Functions: bijective

Error

There was an error loading the page.

Contributors

Daniel Mansfield

Feedback

From users who are members of Discrete Mathematics :

Daniel Mansfield

said

Ready to use

7 years, 9 months ago

History

○

Daniel Mansfield 7 years, 9 months ago

Gave some feedback: Ready to use

○

Daniel Mansfield 7 years, 9 months ago

Published this.

○

Daniel Mansfield 7 years, 9 months ago

Created this.

Name	Status	Author	Last Modified
Functions: bijective	Ready to use	Daniel Mansfield	22/01/2019 02:37
Lois's copy of Functions: bijective	draft	Lois Rollings	19/08/2020 12:46
Gareth's copy of Lois's copy of Functions: bijective	draft	Gareth Woods	31/10/2017 08:07

There are 4 other versions that do you not have access to.

Question statement

A function which is both surjective and injective is said to be bijective.

This is an important property because it guarantees that $f: A \mapsto B$ has an inverse function $f^{-1} : B \mapsto A$ .

Let's prove that the function $f: \mathbb R \mapsto \mathbb R$ defined by

$f(x) = \left\{ \begin{array}{cl} \frac{1}x & x \neq 0 \\ 0 & x = 0.\end{array}\right.$

has an inverse.

Give any introductory information the student needs.

			Name	Type	Generated Value

Automatically regenerate variables?

No variables have been defined in this question.

Parts

Part a) Gap-fill
1. Gap Mathematical expression
  Add
Part b) Gap-fill
1. Gap Number entry
  Add
Part c) Gap-fill
1. Gap Mathematical expression
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Prove that $f(x)$ surjective.

Assume that $b \in \mathbb R$. We must find some $a$ such that $f(a) = b$. There are two cases to consider.

If $b=0$ then choose $a = 0$.

Otherwise, $b \neq 0$ and then choose $a = $ Unnamed gap.

In either case, $f(a) = b$ and so the function is surjective.

Advice

This question summarizes the definitions of surjective and injective, and applies them to prove the existance of an inverse.

For continuous functions it is possible to prove that it injective by using the derivative. But $f$ is not continuous, and nor is it strictly increasing or strictly decreasing. In fact it is strictly decreasing for $x<0$ and strictly increasing for $x>0$.

{myscript()}

So we pretty much have to prove that $f$ is injective directly from the definition.

Since $f$ is a bijection we know that an inverse exists, but this information does not tell us how to find it. This example is special because the inverse is easy to find.

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Mathematical expression
Mathematical expression		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Mathematical expression
Mathematical expression		Hasn't run yet

This question is used in the following exams:

Loading...

Error

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Daniel Mansfield 7 years, 9 months ago

Daniel Mansfield 7 years, 9 months ago

Daniel Mansfield 7 years, 9 months ago

Error in variable testing condition

Parts

Gap-fill