Among the relations S = {(a, b) : a, b in R - | vedclass

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(B) For relation $T = \{(a, b) : a^2 - b^2 \in Z\}$:If $(a, b) \in T$,then $a^2 - b^2 = k$ for some integer $k \in Z$.Then $b^2 - a^2 = -(a^2 - b^2) =

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$T$ is symmetric but $S$ is not

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(B) For relation $T = \{(a, b) : a^2 - b^2 \in Z\}$:If $(a, b) \in T$,then $a^2 - b^2 = k$ for some integer $k \in Z$.Then $b^2 - a^2 = -(a^2 - b^2) = -k$,which is also an integer.Thus,$(b, a) \in T$,so $T$ is symmetric.For relation $S = \{(a, b) : a, b \in R - \{0\}, 2 + \frac{a}{b} > 0\}$:Consider $(a, b) = (1, 1)$. $2 + \frac{1}{1} = 3 > 0$,so $(1, 1) \in S$.Consider $(a, b) = (1, -1)$. $2 + \frac{1}{-1} = 1 > 0$,so $(1, -1) \in S$.Consider $(b, a) = (-1, 1)$. $2 + \frac{-1}{1} = 1 > 0$,so $(-1, 1) \in S$.However,consider $(a, b) = (1, -0.6)$. $2 + \frac{1}{-0.6} = 2 - 1.66 = 0.33 > 0$,so $(1, -0.6) \in S$.Check $(b, a) = (-0.6, 1)$. $2 + \frac{-0.6}{1} = 1.4 > 0$,so $(-0.6, 1) \in S$.Let us check symmetry: If $(a, b) \in S$,then $2 + \frac{a}{b} > 0 \Rightarrow \frac{a}{b} > -2$.For $(b, a) \in S$,we need $2 + \frac{b}{a} > 0 \Rightarrow \frac{b}{a} > -2$.If we take $a = 1, b = -0.6$,then $\frac{a}{b} = -1.66 > -2$ (True).But $\frac{b}{a} = -0.6 > -2$ (True).Wait,let us test $a = 1, b = -0.4$. $2 + \frac{1}{-0.4} = 2 - 2.5 = -0.5 Let us test $a = 1, b = -0.8$. $2 + \frac{1}{-0.8} = 2 - 1.25 = 0.75 > 0$. So $(1, -0.8) \in S$.Now check $(b, a) = (-0.8, 1)$. $2 + \frac{-0.8}{1} = 1.2 > 0$. So $(-0.8, 1) \in S$.Actually,$S$ is not symmetric because if $a=1, b=-0.4$,$(1, -0.4) 
otin S$. If $a=1, b=-0.9$,$(1, -0.9) \in S$ and $(-0.9, 1) \in S$. However,if $a=1, b=-0.1$,$2 + \frac{1}{-0.1} = 2 - 10 = -8 Since $T$ is symmetric and $S$ is not,option $B$ is correct.

Among the relations $S = \{(a, b) : a, b \in R - \{0\}, 2 + \frac{a}{b} > 0\}$ and $T = \{(a, b) : a, b \in R, a^2 - b^2 \in Z\}$,which of the following is true?

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