Question

Show that the relation R defined on the set A=\left\\{ 1,2,3,4,5 \right\\}, given by R=\left\\{ \left( a,b \right):\left| a-b \right|\text{ }is\text{ }even \right\\} is an equivalence relation.

Answer 1

Hint: In order to prove this question as an equivalence relation, we will prove the given relation R is reflexive, symmetric and transitive relation. Also, we need to remember that whenever a number is even, then it is always divisible by 2.

Complete step-by-step answer:
In this question, we have been asked to prove that, R=\left\\{ \left( a,b \right):\left| a-b \right|\text{ }is\text{ }even \right\\} is an equivalence relation. To prove this, we have to prove that the relation is reflexive, symmetric and transitive. If it does not satisfy any of these three conditions, then it 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, {(a, a)}, we have to prove that it satisfies relation R, which is $\left| a-b \right|$. We know that 0 is divisible by all numbers. So, we can say 0 is divisible by 2 also. And on subtracting a with a, we get, $\left| a-a \right|=0$, which is divisible by 2, which is even.
Therefore, relation R is a reflexive relation.
Now, let us consider symmetric relations. Symmetric relation is a relation which is satisfied by its converse relation also, like if $aRb$ is satisfied, then $bRa$ should also be satisfied for a symmetric relation. Here we have been given relation R=\left\\{ \left( a,b \right):\left| a-b \right|\text{ }is\text{ }even \right\\} and if we take $\left| b-a \right|$, then we can say that the negative sign will be removed because of modulus and so, we get, $\left| b-a \right|=\left| a-b \right|$, which means that $\left| b-a \right|$ is even, which is a condition of R=\left\\{ \left( a,b \right):\left| a-b \right|\text{ }is\text{ }even \right\\}.
Hence, we can say that R is a symmetric relation.
Now, we will consider transitive relations. Transitive relation is a relation which states that if $aRb$ and $bRc$, then $aRc$. So, to check if the relation is transitive or not, we will consider (a, b) and (b, c). So, if $aRb$, that is $\left| a-b \right|$ is even and $bRc$, that is $\left| b-c \right|$ is even, then we will check if $\left| a-c \right|$ is even or not. We will first consider $\left| a-b \right|$ is even. So, we can say if $a-b\ge 0$ or $a-b<0$, (a - b) will always be a multiple of 2, which implies that,
a - b = 2m ………… (i)
Similarly, we will check if $\left| b-c \right|$ is even. So, we can say whether, $b-c\ge 0$ or $b-c<0$, (b - c) will always be a multiple of 2, which implies that,
b - c = 2n ………… (ii)
Now, we will add equation (i) and (ii). So, we get,
a - b + b - c = 2m + 2n, which is a - c = 2 (m + n)
And if m + n = x, then we get, a - c = 2x, where x can be positive or negative and so, we can say (a - c) is even which implies that $\left| a-c \right|$ is even.
Hence, we can say that relation R is a transitive relation.
Therefore, we have proved that R is a reflexive, symmetric and transitive relation for $a,b\in Z$, where A=\left\\{ 1,2,3,4,5 \right\\}. Hence, we can say that R=\left\\{ \left( a,b \right):\left| a-b \right|\text{ }is\text{ }even \right\\} is an equivalence relation.

Note: While proving this question, we need to remember that whenever a number is even, it means that the number is divisible by 2. Also, we have to remember that $\left| a-b \right|=\left| b-a \right|$, which we require while proving R as a symmetric relation.