Show that the relation $R = \{ ( x , y ) \mid x - y \text { is an integer } \}$ is an equivalence relation on the set of rational numbers. What are the equivalence classes of 0 and 12 ?

Question

Quizplus · Accepted Answer

The Answer is Since

x - x = 0

is an integer for every rational number

x

it follows that

R

is reflexive. Suppose that

( x , y ) \in R

. Then

x - y

is an integer, which implies that

y - x

is an integer. Hence

( y , x ) \in R

. It follows that

R

is symmetric. Now suppose that

( x , y ) \in R

and

( y , z ) \in R

. Then

x - y

and

y - z

are integers, so

x - z = ( x - y ) + ( y - z )

is also an integer. It follows that

R

is transitive. Hence

R

is an equivalence relation. We have

[ 0 ] _ { R } = \{ x \in \mathbf { Q } \mid x - 0 \in \mathbf { Z } \} = \mathbf { Z }

and

\left[ \frac { 1 } { 2 } \right] _ { R } = \left\{ x \in \mathbf { Q } \mid x - \frac { 1 } { 2 } \in \mathbf { Z } \right\} = \left\{ k + \frac { 1 } { 2 } \mid k \in \mathbf { Z } \right\} =

\left\{ \ldots , - \frac { 5 } { 2 } , - \frac { 3 } { 2 } , - \frac { 1 } { 2 } , \frac { 1 } { 2 } , \frac { 3 } { 2 } , \frac { 5 } { 2 } , \ldots \right\}

.

Show That the Relation R={(x,y)∣x−y is an integer }R = \{ ( x , y ) \mid x - y \text { is an integer } \}R={(x,y)∣x−y is an integer }

Show That the Relation $R = \{ ( x , y ) \mid x - y \text { is an integer } \}$