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Question: Show that each of the relation R in the set\(A = \\{ x \in Z:0 \le x \le 12\\} \) , given by (i)\(...

Show that each of the relation R in the setA=x∈Z:0≤x≤12A = \\{ x \in Z:0 \le x \le 12\\} , given by
(i)R=(a,b):∣a−b∣  isamultipleof4R = \\{ (a,b):\left| {a - b} \right|\;{\rm{is a multiple of 4\\} }}
(ii)R=(a,b):a=bR = \\{ (a,b):a = b\\}
is an equivalence relation. Find the set of all elements related to 1 in each case.

Explanation

Solution

Equivalence relation is described as the relation among elements of a particular set that could be transitive, reflexive or symmetric. To prove any equivalence relation, first we have to prove that it is reflexive relation, symmetric relation and transitive relation.

Complete Step-by-step Solution
(i) According to question, A = \left\\{ {{\rm{ }}0,{\rm{ }}1,{\rm{ }}2,{\rm{ }}3,{\rm{ }}4,{\rm{ }}5,{\rm{ }}6,{\rm{ }}7,{\rm{ }}8,{\rm{ }}9,{\rm{ }}10,11,12} \right\\}.
Now, we will find the multiples of 4 which hare are 0, 4 8, 12 that is,
the value of ∣a−b∣\left| {a - b} \right| can be 0, 4, 8, 12 only.
Reflexive:
The reflexive property says that the variables (for example a) is always equivalent to itself, that is (a,a)∈R(a,a) \in R.
Since, we know that ∣a−a∣=0\left| {a - a} \right| = 0 and 0 is a multiple of 4. So,(a,a)∈R(a,a) \in R, Therefore R is reflexive.
Symmetric:
The symmetric property says that if two variables (for example a and b) are equal that is a=ba = b then, through symmetry b is also equal to a that is b=ab = a.
Since, we know that ∣a−b∣\left| {a - b} \right| is a multiple of 4 and ∣b−a∣\left| {b - a} \right| is also multiple of 4.
Now, we know from symmetric relation,
(a,b)∈R (b,a)∈R\begin{array}{l} (a,b) \in R\\\ {\rm{(b,a)}} \in R \end{array}
Therefore, R is symmetric.
Transitive:
The transitive property says that in three variables if a and b are equal that is a=ba = b and b and c are equal that is b=cb = c, then the first variable and third variables are also equal a=ca = c.
Let us consider that (a,b)∈R(a,b) \in R and (b,c)∈R(b,c) \in R, then through symmetry (b,a)∈R{\rm{(b,a)}} \in R.
Also, ∣a−b∣  and  ∣b−c∣\left| {a - b} \right|\;{\rm{and}}\;\left| {b - c} \right| are multiple of 4.
We have (a−b)=4m  &  (b−c)=4n(a - b) = 4m\;{\rm{\& }}\;(b - c) = 4n for some integers m and n.
On adding above equations, we get the value as,
(a−c)=4(m+n) (a−c)  isamultipleof4. (a,c)∈R\begin{array}{l} (a - c) = 4(m + n)\\\ (a - c)\;{\rm{ is a multiple of 4}}{\rm{.}}\\\ {\rm{(a,c)}} \in {\rm{R}} \end{array}
Hence, R is transitive.
Therefore, being Reflexive, symmetric and transitive, R is an equivalence relation. The set of elements related to 1 is {1, 5, 9}.

(ii)Given:The value of R is R=(a,b):a=bR = \\{ (a,b):a = b\\} .
Reflexive:The reflexive property says that the variables (for example a) is always equivalent to itself, that is (a,a)∈R(a,a) \in R.
Since, a=aa = a, so (a,a)∈R(a,a) \in R{\rm{ }}. Hence, R is reflexive.
Symmetric:
The symmetric property says that if two variables (for example a and b) are equal that is a=ba = b then, through symmetry b is also equal to a that is b=ab = a.
Let us consider, (a,b)∈R(a,b) \in R.
a=b b=a (b,a)∈R\begin{array}{c} a = b\\\ b = a\\\ {\rm{(b,a)}} \in {\rm{R}} \end{array}
Hence, R is symmetric.
Transitive:
The transitive property says that in three variables if a and b are equal that is a=ba = b and b and c are equal that is b=cb = c, then the first variable and third variables are also equal a=ca = c.
Let us consider, (a,b),(b,c)∈R(a,b),(b,c) \in R
a=b b=c a=c (a,c)∈R\begin{array}{c} a = b\\\ b = c\\\ a = c\\\ (a,c) \in R \end{array}
Hence, R is transitive.
Therefore, being reflexive, symmetric and transitive, R is an equivalence relation. The set of elements related to 1 is {1}.

Note: In such types of questions, make sure all b∈Rb \in R will be there in a set of elements related to a∈Ra \in R , which satisfy (a,b)∈R(a,b) \in R.