Discrete Mathematical Structures by Pearson

Functions 3-9

Now, f

y +1

2

1

3





























= 2

y +













1

2

- 1 = y + 1 - 1 = y, showing that each element y ∈ R (range

set) has the pre-image in the domain set R, and hence, the function is ONTO.

Example 4 Prove that the function f: C → R defined by f (z) = |z| is neither one-one nor ONTO.

Solution: Let z = x + iy be any complex number where x, y ∈ R. Then |z| = |x + iy| =

xy

22

+

,

which is a non-negative real number. Thus, |z| ≥ 0.

The function is not one-one as

f (1 + i) = f (-1 + i) = f (1 - i) = f (-1 - i) = √2

which shows that many elements in C are mapped to a single element in R.

The function is not ONTO as all the negative real numbers of co-domain ...