Page 9 - Relations and Functions 19.10.06.pmd

RELATIONS AND FUNCTIONS 9

Example 9 Prove that the function f : R → R, given by f (x) = 2x, is one-one and onto.

Solution f is one-one, as f (x ) = f (x ) ⇒ 2x = 2x ⇒ x = x . Also, given any real
1 2 1 2 1 2
y y y
number y in R, there exists in R such that f ( ) = 2 . ( ) = y. Hence, f is onto.
2 2 2

Fig 1.3

Example 10 Show that the function f : N → N, given by f (1) = f (2) = 1 and f(x) = x – 1,
for every x > 2, is onto but not one-one.
Solution f is not one-one, as f (1) = f (2) = 1. But f is onto, as given any y ∈ N, y ≠ 1,
we can choose x as y + 1 such that f (y + 1) = y + 1 – 1 = y. Also for 1 ∈ N, we
have f (1) = 1.
Example 11 Show that the function f : R → R,
2
defined as f (x) = x , is neither one-one nor onto.
Solution Since f (– 1) = 1 = f (1), f is not one-
one. Also, the element – 2 in the co-domain R is
not image of any element x in the domain R
(Why?). Therefore f is not onto.

Example 12 Show that f : N → N, given by
⎧ x + 1,if x is odd,
fx
() = ⎨
⎩ x − 1,if x is even
is both one-one and onto. Fig 1.4