Explanation Given information: ℤ and 2 ℤ 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 ℤ → ℤ by I ( n ) = n ∀ even integers. The function J : ℤ → 2 ℤ is defined as J ( n ) = 2 [ n 2 ] ∀ n ∈ ℤ . 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 : ℤ → 2 ℤ by H ( n ) = 2 n , for all n ∈ ℤ . First, show that the function H is one-to-one and onto. To verify that H is one-to-one: Let n 1 , n 2 ∈ ℤ such that. H ( n 1 ) = H ( n 2 ) 2 n 1 = 2 n 2 n 1 = n 2 ( Divide both sides by 2 ) Therefore, H is one-to-one. To verify that H is onto: Let p be any element in 2 ℤ . Then p is an even integer. Thus p = 2 k , for some k ∈ ℤ . Therefore H ( k ) = 2 k = p , for some p ∈ 2 ℤ . Thus, for every p ∈ 2 ℤ there exists k in ℤ such that H ( k ) = p . Therefore, H is onto. Hence, there is one-to-one correspondence from ℤ to 2 ℤ . Now show that I is one-to-one but not onto. To verify that I is one-to-one: Let n 1 , n 2 ∈ ℤ such that I ( n 1 ) = I ( n 2 ) n 1 = n 2 Therefore, I is one-to-one

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Author: EPP

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Chapter 7.4, Problem 6ES

To determine

To prove:

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

J is onto but not one-to-one.

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Chapter 7.4, Problem 5ES

Chapter 7.4, Problem 7ES

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Chapter 7 Solutions

Discrete Mathematics With Applications

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To prove: There is a one-to-one correspondence, H from 2 ℤ to ℤ and also show that I is one-to-one but not onto and J is onto but not one-to-one.

Solution Summary: The author explains how to prove that there is a one-to-one correspondence, H from 2mathbbZ

Videos

To prove:

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

J is onto but not one-to-one.

Want to see the full answer?

Chapter 7 Solutions

To prove: There is a one-to-one correspondence, H from 2 ℤ to ℤ and also show that I is one-to-one but not onto and J is onto but not one-to-one.

Solution Summary: The author explains how to prove that there is a one-to-one correspondence, H from 2mathbbZ

Videos

To prove: There is a one-to-one correspondence, H from 2ℤ to ℤ and also show that I is one-to-one but not onto and J is onto but not one-to-one.

Want to see the full answer?

Chapter 7 Solutions

To prove:

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

J is onto but not one-to-one.