R exactly when − equals a multiple of 3.
There exists equivalence classes [0], [1] and [2]. Prove that [1] + [2] is equal to [0].
Not sure about the ambigious part, all I could find from my research was that if it's intepreted different ways the meaning will still be the same e.g. [1] + [2] = [0] is the same as [2] + [1] = [0] or [0] = [2] + [1]. But I am unsure about that.
How do I create these equivalence classes and prove that [1] + [2] is equal to class [0] ?
Here,$$[1]=\{\ldots,-5,-2,1,4,7,\ldots\}=\{3n+1\mid n\in\Bbb Z\}$$and$$\quad[2]=\{\ldots,-4,-1,2,5,8,\ldots\}=\{3n+2\mid n\in\Bbb Z\}.$$And when you add any element of $[1]$ to any element of $[2]$, you get an element of$$[0]=\{\ldots,-6,-3,0,3,6,\ldots\}=\{3n\mid n\in\Bbb Z\}.$$So, if $m\in[1]$ and $n\in[2]$, the class of $m+n$ is independent of the choice of $m$ and $n$, and therefore $[1]+[2]$ is not ambiguous (and it turns out that it is equal to $[0]$).