Let A = {1, 2, 3, ldots, 20}. Let R_1 and R_2 | vedclass

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(B) The set $A = \{1, 2, 3, \ldots, 20\}$.$R_1 = \{(a, b) : b 	ext{ is divisible by } a\}$. The number of elements in $R_1$ is the sum of the number

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$46$

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(B) The set $A = \{1, 2, 3, \ldots, 20\}$.$R_1 = \{(a, b) : b 	ext{ is divisible by } a\}$. The number of elements in $R_1$ is the sum of the number of multiples of each $a \in A$ that are $\leq 20$.For $a=1$,there are $20$ multiples. For $a=2$,there are $10$ multiples. For $a=3$,there are $6$ multiples. For $a=4$,there are $5$ multiples. For $a=5$,there are $4$ multiples. For $a=6$,there are $3$ multiples. For $a=7, 8, 9, 10$,there are $2$ multiples each. For $a=11, 12, \ldots, 20$,there is $1$ multiple each (the number itself).$n(R_1) = 20 + 10 + 6 + 5 + 4 + 3 + (4 	imes 2) + (10 	imes 1) = 48 + 8 + 10 = 66$.$R_2 = \{(a, b) : a 	ext{ is an integral multiple of } b\}$. This is equivalent to saying $b$ is a divisor of $a$,which is the same as $R_1$.However,the definition of $R_2$ given is $a = kb$ for some integer $k$. Since $a, b \in A$,$k$ must be a positive integer. This is the same as $R_1$.Wait,$R_1 = \{(a, b) : b = ka, k \in \mathbb{Z}^+\}$ and $R_2 = \{(a, b) : a = kb, k \in \mathbb{Z}^+\}$.$R_1 \cap R_2 = \{(a, b) : b = ka 	ext{ and } a = mb\} = \{(a, b) : a = mka \implies mk = 1\}$. Since $m, k \in \mathbb{Z}^+$,$m=1, k=1$. Thus $a=b$.$R_1 \cap R_2 = \{(1, 1), (2, 2), \ldots, (20, 20)\}$.$n(R_1 \cap R_2) = 20$.$n(R_1 - R_2) = n(R_1) - n(R_1 \cap R_2) = 66 - 20 = 46$.

Let $A = \{1, 2, 3, \ldots, 20\}$. Let $R_1$ and $R_2$ be two relations on $A$ such that $R_1 = \{(a, b) : b \text{ is divisible by } a\}$ and $R_2 = \{(a, b) : a \text{ is an integral multiple of } b\}$. Then,the number of elements in $R_1 - R_2$ is equal to . . . . . . .

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