Answer to Question #167231 in Linear Algebra for Nikhil Singh

Question

Consider the relation ~ in R^2 given by (x1,X2)~ (y1,y2) if x1x2~y1y2=1. Check if ~ is an equivalence relation

Answer 1

On the set $\R^2$

From the definition (x1, y1) ∼ (x2, y2) if x1y1 = x2y2. Showing that ∼ is an equivalence

relation,

Solution: For any (a, b) ∈ $\ R^2$

, we have (a, b) ∼ (a, b) since ab = ab, which shows

that ∼ is reflexive. If (a, b) ∼ (c, d), then ab = cd, and so cd = ab, which implies

that (c, d) ∼ (a, b), and so ∼ is symmetric. If (a, b) ∼ (c, d) and (c, d) ∼ (e, f), then

ab = cd and cd = ef. The transitive law for equality implies that ab = ef, and

therefore (a, b) ∼ (e, f), so ∼ is transitive. We conclude that ∼ is an equivalence

relation.