Answered: Find a function whose domain is the set…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter1: Fundamentals

Section1.4: Binary Operations

Problem 11E

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Find a function whose domain is the set of all integers and whose target is the set of all positive integers that satisfies each set of properties.

(b)
One-to-one, but not onto.
(c)
Onto, but not one-to-one.
(d)
One-to-one and onto.

Expert Solution

Step 1

Step:-1

Set of integers = $ℤ = \{. . . . . . . . - 3, - 2, - 1, 0, 1, 2, 3, . . . . . . . .\}$ ,

set of positive integers= $ℕ = \{1, 2, 3, 4, . . . . . . .\}$

Note:- 0 is not included in set of all positive integers.

Part (b):- One-to-one but not onto.

Let f be a function $f : ℤ \to ℕ$ such that

$f (n) = \{\begin{cases} 3 n + 1 & n \geq 0 \\ 3 n^{2} & n < 0 \end{cases}$

This is one-to-one function because

$f (a) = f (b) \Rightarrow a = b; a, b \in ℤ$

but not onto because there is no $n \in ℤ$ such that f(n)=2, 8, 14,.... that is, 2, 8, 14,... do not have a pre-image.

Step:-2

Part (c):- onto but not one-to-one.

Let f be a function $f : ℤ \to ℕ$ such that

$f (n) = |n|$ +1

This is not one-to-one function because

$f (a) = f (b) does not imply a = b; a, b \in ℤ$ as

$f (- 1) = f (1) = 2 f (- 2) = f (2) = 3$

and so on.

but onto because every positive integer in $ℕ$ have a pre-image in $ℤ$ .

Step:-3

Part (d):- one-to-one and onto.

Let f be a function $f : ℤ \to ℕ$ such that

$f (n) = \{\begin{cases} 2 n + 1 & n \geq 0 \\ - 2 n & n < 0 \end{cases}$

This is one-to-one function because

$f (a) = f (b) \Rightarrow a = b; a, b \in ℤ$

and it is onto because every positive integer in $ℕ$ have a pre-image in $ℤ$ .

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