1.A relation R is defined on the set Z of integers as follows:
(x,y)∈R iff x 2 + y 2 = 25.
Express R and R-1 as the set of odered pairsand hence find the respective domains.
2.Let f : R → R . be defined as { 2x+1, x<= 4 Show that f is not a function
x+1 , x>=4 }
We have,
(x,y) ∈ R ⇔ x2 + y2 = 25 ⇔ y = ±√(25-x2)
We observe that x = 0 ⇒ y = ± 5
∴ (0,5) ∈ R and (0,-5) ∈ R
x = ± 3 ⇒y = ±√(25-9) = ± 4
∴ (3,4) ∈ R, (-3,4) ∈ R, (3,-4) ∈ R and (-3,-4) ∈ R
x = ± 4 ⇒y = ±√(25-16) = ± 3
∴ (4,3) ∈ R, (-4,3) ∈ R, (4,-3) ∈ R and (-4,-3) ∈ R
x = ± 5 ⇒y = ±√(25-25) = 0
∴ (5,0) ∈ R and (-5,0) ∈ R
We also notice that for any other integral value of x,y is not an integer.
∴ R = {(0,5), (0,-5),(3,4),(-3,4),(3,-4),(-3,-4),(4,3),(-4,3),(4,-3),(-4,-3),(5,0),(-5,0)}
and R-1 = {(5,0), (-5,0),(4,3),(4,-3),(-4,3),(-4,-3),(3,4),(3,-4),(-3,4),(-3,-4),(0,5),(0,-5)}
Clearly domain [R] = {0,3,-3,4,-4,5,-5} = domain (R-1)