Consider the relations R_1 and R_2 defined as a R_1 b a^2+b^2=1 for all a, b R and (a, b) …

Question

Consider the relations $R_1$ and $R_2$ defined as $a R_1 b \Leftrightarrow a^2+b^2=1$ for all $a, b \in \mathbf{R}$ and $(a, b) R_2(c, d) \Leftrightarrow$ $a+d=b+c$ for all $(a, b),(c, d) \in \mathbf{N} 	imes \mathbf{N}$. Then :

Accepted Answer

To determine if the given relations ${R}_{1}$ and ${R}_{2}$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Let's start by analysing ${R}_{1}$:

Reflexivity: A relation $R$ on a set $S$ is reflexive if every element is related to itself, that is, for every $a\in S$, the pair $(a,a)$ is in $R$. In the case of ${R}_{1}$, for an arbitrary $a\in R$, we need to check if $a{R}_{1}a$ holds, which translates to checking if ${a}^{2}+{a}^{2}=1$. This would mean $2{a}^{2}=1$ or ${a}^{2}=\frac{1}{2}$. Since this is not true for every real number $a$, ${R}_{1}$ is not reflexive.

Symmetry: A relation $R$ is symmetric if whenever $aRb$, then $bRa$. For ${R}_{1}$, if ${a}^{2}+{b}^{2}=1$, then it is also true that ${b}^{2}+{a}^{2}=1$. Thus, ${R}_{1}$ is symmetric.

Transitivity: A relation $R$ is transitive if whenever $aRb$ and $bRc$, then $aRc$. For ${R}_{1}$, suppose $a{R}_{1}b$ and $b{R}_{1}c$, this means ${a}^{2}+{b}^{2}=1$ and ${b}^{2}+{c}^{2}=1$. However, if we add these two, we get ${a}^{2}+2{b}^{2}+{c}^{2}=2$. There is no guarantee that ${a}^{2}+{c}^{2}=1$. Therefore, ${R}_{1}$ is not transitive.

Conclusion: Relation ${R}_{1}$ is not reflexive or transitive, and hence it is not an equivalence relation.

Now let's consider ${R}_{2}$:

Reflexivity: For any $(a,b)\in N	imes N$, it's clear that $a+b=b+a$, which is just a reiteration of the commutative property of addition. Therefore, $(a,b){R}_{2}(a,b)$, and ${R}_{2}$ is reflexive.

Symmetry: If $(a,b){R}_{2}(c,d)$, meaning $a+d=b+c$, then by reordering the terms we can similarly have $c+b=d+a$, which means $(c,d){R}_{2}(a,b)$, so ${R}_{2}$ is symmetric.

Transitivity: Suppose $(a,b){R}_{2}(c,d)$ and $(c,d){R}_{2}(e,f)$, i.e., $a+d=b+c$ and $c+f=d+e$, we want to show that $(a,b){R}_{2}(e,f)$. Adding the two equations, we get $a+d+c+f=b+c+d+e$. By rearranging and simplifying, we get $a+f=b+e$, thus, $(a,b){R}_{2}(e,f)$.

So ${R}_{2}$ is reflexive, symmetric, and transitive, and therefore it is an equivalence relation.

Conclusion: According to the above examination, only ${R}_{2}$ is an equivalence relation. Thus, the correct answer is:

Option C: Only ${R}_{2}$ is an equivalence relation.

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