First slide

Relations XII

Question

Let R be the real line. Consider the following subsets of the plane $R \times R$ .
$\begin{array}{l} S = {(x, y) : y = x + 1 and 0 < x < 2} \\ T = {(x, y) : x - y is an integer} \end{array}$
Statement-1 : T is an equivalence relation on R but S is not an equivalence relation on R.
Statement-2 : S is neither reflexive nor symmetric but T is reflexive, symmetric and transitive

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

Since x $\neq$ x + 1 $(x, x) \notin S$ , so S is not reflexive
Next $x, y \in S, \Rightarrow y = x + 1 \Rightarrow x = y - 1 \Rightarrow (y, x) \notin S$ so is not symmetric
Since x – x = 0 0 is an integer $(x, x) \in T \forall x \in T$
Again $(x, y) \in T \Rightarrow x - y$ is an integer
$\Rightarrow$ y – x is also an integer $\Rightarrow (y, x) \in T$
So T is symmetric.
Also , $(x, y) \in T, (y, z) \in T$
$\Rightarrow$ x – y and y – z are integers
$\Rightarrow$ x – z = (x – y) – (y – z) is also an integer
$\Rightarrow (x, z) \in T$
so T is s Transitive .
Which shows that statement-2 is true and hence statement-1 is also true.

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