Let R1,2,3,41,2,3,4 be a relation defined on the set A=1,2,3,4 by [R = ((a,b),(c,d)) : 2a+…

Question

Let $R\subseteq\{1,2,3,4\}\times\{1,2,3,4\}$ be a relation defined on the set $A=\{1,2,3,4\}$ by
$$R = \{((a,b),(c,d)) : 2a+3b = 3c+4d\}.$$
Then the number of elements in $R$ is:

Accepted Answer

Let $L_1 = 2a+3b$ and $L_2 = 3c+4d$ for $a,b,c,d\in\{1,2,3,4\}$.

Values of $L_1$ (over all $(a,b)$) with multiplicities: $5,8,11,14$ (from $a=1$); $7,10,13,16$ (from $a=2$); $9,12,15,18$ (from $a=3$); $11,14,17,20$ (from $a=4$). So $11$ and $14$ each occur twice; all others occur once.

Values of $L_2$ (over all $(c,d)$, same functional form as $3c+4d$): $7,11,15,19,10,14,18,22,13,17,21,25,16,20,24,28$ --- each value occurs once.

Common values between $L_1$ and $L_2$: $7,10,11,13,14,15,16,17,18,20$.

For each common value $v$, the number of matching pairs is $(	ext{freq of } v 	ext{ in } L_1)	imes(	ext{freq of } v 	ext{ in } L_2)$:

- $v=7,10,13,15,16,17,18,20$: each contributes $1	imes1=1$ 	o  subtotal $8	imes1=8$
- $v=11$: $2	imes1=2$
- $v=14$: $2	imes1=2$

Total: $8+2+2 = 12$

Solutions · 1