Consider the relations R_1 and R_2 defined as | vedclass

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(B) For relation $R_1$: $a R_1 b \Leftrightarrow a^2+b^2=1$ for $a, b \in R$. $1$. Reflexivity: For $a=0.5$,$a^2+a^2 = 0.25+0.25 = 0.5 
eq 1$. So,$R_

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Only $R_2$ is an equivalence relation

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(B) For relation $R_1$: $a R_1 b \Leftrightarrow a^2+b^2=1$ for $a, b \in R$. $1$. Reflexivity: For $a=0.5$,$a^2+a^2 = 0.25+0.25 = 0.5 
eq 1$. So,$R_1$ is not reflexive. $2$. Symmetry: If $a^2+b^2=1$,then $b^2+a^2=1$,so $b R_1 a$. $R_1$ is symmetric. $3$. Transitivity: If $a R_1 b$ and $b R_1 c$,then $a^2+b^2=1$ and $b^2+c^2=1$. This does not imply $a^2+c^2=1$. For example,$a=1, b=0, c=1$. $1^2+0^2=1$ and $0^2+1^2=1$,but $1^2+1^2=2 
eq 1$. Thus,$R_1$ is not transitive. For relation $R_2$: $(a, b) R_2 (c, d) \Leftrightarrow a+d=b+c$ for $(a, b), (c, d) \in N 	imes N$. $1$. Reflexivity: $a+b=b+a$ is true,so $(a, b) R_2 (a, b)$. $2$. Symmetry: If $a+d=b+c$,then $c+b=d+a$,so $(c, d) R_2 (a, b)$. $3$. Transitivity: If $(a, b) R_2 (c, d)$ and $(c, d) R_2 (e, f)$,then $a+d=b+c$ and $c+f=d+e$. Adding these,$a+d+c+f = b+c+d+e \Rightarrow a+f=b+e$. Thus,$(a, b) R_2 (e, f)$. Since $R_2$ is reflexive,symmetric,and transitive,it is an equivalence relation. Therefore,only $R_2$ is an equivalence relation.

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

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