Given the relation R on Z−{0} defined by

(x,y) \(\displaystyle\in\) R if and only if xy > 0.

a) A relation R on a set A is an equivalence relation if and only if R is reflexive, symmetric and transitive.

For any x \(\displaystyle\in\) Z−{0}, x \(\displaystyle\cdot\) x = \(\displaystyle{x}^{{2}}\) > 0. So (x,x) \(\displaystyle\in\) R for all x and hence the relation R is symmetric.

To prove R is symmetric.

Let (x,y) \(\displaystyle\in\) R. Then xy > 0.

Since Z−{0} is commutative, xy = yx. Then yx > 0 and so (y,x) \(\displaystyle\in\) R.

Therefore, R is symmetric.

Prove R is transitive.

Let (x,y) \(\displaystyle\in\) R and (y,z) \(\displaystyle\in\) R. Then xy > 0 and yz > 0.

Therefore, xy \(\displaystyle\cdot\) yz = \(\displaystyle{x}{y}^{{{2}}}{z}\) > 0.

Let y=1. Then xz>0 implies (x,z) \(\displaystyle\in\) R.

Hence R is transitive. Thus, R is an equivalence relation.

b) Find the equivalence class of 1.

[1]={x \(\displaystyle\in\) Z−{0} . (1,x) \(\displaystyle\in\) R}

={x \(\displaystyle\in\) Z−{0} . 1 \(\displaystyle\cdot\) x > 0}

={x \(\displaystyle\in\) Z−{0} . x>0}

\(=Z^+\)

Similarly, [−1]=Z−. There are only two distinct equivalence classes Z+ and Z−.