Question ID: #835
Let $A = \{0, 1, 2, …, 9\}$. Let $R$ be a relation on $A$ defined by $(x, y) \in R$ if and only if $|x-y|$ is a multiple of 3.
Given below are two statements:
Statement I: $n(R) = 36$
Statement II: $R$ is an equivalence relation.
In the light of the above statements, choose the correct answer from the options given below
- (1) Both Statement I and Statement II are correct
- (2) Statement I is incorrect but Statement II is correct
- (3) Statement I is correct but Statement II is incorrect
- (4) Both Statement I and Statement II are incorrect
Solution:
We analyze the sets based on remainders modulo 3 (congruence classes).
$A = \{0, 1, 2, …, 9\}$.
Class $C_0$ (remainder 0): $\{0, 3, 6, 9\}$ (4 elements)
Class $C_1$ (remainder 1): $\{1, 4, 7\}$ (3 elements)
Class $C_2$ (remainder 2): $\{2, 5, 8\}$ (3 elements)
For $|x-y|$ to be a multiple of 3, both $x$ and $y$ must belong to the same class.
Number of elements in $R$:
$$ n(R) = n(C_0 \times C_0) + n(C_1 \times C_1) + n(C_2 \times C_2) $$
$$ n(R) = 4^2 + 3^2 + 3^2 = 16 + 9 + 9 = 34 $$
So, Statement I says $n(R) = 36$, which is **Incorrect**.
Check Statement II:
Reflexive: $|x-x| = 0$ (multiple of 3). True.
Symmetric: If $|x-y| = 3k$, then $|y-x| = 3k$. True.
Transitive: If $x \equiv y \pmod 3$ and $y \equiv z \pmod 3$, then $x \equiv z \pmod 3$. True.
So, $R$ is an equivalence relation. Statement II is **Correct**.
Ans. (2)
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