Question

Let R be the real line. Consider the following subsets of the plane R2-
S = {(x, y): $y = x + 1$ and 0 < x < 2}
T = {(x, y): $x - y$ is an integer}
Which of the following is true-
A. Both S and T are equivalence relations on R
B. S is an equivalence relation but T is not
C. T is an equivalence relation but S is not
D. Neither S nor T are equivalence relations

Answer 1

Hint: Any relation can be classified as reflexive, symmetric and transitive. If aRa exists in the relation, then it is said to be reflexive. If aRb and bRa both exist in the relation, then it is said to be symmetric. If aRb and bRc exist implies that aRc also exists, the relation is transitive. Using these definitions we will check the relation R.

Complete step-by-step answer:
First we will check if the two relations are reflexive or not. For a relation to be reflexive, (x, x) should be an element of the relation.
For relation S,
$x = x + 1$
0 = 1 which is absurd.
Hence, S is not a reflexive relation. An equivalence relation should be reflexive, symmetric and transitive. So, S is not an equivalence relation.

For relation T,
x - x is an integer
= 0 is an integer
This statement is true. Hence, T is a reflexive relation.

For a relation to be symmetric, if (x, y) is an element then (y, x) should also be an element of that relation.
Let y = x + 1 where 0 < x < 2 …(1)
Now, we will check x = y + 1 is true.
Substituting the value of equation (1),
x = (x + 1) + 1
x = x + 2
0 = 2 , which is absurd.
Hence, S is not symmetric.

For relation T,
Let $x - y = k$ where k is an integer.
We need to check if (y, x) is also an element of T,
$y - x = -(x - y) = -k$
If k is an integer, then its negative will also be an integer. This means that y - x is an integer.
Hence, T is a symmetric relation.

For a relation to be transitive, if (x, y) and (y, z) are elements of the relation, then (x, z) is also an element.
For relation S,
Let y = x + 1 and z = y + 1 exist,
We need to check if z = x + 1 is true or not.
y = x + 1 …(2)
z = y + 1 …(3)
Adding (2) and (3) we get-
z = x + 1 + 1 = x + 2
This means that z = x + 1 is false, the relation S is not transitive.

For relation T,
Let $x - y = m$ and $y - z = n$, where m, n are integers.
We need to check if (x, z) is also an element of T,
Adding the two equations-
$x - y + y - z = m + n$
$x - z = m + n$
We know that the sum of two integers is an integer. So, x - z is also an integer.
Hence, T is a transitive relation.

T is reflexive, symmetric and transitive, so it is an equivalence relation but S is not. The correct option is C.

Note: It is important to check carefully for each condition. It is also recommended to check and verify each condition using a suitable example. Even if one case is false, the condition is not verified. Also if it is not possible to prove that relation is symmetric, reflexive or transitive, then use a suitable example to show that it is not.