how is the relation defined by (x,y)$\in$ R iff and only if $x^2-4xy+3y^2$ not symmetric?

Question

A Relation R on the set N of Natural numbers be defined as (x,y) $\in$R if and only if $x^2-4xy+3y^2=0$ for allx,y $\in$N then show that the relation is reflexive,transitive but not SYMMETRIC.

i got how this relation is reflexive or transitive but i am not able to think of any reason of why this relation is not symmetric.

score 2 · Accepted Answer · answered Jul 23 '16 at 08:08

2

R is not symmetric because $(3,1)\in R$ but $(1,3)\not\in R$. Note that $x^2-4xy+3y^2=(x-y)(x-3y)$.

answered Jul 23 '16 at 08:08

Robert Z

145,942

score 0 · Answer 2 · answered Jul 23 '16 at 08:10

If the relation is symmetric, then for all $(x,y) \in R$ that satisfy $x^2 - 4xy + 3y^2 = 0 \rightarrow (y,x) \in R$, so $y^2 - 4yx + 3x^2 = 0$.

For $x^2 - 4xy + 3y^2 = (x - 3y)(x - y) = 0$, x must be equal to $3y$ or $y$, but does this ALWAYS imply that $y^2 - 4yx + 3x^2 = (y - 3x)(y - x) = 0$?

how is the relation defined by (x,y)$\in$ R iff and only if $x^2-4xy+3y^2$ not symmetric?

2 Answers2