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In the relation ‘is the square of’ for the set of natural numbers N an equivalence relation?

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Last updated date: 13th Jun 2024
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Answer
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Hint: We take a relation from a set of natural numbers to the set of natural numbers. We define a relation using the function square of a natural number. We check if the relation is an equivalence relation by checking for the reflexive, symmetric, and transitive nature of the relation.
* Reflexive: A relation R is said to be reflexive if \[(a,a) \in R\] for every \[a \in R\].
* Symmetric: A relation R is said to be symmetric if \[(a,b) \in R\] then \[(b,a) \in R\] for every \[a,b \in R\].
* Transitive: A relation R is said to be transitive if \[(a,b) \in R;(b,c) \in R\] then \[(a,c) \in R\] for every \[a,b,c \in R\].

Complete step by step answer:
Let us take two sets of natural numbers A and B
We define a relation \[R:A \to B\]such that for every x in A there is y in B defined by the function\[y = {x^2}\]for every\[x,y \in N\].
We can write the first few elements of set A and B
\[A = \left\{ {1,2,3,4...} \right\}\]
\[B = \left\{ {1,4,9,16,...} \right\}\]
Now we assume three elements of the set R say x, y, and z.
Reflexive:
We know the function \[y = {x^2}\]gives the square value of the element of A.
So, for \[x = 1\]we get \[{x^2} = 1\]
But for \[x = 2\]we get \[{x^2} = 4\]
\[ \Rightarrow (x,x) \notin R\]

So the relation is not reflexive.

Symmetric:
We know the function \[y = {x^2}\]gives the square value of the element of A.
For elements \[x,y \in A\]we know\[(x,y) \in R\]
\[ \Rightarrow y = {x^2}\]
We check if \[(y,x) \in R\]
\[ \Rightarrow x = {y^2}\]
Since both\[x,y \in N\], then \[x,{y^2} \in N\]
\[ \Rightarrow (y,x) \in R\]

So the relation is symmetric.

Transitive:
We know the function \[y = {x^2}\]gives the square value of the element of A.
For elements \[x,y,z \in A\]we know\[(x,y) \in R;(y,z) \in R\]
\[ \Rightarrow y = {x^2}\]and \[z = {y^2}\]
We check if \[(x,z) \in R\]i.e. check if \[z = {x^2}\]
Substitute the value of \[y = {x^2}\]in place of y in equation \[z = {y^2}\]
\[ \Rightarrow z = {({x^2})^2}\]
\[ \Rightarrow z = {x^4}\]
So, \[z \ne {x^2}\]
\[ \Rightarrow (x,z) \notin R\]

So the relation is not transitive.

Since the relation is not symmetric and not transitive, we can say the relation R defined by ‘is square of’ is not an equivalence relation.

Note:
Students might try to prove the relation is an equivalence relation by giving examples of (1,1) etc. Keep in mind we give examples to prove the contradiction. To prove the relation is equivalence we will have to give a general proof.