Question

Que 1.22. Let R be a relation on R, the set of real numbers, such
that R = {(x, y)||*- y < 1. Is R an equivalence relation ? Justify.​

Answer 1

Question

Let R be a relation on R, the set of real numbers, such that R = {(x, y)|| x - y < 1}. Is R an equivalence relation?

Solution

Given that,

R = {(x, y)|| x - y < 1}

To prove that R is an equivalence relation.

A function is said to be equivalent if it is Reflexive, Transitive and Symmetric in nature.

Reflexivity of R : A function in which the pre images and images are similar.

Since, x belongs to R. (x,x) € R

Now, x - x = 0 < 1

Thus, R is reflexive.

Symmetry of R : A function is said to be symmetric if (x,y) € R and (y,x) € R.

Now, (x,y) € R and x - y < 1

By intuition, if the difference of x and y is less than one, so should be the difference between y and x.

Thus, (y,x) € R and y - x < 1.

Thus, R is Symmetric.

Transitivity of R : A function is said to be transitive if (x,y) € R, (y,z) € R and (z,x) € R.

Now,

x - y < 1

y - z < 1

z - x < 1.

Thus, R is Transitive.

Henceforth, R is an equivalence relation.

Answer 2

Given

R = {(x, y)|| x - y < 1}

To prove:-

That R is an equivalence relation.

A function is said to be equivalent if it is Reflexive, Transitive and Symmetric in nature.

Reflexivity of R : A function in which the pre images and images are similar.

Since, x belongs to R. (x,x) € R

Now,

x - x = 0 < 1

Thus, R is reflexive.

Symmetry of R : A function is said to be symmetric if (x,y) € R and (y,x) € R.

Now, (x,y) € R and x - y < 1

By intuition, if the difference of x and y is less than one, so should be the difference between y and x.

Thus, (y,x) € R and y - x < 1.

Thus, R is Symmetric.

Transitivity of R : A function is said to be transitive if (x,y) € R, (y,z) € R and (z,x) € R.

Now,

x - y < 1

y - z < 1

z - x < 1.

Thus, R is Transitive.

Henceforth, R is an equivalence relation.