Question

Let A=\left\\{ x\in Z:0\le x\le 12 \right\\} . Show that R=\left\\{ \left( a,b \right):\left| a-b \right|\text{ is a simple multiple of 4} \right\\} is (i) reflexive, (ii) Symmetric and (iii) transitive. Find the set of elements related to 1.

Answer 1

Hint:To show that given relation R is reflexive, Find an example such that $\left( a,a \right)\in R$ for all $a\in S$ . To show that relation R is symmetric, find an example such that $\left( a,b \right)\in R$ but $\left( b,a \right)\in R$ for all $a,b\in S$ . To show that relation R is transitive, , find an example such that $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ but $\left( a,c \right)\in R$ . After that to find set of elements related to 1.Here a=1 and b=\left\\{ 1,2,3.........12 \right\\} .Find $\left( a-b \right)$ in each case and find the set of elements for which $\left( a-b \right)$ is a multiple of 4.

Complete step-by-step answer:
Let ‘A’ be a set then Reflexivity, Symmetry and transitivity of a relation on set ‘A’ is defined as follows:
Reflexive relation: - A relation R on a set ‘A’ is said to be reflexive if every element of A is related to itself i.e. for every element say (a) in set A, $\left( a,a \right)\in R$ .
Thus, R on a set ‘A’ is not reflexive if there exists an element $a\in A$ such that $\left( a,a \right)\notin R$.
Symmetric Relation: - A relation R on a set ‘A’ is said to be symmetric relation if $\left( a,b \right)\in R$ then $\left( b,a \right)$must be belong to R. i.e. $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$ For all $a,b\in A$.
Transitive relation:- A relation R on ‘A’ is said to be a transitive relation If $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$.
I.e. $\left( a,b \right)\in R$and $\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
Given relation R=\left\\{ \left( a,b \right):\left| a-b \right|\text{ is a simple multiple of 4} \right\\}.
Since,
\begin{aligned} & A=\left\\{ x\in Z;0\le x\le 12 \right\\} \\\ & \therefore A=\left\\{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right\\} \\\ \end{aligned}
Multiples of 4 are 0,4,8,12 only
Check Reflexivity: If $\left( a,a \right)\in R$ , foe $a\in S$ , then R is reflexive.
It is obvious that the difference between any number and itself is always zero. That is, $\left| a-a \right|=\left| 0 \right|=0$ and 0 is a multiple of 4.
So, $\left| a-a \right|$ is a multiple of 4.
$\therefore \left( a,a \right)\in R$ .
Therefore, R is reflexive.
Check symmetricity: If $\left( a,b \right)\in R$ for $a.b\in S$ and $\left( b,a \right)\in R$ , Then R is not symmetric.
Since, $\left| a-b \right|=\left| -\left( b-a \right) \right|$
$\therefore \left| a-b \right|=\left| b-a \right|$ .
If $\left( a.b \right)\in R$ , that means$\left| a-b \right|$ is a multiple of 4.
Since, $\left| a-b \right|=\left| b-a \right|$ .
Therefore, $\left| b-a \right|$ is also a multiple of 4.
Hence, If $\left( a,b \right)\in R$ , then $\left( b,a \right)\in R$ ,
Therefore, R is symmetric.
Check transitivity: If $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ , then $\left( a,c \right)\in R$ , then we can say that R is transitive.
If $\left( a,b \right)\in R$ , that means $\left| a-b \right|$ is a multiple of 4 then $\left( a-b \right)$ is a multiple of 4…..(1)
Similarly, if $\left( b,c \right)\in R$ . that means $\left| b-c \right|$ is a multiple of 4, then $\left( b-c \right)$ is a multiple of 4……(2)
Now, we can think of 8 and 12 as being multiple of 4. And $\left( 8+12 \right)=20$ is also a multiple of 4.
Therefore, the sum of multiple of 4 is also a multiple of 4.
From (1) and (2) we can write
$\left( a-b+b-c \right)$ is a multiple of 4.
$\Rightarrow \left( a-c \right)$ is a multiple of 4.
Hence, $\left| a-c \right|$ is a multiple of 4.
$\therefore \text{ if }\left| a-b \right|\And \left| b-c \right|\text{ is a multiple of 4, then }\left| a-c \right|\text{ is a multiple of 4}\text{.}$
i.e. if $\left( a,b \right)\in R\text{ }{ and }\text{ }\left( b,c \right)\in R,\text{ then }\left( a,c \right)\in R$ .
Hence, R is transitive.
Therefore, R is reflexive, symmetric and transitive.
Now, we need to find a set of elements related to 1.
R=\left\\{ \left( a,b \right):\left| a-b \right|\text{ is a simple multiple of 4} \right\\}
\And A=\left\\{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right\\} .
Here, a=1.

a| b| $\left| a-b \right|$ | Is $\left| a-b \right|$ a multiple of 4
---|---|---|---
1| 1| $\left| 1-1 \right|=\left| 0 \right|=0$ | Yes
1| 2| $\left| 1-2 \right|=\left| -1 \right|=1$ | No
1| 3| $\left| 1-3 \right|=\left| -1 \right|=2$| No
1| 4| $\left| 1-4 \right|=\left| -3 \right|=3$| No
1| 5| $\left| 1-5 \right|=\left| -4 \right|=4$| Yes
1| 6| $\left| 1-6 \right|=\left| -5 \right|=5$| No
1| 7| $\left| 1-7 \right|=\left| -6 \right|=6$| no
1| 8| $\left| 1-8 \right|=\left| -7 \right|=7$| No
1| 9| $\left| 1-9 \right|=\left| -8 \right|=8$| Yes
1| 10| $\left| 1-10 \right|=\left| -9 \right|=9$| No
1| 11| $\left| 1-11 \right|=\left| -10 \right|=10$| No
1| 12| $\left| 1-12 \right|=\left| -11 \right|=11$| No

Hence, The set of elements related to 1 are \left\\{ 1,5,9 \right\\} .

Note: The relation which is reflexive, symmetric and transitive is known as equivalence relation. To prove that the given relation is equivalence i.e. reflexive, symmetric and transitive, we can’t use just one example to prove each, instead we need to prove the conditions of reflexivity, symmetricity and transitivity for all possible cases.