Let A = {-4, -3, -2, 0, 1, 3, 4} and R = {(a, | vedclass

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(C) Given $A = \{-4, -3, -2, 0, 1, 3, 4\}$.First,we find the elements of $R$ based on the conditions $b = |a|$ or $b^2 = a + 1$:For $b = |a|$: $(-4, 4

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$6$

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(C) Given $A = \{-4, -3, -2, 0, 1, 3, 4\}$.First,we find the elements of $R$ based on the conditions $b = |a|$ or $b^2 = a + 1$:For $b = |a|$: $(-4, 4), (-3, 3), (-2, 2), (0, 0), (1, 1), (3, 3), (4, 4)$. Note that $(-2, 2)$ is not in $A 	imes A$ because $2 
otin A$. So,we have $(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4)$.For $b^2 = a + 1$: If $a = -4, b^2 = -3$ (no); $a = -3, b^2 = -2$ (no); $a = -2, b^2 = -1$ (no); $a = 0, b^2 = 1 \Rightarrow b = 1, -1$ (only $1 \in A$); $a = 1, b^2 = 2$ (no); $a = 3, b^2 = 4 \Rightarrow b = 2, -2$ (no); $a = 4, b^2 = 5$ (no).Thus,$R = \{(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4), (0, 1)\}$.For $R$ to be reflexive,we need $(a, a) \in R$ for all $a \in A$. Missing elements: $(-4, -4), (-3, -3), (-2, -2)$. ($3$ elements).Now $R' = R \cup \{(-4, -4), (-3, -3), (-2, -2)\} = \{(-4, 4), (-3, 3), (0, 0), (1, 1), (3, 3), (4, 4), (0, 1), (-4, -4), (-3, -3), (-2, -2)\}$.For $R'$ to be symmetric,if $(a, b) \in R'$,then $(b, a) \in R'$.Pairs to add: $(4, -4), (3, -3), (1, 0)$. ($3$ elements).Total elements added = $3 + 3 = 6$.

Let $A = \{-4, -3, -2, 0, 1, 3, 4\}$ and $R = \{(a, b) \in A \times A : b = |a| \text{ or } b^2 = a + 1\}$ be a relation on $A$. Then the minimum number of elements that must be added to the relation $R$ so that it becomes reflexive and symmetric is $........$.

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