Let R be the real line. Consider the following subsets of the plane R × R:
$S=\{(x, y): y=x+1\, and\, 0 < x <2\}$,
$T=\{(x, y): x-y\, is\, an\, integer\}$
Which one of the following is true?
Answer & explanation
Correct answer: option 2
For any $x ∈ (0, 2)$, we find that $x ≠ x + 1$.
So, $(x, x) ∉ S$.
∴ S is not a reflexive relation.
Hence, S is not an equivalence relation.
Reflexivity of T: For any $x ∈ R$, we have
$x-x=0$, which is an integer.
$⇒(x, x) ∈T$
∴ T is reflexive on R
Symmetry of T: Let $(x, y) ∈ T$. Then,
$x-y$ is an integer
$⇒y-x$ is an integer
$⇒(y, x) ∈T$
T is symmetric on R.
Transitivity of T: Let $(x, y) ∈ T$ and $(y, z) ∈ T$. Then,
$x-y$ is an integer and $y-z$ is an integer
$⇒x-z$ is an integer
$⇒(x, z) ∈T$
∴ T is transitive on R.
Hence, T is an equivalence relation on R.