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

Question

Let $R$ be a relation defined on the set ${1,2,3,4}	imes {1,2,3,4}$ by

$R={((a,b),(c,d)):2a+3b=3c+4d}.$

Then the number of elements in R is

Accepted Answer

let $A={1,2,3,4}$

$R$ is defined on $A	imes A$ such that

$\begin{gathered} R={((a,b),(c,d)):2a+3b=3c+4d} \ a,b,c,d\in A \end{gathered}$

$2a+3b=L13c+4d=L2$

$\begin{gathered} a \quad  b \quad  2a+3b \quad  3c+4d \ 1 \quad  1 \quad  5 \quad  7 \ 1 \quad  2 \quad  8 \quad  11 \ 1 \quad  3 \quad  11 \quad  15 \ 1 \quad  4 \quad  14 \quad  19 \ 2 \quad  1 \quad  7 \quad  10 \ 2 \quad  2 \quad  10 \quad  14 \ 2 \quad  3 \quad  13 \quad  18 \ 2 \quad  4 \quad  16 \quad  22 \ 3 \quad  1 \quad  9 \quad  13 \ 3 \quad  2 \quad  12 \quad  17 \ 3 \quad  3 \quad  15 \quad  21 \ 3 \quad  4 \quad  18 \quad  25 \ 4 \quad  1 \quad  11 \quad  16 \ 4 \quad  2 \quad  14 \quad  20 \ 4 \quad  3 \quad  17 \quad  24 \ 4 \quad  4 \quad  20 \quad  28 \end{gathered}$

$\begin{gathered} 2a+3b=3c+4d \ 7,10,11,13,14,15,16,17,18,20 \end{gathered}$

value $7,10,13,15,16,17,18,20$

occurs 1 time in L1 and 1 time in L2

e.g value 7 occurs 1 time in L1 $\&$ 1 time in L2 total $=1	imes 1=1$

so, total number of these types of elements

in $R=8$--- (1)

value 11 and 14 occurs 2 times in L1

and 1 time in L2

e.g value 11 occurs 2 time in L1 $\&$ 1 time in L2 total

elements in $R$ is $2	imes 1=2$

so total number these type of elements $=2+2=4$--- (2)

total number of elements in R = 8+4 = 12

Solutions · 1