Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Suppose that f : A → B is a function, viewed as a relation (set of ordered pairs). Let f* denote the “reverse" relation: f* {(b, а) € В х A|(а, b) € f} For example, if f : {1,2, 3} → {*, ♡, A} is the function f = {(1, *), (2, ♡), (3, *)}, then f* = {(*, 1), (♡, 2), (*, 3)}. We proved in class that if f is a bijection, then f* is also a bijection (from B back onto A). In general, f* need not even be a function, as we can see from the example above. In fact, f* is a function if and only if f is a bijection, as you will prove below. (a) Prove that if a function f : A → B is not a bijection, then f* is not a function. (To do this, prove that if f is not an injection, then f* is not a function, and also prove that if f is not a surjection, then f* is not a function.) (b) Consider the function f : {1,2,3} function is injective, but not surjective. + {1, 2, 3, 4, 5} defined the f(n) = n + 1. Observe that this Define a surjection g : {1,2, 3, 4, 5} → {1,2,3} such that g Ɔ f*. (In class we proved such a surjection always exists.) (c) Consider the function f : Z → NU{0} defined by f(n) = |n|. Observe that this function is surjective, but not injective. Define an injection g : N → Z such that g C f*. (In class we proved such an injection always exists.)

Suppose that f : A → B is a function, viewed as a relation (set of ordered pairs). Let f* denote the “reverse" relation: f* {(b, а) € В х A|(а, b) € f} For example, if f : {1,2, 3} → {*, ♡, A} is the function f = {(1, *), (2, ♡), (3, *)}, then f* = {(*, 1), (♡, 2), (*, 3)}. We proved in class that if f is a bijection, then f* is also a bijection (from B back onto A). In general, f* need not even be a function, as we can see from the example above. In fact, f* is a function if and only if f is a bijection, as you will prove below. (a) Prove that if a function f : A → B is not a bijection, then f* is not a function. (To do this, prove that if f is not an injection, then f* is not a function, and also prove that if f is not a surjection, then f* is not a function.) (b) Consider the function f : {1,2,3} function is injective, but not surjective. + {1, 2, 3, 4, 5} defined the f(n) = n + 1. Observe that this Define a surjection g : {1,2, 3, 4, 5} → {1,2,3} such that g Ɔ f*. (In class we proved such a surjection always exists.) (c) Consider the function f : Z → NU{0} defined by f(n) = |n|. Observe that this function is surjective, but not injective. Define an injection g : N → Z such that g C f*. (In class we proved such an injection always exists.)

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:HOUGHTON MIFFLIN HARCOURT
Chapter4: Writing Linear Equations
Section: Chapter Questions
Problem 7CA
See similar textbooks
icon
Related questions
Question
1
Suppose that f : A → B is a function, viewed as a relation (set of ordered pairs). Let f* denote the “reverse" relation: f* {(b, а) € В х A|(а, b) € f} For example, if f : {1,2, 3} → {*, ♡, A} is the function f = {(1, *), (2, ♡), (3, +)}, %3| then f* = {(*, 1), (♡, 2), (*, 3)}. We proved in class that if f is a bijection, then f* is also a bijection (from B back onto A). In general, f* need not even be a function, as we can see from the example above. In fact, f* is a function if and only if f is a bijection, as you will prove below. (a) Prove that if a function f : A → B is not a bijection, then f* is not a function. (To do this, prove that if f is not an injection, then f* is not a function, and also prove that if f is not a surjection, then f* is not a function.) (b) Consider the function f : {1,2,3} function is injective, but not surjective. + {1, 2, 3, 4, 5} defined the f(n) = n + 1. Observe that this Define a surjection g : {1,2,3, 4, 5} → {1,2,3} such that g 2 f*. (In class we proved such a surjection always exists.) (c) Consider the function f : Z → NU {0} defined by f(n) = |n|. Observe that this function is surjective, but not injective. Define an injection g : N → Z such that g C f*. (In class we proved such an injection always exists.)
Transcribed Image Text:Suppose that f : A → B is a function, viewed as a relation (set of ordered pairs). Let f* denote the “reverse" relation: f* {(b, а) € В х A|(а, b) € f} For example, if f : {1,2, 3} → {*, ♡, A} is the function f = {(1, *), (2, ♡), (3, +)}, %3| then f* = {(*, 1), (♡, 2), (*, 3)}. We proved in class that if f is a bijection, then f* is also a bijection (from B back onto A). In general, f* need not even be a function, as we can see from the example above. In fact, f* is a function if and only if f is a bijection, as you will prove below. (a) Prove that if a function f : A → B is not a bijection, then f* is not a function. (To do this, prove that if f is not an injection, then f* is not a function, and also prove that if f is not a surjection, then f* is not a function.) (b) Consider the function f : {1,2,3} function is injective, but not surjective. + {1, 2, 3, 4, 5} defined the f(n) = n + 1. Observe that this Define a surjection g : {1,2,3, 4, 5} → {1,2,3} such that g 2 f*. (In class we proved such a surjection always exists.) (c) Consider the function f : Z → NU {0} defined by f(n) = |n|. Observe that this function is surjective, but not injective. Define an injection g : N → Z such that g C f*. (In class we proved such an injection always exists.)
Expert Solution
steps

Step by step

Solved in 3 steps with 5 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Big Ideas Math A Bridge To Success Algebra 1: Stu…
Big Ideas Math A Bridge To Success Algebra 1: Stu…
Algebra
ISBN:
9781680331141
Author:
HOUGHTON MIFFLIN HARCOURT
Publisher:
Houghton Mifflin Harcourt
Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
Algebra
ISBN:
9780395977224
Author:
Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:
McDougal Littell
Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
Geometry
ISBN:
9781337614085
Author:
Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:
Cengage,
College Algebra
College Algebra
Algebra
ISBN:
9781305115545
Author:
James Stewart, Lothar Redlin, Saleem Watson
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Algebra for College Students
Algebra for College Students
Algebra
ISBN:
9781285195780
Author:
Jerome E. Kaufmann, Karen L. Schwitters
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS