   Chapter 7.4, Problem 6ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Use the functions I and J defined in the paragraph following Example 7.4.1 to show that even though following Example 7.4.1 to show that even though there is a one-to-one correspondence, H, from 2Z to Z, there is also a function from 2Z to Z that is one-to-one but not and a function from Z to show I is one-to-one. In other words, show that I is one-to-one but not onto, and show that J is onto but not one-to-one.

To determine

To prove:

There is a one-to-one correspondence, H from 2 to  and also show that I is one-to-one but not onto and

J is onto but not one-to-one.

Explanation

Given information:

and 2

Concept used:

A function is said to be one-to-one function if the distinct elements in domain must be mapped with distinct elements in codomain.

A function is said to be onto function if each element in codomain is mapped with at least one element in domain.

Proof:

Define the function I:2 by I(n)=n even integers.

The function J:2 is defined as J(n)=2[n2]n.

The objective is to show that I is one-to-one but not onto and J is onto but not one-to-one.

Define a function H:2 by H(n)=2n, for all n.

First, show that the function H is one-to-one and onto.

To verify that H is one-to-one:

Let n1,n2 such that.

H(n1)=H(n2)2n1=2n2n1=n2       (Divide both sides by 2)

Therefore, H is one-to-one.

To verify that H is onto:

Let p be any element in 2.

Then p is an even integer.

Thus

p=2k, for some k.

Therefore H(k)=2k=p, for some p2.

Thus, for every p2 there exists k in  such that H(k)=p.

Therefore, H is onto.

Hence, there is one-to-one correspondence from

to 2.

Now show that I is one-to-one but not onto.

To verify that I is one-to-one:

Let n1,n2 such that

I(n1)=I(n2)n1=n2

Therefore, I is one-to-one

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