Let [A=2,3,6,7] and [B=4,5,6,8].

Question

Let $$A=\{2,3,6,7\}$$ and $$B=\{4,5,6,8\}$$. Let $R$ be a relation defined on $$A \times B$$ by $(a_1, b_1) R(a_2, b_2)$ if and only if $a_1+a_2=b_1+b_2$. Then the number of elements in $R$ is \_\_\_\_\_\_\_\_\_\_.

Answer: 25

Accepted Answer

To find the number of elements in the relation $R$ defined on $A	imes B$, we need to determine all pairs $(({a}_{1},{b}_{1}),({a}_{2},{b}_{2}))$ such that ${a}_{1}+{a}_{2}={b}_{1}+{b}_{2}$, where ${a}_{1},{a}_{2}\in A$ and ${b}_{1},{b}_{2}\in B$.

First, consider all possible sums of pairs from set $A$ and set $B$.

Possible sums from set $A={2,3,6,7}$:

$2+2=4$

$2+3=5$

$2+6=8$

$2+7=9$

$3+2=5$

$3+3=6$

$3+6=9$

$3+7=10$

$6+2=8$

$6+3=9$

$6+6=12$

$6+7=13$

$7+2=9$

$7+3=10$

$7+6=13$

$7+7=14$

Possible sums from set $B={4,5,6,8}$:

$4+4=8$

$4+5=9$

$4+6=10$

$4+8=12$

$5+4=9$

$5+5=10$

$5+6=11$

$5+8=13$

$6+4=10$

$6+5=11$

$6+6=12$

$6+8=14$

$8+4=12$

$8+5=13$

$8+6=14$

$8+8=16$

Now, identify the common sums from both sets:

Common sums: $8,9,10,12,13,14$

For each common sum, count the pairs from set $A$ and set $B$ that produce these sums:

Sum = 8: From $A$: {(2,6), (6,2)} - 2 pairs; From $B$: {(4,4)} - 1 pair; Hence, 2 * 1 = 2 pairs

Sum = 9: From $A$: {(2,7), (3,6), (6,3), (7,2)} - 4 pairs; From $B$: {(4,5), (5,4)} - 2 pairs; Hence, 4 * 2 = 8 pairs

Sum = 10: From $A$: {(3,7), (7,3)} - 2 pairs; From $B$: {(4,6), (5,5), (6,4)} - 3 pairs; Hence, 2 * 3 = 6 pairs

Sum = 12: From $A$: {(6,6)} - 1 pair; From $B$: {(4,8), (6,6), (8,4)} - 3 pairs; Hence, 1 * 3 = 3 pairs

Sum = 13: From $A$: {(6,7), (7,6)} - 2 pairs; From $B$: {(5,8), (8,5)} - 2 pairs; Hence, 2 * 2 = 4 pairs

Sum = 14: From $A$: {(7,7)} - 1 pair; From $B$: {(6,8), (8,6)} - 2 pairs; Hence, 1 * 2 = 2 pairs

Adding all these, we get the number of elements in the relation $R$:

2 + 8 + 6 + 3 + 4 + 2 = 25

Thus, the number of elements in $R$ is 25.

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