Question

Consider a relation R on Z — {0} defined by the rule that (z,y) \in R if and only if xy > 0. How many distinct equivalence classes are there?

Discrete math
ANSWERED
asked 2021-07-30
Consider a relation R on Z — {0} defined by the rule that (z,y) \(\displaystyle\in\) R if and only if xy > 0. a) Prove that R is an equivalence relation. b) Determine all elements in the equivalence class containing 1. How many distinct equivalence classes are there?

Expert Answers (1)

2021-07-31

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−.

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