Question

Let R=\left\\{ \left( a,b \right):a,b\in Z\text{ }and\text{ }\left( a+b \right)\text{ }is\text{ }even \right\\}. Show that R is an equivalence relation on Z.

Answer 1

Hint: In this question, we are asked to prove a relation R as equivalence relation. For that we will prove that the relation is reflexive, symmetric, and transitive. And we know, reflexive relation is $aRa$, symmetric relation is $aRb$, and transitive relation is if $aRb$ and $bRc$, then $aRc$.

Complete step-by-step answer:
In this question, we are asked to prove that, R=\left\\{ \left( a,b \right):a,b\in Z\text{ }and\text{ }\left( a+b \right)\text{ }is\text{ }even \right\\} is an equivalence relation on Z. To prove this, we have to prove that R is a reflexive relation, symmetric relation and transitive relation. If any of the three does not satisfy, then R will not be an equivalence relation.
So, let us first go for the reflexive relation. Reflexive relation is a relation in which each element maps for itself. So, for, \left\\{ \left( a,a \right):a\in Z \right\\}, we have to prove that it satisfies relation R, which is (a + b) is even. We know that when 2 equal numbers are added, then we get, a + a = 2a. And we know that, if 2 is a factor of any number, then the number is even. So, we can say 2a is even. Therefore, we can write a + a is even.
Hence, we can say that R is reflexive.
Now, let us go with a symmetric relation. Symmetric relation is a relation which is satisfied by the converse relation also. For example, if $aRb$ is satisfied, then for symmetric relation, $bRa$ should also be satisfied. Here, we have been given a relation, R=\left\\{ \left( a,b \right):a,b\in Z\text{ }and\text{ }\left( a+b \right)\text{ }is\text{ }even \right\\}. Now, if we take a + b, then we know that in addition there is a commutative property, which means that a + b =b + a. So, if a + b is even, then definitely b + a will also be even.
Hence, we can say that R is a symmetric relation.
Now, let us consider the transitive relation. Transitive relation is a relation which states that if $XRY$ and $YRZ$ are there, then $XRZ$. So, to check the transitive relation of R, we will consider (a, b) and (a, c). So, if $aRb$, that is (a + b) is even, and $bRc$, that is (b + c) is even, then we will try to find whether (a + c) is even or not. We know that the sum of 2 even numbers is even. So, we can say that (a + b) + (b + c) is even, which can also be written as, (a +c) + 2b. Now, for (a + c) + 2b to be even, (a + c) should be even because 2b is even and an even number is produced only when 2 even numbers are added or 2 odd numbers are added. Hence, we can say that (a + c) is even.
Therefore, we can say that R is a transitive relation too.
Now, we have proved that R is reflexive, symmetric, and transitive relation, which are the conditions of equivalence relation. Hence, we can say that, R=\left\\{ \left( a,b \right):a,b\in Z\text{ }and\text{ }\left( a+b \right)\text{ }is\text{ }even \right\\} is an equivalence relation.

Note: While proving the question, we need to remember that whenever 2 is a factor of a number, then the number is even. Also, we should keep in mind that even numbers are produced either by the sum of 2 evens or by the sum of 2 odd numbers. We may make silly mistakes, so we should be careful while solving the questions.