If A = {x in Z^+ : x

Question

(B) relation $R$ on a set $A$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$.First,we identify the elements of set $A$: $A

Vedclass · Accepted Answer

$2^{15}$

Vedclass · Answer

(B) relation $R$ on a set $A$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$.First,we identify the elements of set $A$: $A = \{3, 4, 6, 8, 9\}$. The number of elements $n = 5$.The total number of ordered pairs in $A 	imes A$ is $n^2 = 5^2 = 25$.These $25$ pairs can be categorized into:$1$. Pairs of the form $(a, a)$: There are $n = 5$ such pairs: $(3, 3), (4, 4), (6, 6), (8, 8), (9, 9)$.$2$. Pairs of the form $(a, b)$ where $a 
eq b$: There are $n^2 - n = 25 - 5 = 20$ such pairs.For a symmetric relation,if $(a, b)$ is included,$(b, a)$ must also be included. These $20$ pairs form $10$ pairs of the form $\{(a, b), (b, a)\}$.For each of the $5$ diagonal pairs $(a, a)$,we have $2$ choices (include or exclude).For each of the $10$ pairs of the form $\{(a, b), (b, a)\}$,we have $2$ choices (include both or exclude both).Thus,the total number of symmetric relations is $2^5 	imes 2^{10} = 2^{5+10} = 2^{15}$.

If $A = \{x \in Z^+ : x < 10\}$ and $x$ is a multiple of $3$ or $4$,where $Z^+$ is the set of positive integers,then the total number of symmetric relations on $A$ is

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