The problem is : define relation equivalence on Z by $m=n$ in case $m^2=n^2$.
a)Show that its an equivalence relation on Z.
b)Describe the equivalence classes for = how many are there.
For part a, I proved it to be true by showing that it's reflexive, symmetric and transitive. I have used matrix to do that however I cant figure out part b. Can someone explain part b to me please.
On
In general if $f:X\rightarrow Y$ is a function then $\sim$ defined by $x\sim x'\iff f(x)=f(x')$ is always an equivalence relation on $X$ and this is not difficult to prove. Note that:
1) $f(x)=f(x)$ reflexive
2) $f(x)=f(y)\Rightarrow f(y)=f(x)$ symmetric
3) $f(x)=f(y)\wedge f(y)=f(z)\Rightarrow f(x)=f(z)$ transitive
In your case we deal with $f(n)=n^2$ on $\mathbb Z$.
For b) see the answer of Goos.
Hint
$$ m^2 = n^2 \iff m^2 - n^2 = 0 \iff (m - n)(m + n) = 0 \iff m = n \text{ or } m = -n $$
So $m$ and $n$ are equivalent if and only if either $m = n$ or $m = -n$.