Question

If ${{A}_{1}},{{A}_{2}}$ and ${{A}_{3}}$ are subsets of Z, whose union is Z defined by
{{A}_{1}}=\left\\{ x\in Z:x\text{ is a multiple of 3} \right\\}
{{A}_{2}}=\left\\{ x\in Z:x-1\text{ is a multiple of 3} \right\\}
{{A}_{3}}=\left\\{ x\in Z:x-2\text{ is a multiple of 3} \right\\}
And R is the relation in Z given by R=\left\\{ \left( a,b \right):3\text{ divides }a-b \right\\}.
${{A}_{1}}$ coincides with the set of all integers in Z which are related to …..A….., ${{A}_{2}}$ coincides with the set of all integers in Z which are related to …..B….., ${{A}_{3}}$ coincides with the set of all integers in Z which are related to ….C…. . Here A, B, and C are respectively
(a)One, zero, two
(b)Zero, one, two
(c)Two, one, zero
(d)One, two, zero

Answer 1

Hint: First observe the relation given in question. Find all the properties related to the relation. Try to substitute values of variables to check their properties. Try to use definitions of reflexive, transitive, symmetry, Anti-symmetry. First and the foremost thing to do is use the definition of relation and say that the given relation satisfies all the conditions needed for it to be a relation. Next see the range on which relation is defined. While checking properties, substitute the few elements only which satisfy the range given in the question.

Complete step-by-step answer:
Relation: The relations in maths are also called a binary relation over set X, Y is a subset of the cartesian product of the sets X, Y that is the relation contains the elements of the set of $X\times Y$ . $X\times Y$is a set of ordered pairs consisting of elements x in X and y in Y. It encodes the information, $\left( x,y \right)$ will be in the Relation set if and only if x is related to the elementary in the relation property given.
Reflexive Relation: A relation R is said to be reflexive if a is an element in the range. If the pair $\left( a,a \right)$ belongs to relation, then relation is reflexive. In other words, if a is related to itself by given relation then relation is reflexive.
Symmetric Relation: A relation R is said to Symmetric, if a, b are elements in the range. If the pair $\left( a,b \right)$ belongs to relation then $\left( b,a \right)$ must belong to relation. In other words if a is related to b, then b must also be related to a for a relation to be symmetric.
Transitive Relation: If $\left( a,b \right)$ belongs to relation and $\left( b,c \right)$ also belongs to the relation, then $\left( a,c \right)$ must belong to relation to make it transitive.
Anti-symmetric Relation: If $\left( a,b \right)$ and $\left( b,a \right)$ are given belong to the relation, then a must be equal to b to make the relation as anti-symmetric.
If a relation is reflexive, transitive and symmetric then it is an equivalence relation.
Congruency: Two integers a, b are in congruent modulo n, if and only if they have the same remainder when divided by n. In other words, for some integers K (positive or negative) it must follow $a=b+kn$ .
Mathematical representation: - $a\equiv b$ mod n. from this we can say n divides $a-b$ .
{{A}_{1}}=\left\\{ x:x\text{ is a multiple of 3} \right\\}
$x$ is multiple of 3 also means 3 divides $x-0$ .
By congruence method we get: $x\equiv 0$ mod 3.
{{A}_{2}}=\left\\{ x-1:x-1\text{ is a multiple of 3} \right\\}
$x-1$is multiple of 3 also means 3 divides $x-1$ .
By congruence method, we get: $x\equiv 1$ mod 3
{{A}_{3}}=\left\\{ x-2:x-2\text{ is a multiple of 3} \right\\}
$x-2$ is multiple of 3 also means 3 divides $x-2$ .
By congruence method we get: $x\equiv 2$ mod 3.
By all the above 3 congruence conditions we get that x’s divisibility rule by 3.
So, relation $\left( a,b \right)$ such that 3 divides a, b is given.
It is given $a=x$ . Then b will be 0 in ${{A}_{1}}$, 1 in ${{A}_{2}}$, 2 in ${{A}_{3}}$. These are said by congruence conditions above.
So, 0,1,2 are answers for the blanks.
Option (b) is the correct match for the question.

Note: Whenever we see a difference of 2 numbers divisible by a number we must get the idea of using congruence properties. As this makes any divisibility conditions easy to solve. Here 3 functions are solved by 3 equations with congruency. If we do not use congruence then we will solve in a very long method, which is not liable.