Question

The relation “congruence modulo $m$” is
A.Reflexive only
B.Transitive only
C.Symmetric only
D.An equivalence relation

Answer 1

Hint: The congruence modulo $m$ is defined as the relation between $x$ and $y$ such that $x - y$ is divisible by m. Use the properties of set theory to prove that the congruence modulo $m$ is an equivalence relation.

Complete step-by-step answer:
The congruence modulo $m$ is defined as the relation between $x$ and $y$ such that $x - y$ is divisible by $m$. $xRy = x - y$is divisible by $m$. Or $x - y = km$ where $k$ is an integer.

For an relation to be reflexive , it should satisfy $xRx$.
For the given relation, $xRx = x - x = 0$, which is divisible by $m$. Therefore the relation is true for all values $\left( {x,x} \right)$. Therefore the given relation is reflexive.

For in relation to be symmetric, if $\left( {x,y} \right)$satisfies the relation then $\left( {y,x} \right)$ must also satisfy the relation.
For pair $\left( {x,y} \right)$ satisfying the given relation,
$xRy = x - y$ is divisible by $m$ or $x - y = km$
For the pair $\left( {y,x} \right)$, $yRx$ gives $y - x$ which is also divisible by $m$ as
$x - y = km$,
On taking $ - 1$ common from both sides, we get,
$y - x = - km$
$ - k$ is also an integer.
Therefore $yRx$ is true.
Thus the given relation is symmetric.

For a relation to be transitive, if $x,y$ satisfy the relation and $y,z$ satisfy the relation then $x,z$ must also satisfy the relation.
For $x,y$ and $y,z$ satisfy relation we can say
$x - y$ is divisible by $m$ and \[y - z\] is also divisible by $m$.
\[x - y = {k_1}m\] and \[y - z = {k_2}m\].
Adding both equations, we get
\[x - y + y - z = \left( {{k_1} + {k_2}} \right)m\]
\[x - z = {k_3}m\], where \[{k_3}\] is an integer.
Thus \[x,z\] also satisfies the relation.
The given relation is also transitive.

And since the given relation is reflexive, symmetric and transitive in nature, it is an equivalence relation.

Note: The relation congruence modulo $m$ for the ordered pair \[x,y\] means that the value \[x - y\] is divisible by $m$. For in relation to be equivalence, the relation must be reflexive that is it should satisfy $xRx$, symmetric that is if $\left( {x,y} \right)$ satisfy the relation then $\left( {y,x} \right)$ must also satisfy the relation and transitive , which implies, for in relation to be transitive, if $x,y$ and $y,z$ satisfy the relation then $x,z$ must also satisfy the relation.