Yes, proving that something is a *** requires showing that the definition of *** applies! The definition of "equivalence relation" is that it is a relation that is "reflexive", "symmetric", and "transitive". So we move on to the definitions of those. Given a relation, ~, on a set, X:
1) ~ is reflexive if and only if, for every x in X, x~ x is true.
2) ~ is symmetric if and only if, whenever x~y is true, so is y~x.
3) ~ is transitive if and only if, whenever both x~y and y~z are true, so x~z.
Here X is the set of all real numbers and the relation is defined as "x~ y if and only of there exist positive integers, i and j, such that \(\displaystyle \frac{x}{i}= \frac{y}{j}\).
1. Reflexive: take i= j= 1. \(\displaystyle \frac{x}{1}= \frac{x}{1}\) so x~ x for any real number, x.
2. Symmetric. If x~ y then there exist positive integers, i and j, such that \(\displaystyle \frac{x}{i}= \frac{y}{j}\). Since "=" itself is an equivalence relation, it is symmetric and \(\displaystyle \frac{y}{j}= \frac{x}{i}\) so y~ x.
3. Transitive. If x~ y then there exist positive integers, i and j, such that \(\displaystyle \frac{x}{i}= \frac{y}{j}\). If, also, y~ z then there exist positive integers, m and n, such that \(\displaystyle \frac{y}{m}= \frac{z}{n}\). Since \(\displaystyle \frac{x}{i}= \frac{y}{j}\), \(\displaystyle \frac{x}{mi}= \frac{y}{mj}\). Since \(\displaystyle \frac{y}{m}= \frac{z}{n}\), \(\displaystyle \frac{y}{mj}= \frac{z}{nj}\). So, since "=" itself is an equivalence relation, it is transitive and \(\displaystyle \frac{x}{mi}= \frac{z}{nj}\). And, since i, j, m, and n are all positive integers, so are mi and nj.
