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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.

I hate math
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Garmagon117
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2 Answers2

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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 $m = -n$.

Caleb Stanford
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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.

drhab
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