Question

For real numbers \[x\] and \[y\] , we write \[xRy \Rightarrow x - y + \sqrt 2 \] is an irrational number. Then the relation \[R\] is
A.Reflective
B.Symmetric
C.Transitive
D.Equivalence relation
Let \[A = N*N\] be the Certesian product of \[N\] and \[N\] .Let

Answer 1

Hint: In order to determine the irrational number \[xRy \Rightarrow x - y + \sqrt 2 \] , for the real numbers \[x\] and \[y\] . Here, the relation operator \[R\] says for real numbers \[x\] and \[y\] and no further relation can be developed. Now, if either or both \[x\] and \[y\] , if an irrational or imaginary is operable or unable to say. Also the reversibility o

Complete step by step solution:
we are given equation \[xRy \Rightarrow x - y + \sqrt 2 \] are the real numbers \[x\] and \[y\]
We have to check the relations on comparing with the irrational number.
Reflexive: A reflexive relation is the one in which every element maps to itself.
Let us check if the number is reflective or not.
Since \[xRy \Rightarrow x - y + \sqrt 2 \] is an irrational number.
Let us consider the real number \[x = y = a\]
So, \[aRa \Rightarrow a - a + \sqrt 2 = \sqrt 2 \]
Therefore, \[\sqrt 2 \] is an irrational number.
Hence, \[R\] is a reflexive relation.

Symmetric: A symmetric relation is a type of binary relation. An example is the relation "is equal to", because if \[x = y\] is true then \[y = x\] is also true.
Let us check the real numbers are symmetric or asymmetric
In Symmetric to get \[yRx \Rightarrow - y + x + \sqrt 2 \] but we needed plus root 2. also try the example \[x = 1\] and \[y = \sqrt 2 \] .
For \[xRy\] , \[x - y + \sqrt 2 \] is irrational.
So, \[y - x + \sqrt 2 \] is also an irrational number.
Since \[\left( {x - y} \right) = - \left( {y - x} \right)\] .which implies that \[yRx\] .
So, \[R\] is a symmetric relation.

Transitive: In mathematics, a homogeneous relation \[R\] over a set \[X\] is transitive if for all elements \[x,y,z\] in \[X\] , whenever \[R\] relates \[x\] to \[y\] and \[y\] to \[z\] , then \[R\] also relates \[x\] to \[z\] .
Let us check if the number is transitive or not.
Given, For real numbers \[x\] and \[y\] ,
 \[x - y + \sqrt 2 \] is an irrational number.
Let \[R\] is a binary relation for the real numbers \[x\] and \[y\] .
Now, let \[R\] is transitive if for all \[(x,y) \in R\] and \[(y,z) \in R\] implies (x, z) ∈ R
Given, \[xRy \Rightarrow x - y + \sqrt 2 \] is irrational -----(1)
And \[yRz \Rightarrow - y + z + \sqrt 2 \] is irrational -----(2)
Add equation (1) and (2), we get
 \[\;{\text{ (}}x - {\text{ }}y + \sqrt 2 ){\text{ }} + {\text{ }}(y - {\text{ }}z + \sqrt 2 )\] is irrational number..
 $x-y+ \sqrt 2$ is irrational, which implies \[xRz\] is an irrational number.
Therefore, the difference of two irrational numbers is also irrational
As a result, \[R\] is a transitive relation.
Equivalence relation: A relation \[R\] is an equivalence relation if and only if R is reflexive, symmetric, and transitive.
Since, \[R\] satisfies all the above relations.
Therefore all the relation is true.
Hence, \[R\] is an equivalence relation.
So, the correct answer is “Option D”.

Note: We note that the relation of the irrational number satisfies \[R\] as follows the rule.
A reflexive relation is the one in which every element maps to itself.
A symmetric relation is a type of binary relation.
Transitive is a homogeneous relation \[R\] over a set \[X\] is transitive if for all elements \[x,y,z\] in \[X\] , whenever \[R\] relates \[x\] to \[y\] and \[y\] to \[z\] , then \[R\] also relates \[x\] to \[z\] .
Equivalence relation: A relation \[R\] is an equivalence relation if and only if R is reflexive, symmetric, and transitive.