Question

Write the following relations as sets of ordered pairs and find which of them are functions :
(i) $\left\{\right(x,y):y=3x,xϵ(1,2,3),yϵ(3,6,9,12\left)\right\}$
(ii) $\left\{\right(x,y):y>x+1,x=1,2\}$ and y = 2, 4, 6
(iii) $\left\{\right(x,y):x+y=3,x,yϵ\{0,1,2,3\left\}\right\}$

Answer 1

(i) We have,
$\left\{\right(x,y)=y=3x,xϵ(1,2,3),yϵ(3,6,9,12\left)\right\}$
Putting x = 1, 2, 3 in y = 3x, we get
y = 3, 6, 9 respectively
$\therefore$ R = {(1, 3), (2, 6), (3, 9)}
Yes, it is a function.

(ii) We have,
$(x,y):y>x+1,x=1,2$ and y = {2, 4, 6}
Putting x = 1, 2 in y > x + 1, we get
y > 2, y > 3 respectively.
$\therefore R=\left\{\right(1,4),(1,6),(2,4),(2,6\left)\right\}$
It is not a function from A to B because two ordered pairs in R have the same first element.

(iii) $\left\{\right(x,y):x+y=3,x,yϵ\{0,1,2,3\left\}\right\}$
We have,
$\left\{\right(x,y):x+y=3,x,yϵ\{0,1,2,3\left\}\right\}$
Now,
$y=3-x$
Putting x = 0, 1, 2, 3 we get
y = 3, 2, 1, 0 respectively
R = {(0, 3), (1, 2), (2, 1), (3, 0)}
Yes, this relation is a function.