Question

Write the following relations as the sets of ordered pairs :

(i) A relation R from the set {2, 3, 4, 5, 6} to the set {1, 2, 3} defined by x = 2y.

(ii) A relation R from the set {1, 2, 3, 4, 5, 6, 7} defined by $(x,y)ϵR\iff x$ is relatively prime to y.

(iii) A relation R on the set {0, 1, 2, ...., 10} defined by 2x + 3y = 12.

(iv) A relation R from a set A = {5, 6, 7, 8} to the set B = {10, 12, 15, 16, 18} defined by $(x,y)ϵR\iff x$ divides y.

Answer 1

(i) We have,

x = 2y

Putting y = 1, 2, 3 we get x = 2, 4, 6 respectively,

$\therefore R=\left\{\right(2,1),(4,2),(6,3\left)\right\}$

(ii) We have,

It is given that relation R on the set {1, 2, 3, 4, 5, 6, 7} defined by $(x,y)ϵR\iff x$ is relatively prime to y

$\therefore (2,3)ϵR,(2,5)ϵR,(2,7)ϵR,(3,2)ϵR,(3,4)ϵR,(3,5)ϵR,(3,7)ϵR,(4,3)ϵR,(4,5)ϵR,(4,7)ϵR,(5,2)ϵR,(5,3)ϵR,(5,4)ϵR,(5,6)ϵR,(5,7)ϵR,(6,7)ϵR,(7,2)ϵR,(7,3)ϵR,(7,4)ϵR,(7,5)ϵR,$ and $(7,6)ϵR,$

Thus,

$\left\{\begin{array}{c}(2,3),(2,5),(2,7),(3,2),(3,4),(3,5),(3,7),(4,3),(4,5),(4,7),(5,2),\\ (5,3),(5,4),(5,6),(5,7),(6,5),(6,7),(7,2),(7,3),(7,4),(7,5),(7,6),\end{array}\right\}$

(iii) We have,

2x + 3y = 12

$\Rightarrow 2x=12-3y\Rightarrow x=\frac{12-3y}{2}$

Putting y = 0, 2, 4 we get x = 6, 3, 0 respectively

For $y=1,3,5,6,7,8,9,10,xϵ$ give set

R = {(6, 0), (3, 2), (0, 4)}

= {(0, 4), (3, 2), (6, 0)}

(iv) We have,

A = (5, 6, 7, 8) and B = (10, 12, 15, 16, 18)

Now, $\frac{a}{b}$ stands for 'a divides b'. For the elements of the given set A and B, we find that $\frac{5}{10},\frac{5}{15},\frac{6}{12},\frac{6}{18}$ and $\frac{8}{16}$

$\therefore (5,10)ϵR,(5,15)ϵR,(6,12)ϵR,(6,18)ϵR,$ and $(8,16)ϵR$

Thus,

R = {(5, 10), (5, 15), (6, 2), (6, 18), (8, 16)}