12 ∈ [0] so[12] = [0]. Thus [3] · [4] = [0] and this statement is false.

However, my problem is with salvaging it. I've been able to come up with what I believe to be the correct statement:

For [a], ∈ Zn for a positive integer n, if [a] · = [0], then ab ≡ 0 (mod n).

But I have no idea how to prove it, I don't even know where to start. I would really appreciate any help on this,

thanks.

 

[HallsofIvy]

What you have, [m][n]= 0 implies mn= 0 (mod n) is just the definition of [m][n]= 0.

Let n be a prime integer, then ...

 

[jcfaul01]

Thanks! I think I've got it now!