Equivalence Relations

Example #3: Consider the set of rational numbers: $\mathbb{Q}$. Each number can be written in more than one way, e.g., $\frac{1}{2} \;$$= \frac{2}{4} \;$$= 0.5 \;$$= 50:100 \;$$= 50\%$, etc.

The relation that relates ways of writing out the same number:

$\text{Same-$\mathbb{Q}$-Num} \;$$= \{(\frac{1}{2}, \,$$\frac{1}{2}), \,$$(\frac{1}{2}, 0.5), \,$$(0.5, \frac{1}{2}), \,$$(0.5, 0.5), \,$$(\frac{1}{2}, 50:100), \,$$\dots\}$

is an equivalence relation. This is true because:

• For every $x \in \mathbb{Q}$, we have $x = x$ (e.g., $0.5 = 0.5$), so the relation is reflexive.
• Also, since $x = y$ (e.g., $\frac{1}{2} = 0.5$) implies that $y = x$ ($0.5 = \frac{1}{2}$) for any $x, y \in \mathbb{Q}$, this relation is also symmetric.
• Finally, for any $x, y, z \in \mathbb{Q}$, if $x = y$ (e.g., $\frac{1}{2} = 0.5$,) and $y = z$ (e.g., $0.5 = 50\%$,) this means that $x = z$ (so $\frac{1}{2} = 50\%$,) so the relation is transitive.