49 views
Suppose that $A$ is a nonempty set, and $f$ is a function that has $A$ as its domain. Let $R$ be the relation on $A$ consisting of all ordered pairs $(x, y)$ such that $f (x)=f (y)$

$a)$ Show that $R$ is an equivalence relation on $A$

$b)$ What are the equivalence classes of $R?$

edited | 49 views

$a)$ For $\left ( x,y \right )$ pair , if the relation $R$ given as $f\left ( x \right )=f\left ( y \right )$

and  if we suppose $x=y$ present in relation, then pair $\left ( x,x \right )$ in $R$ is a reflexive relation.

Now, in pair $\left ( x,y \right )$ if $f\left ( x \right )=f\left ( y \right )$ exists, then$f\left ( y \right )=f\left ( x \right )$ also exists

So, we can say $\left ( x,y \right )$ and $\left ( y,x \right )$ both present in relation $R$.Which implies relation is symmetric

Now, finally suppose $f\left ( x \right )=f\left ( y \right )=f\left ( z \right )$

In the set $A$ if $\left ( x,y \right ),$$\left ( y,z \right )$ both is present implies $\left ( x,z \right )$ also present in $A$ That means relation is transitive
So, it $R$ is an equivalence relation on set $A=\left \{ 1,2,3 \right \}$
$b)$ In our example ordered pair is $\left ( x,y,z \right )$
by Veteran (117k points)
edited by

+1 vote