Functions: definition of a function

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Daniel Mansfield

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Daniel Mansfield 7 years, 9 months ago

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Daniel Mansfield 7 years, 9 months ago

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Daniel Mansfield 7 years, 9 months ago

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Name	Status	Author	Last Modified
Functions: definition of a function	Ready to use	Daniel Mansfield	02/06/2019 06:57
Simon's copy of Functions: definition of a function	draft	Simon Thomas	07/03/2019 15:45
Functions: definition of a function	draft	Xiaodan Leng	10/07/2019 21:41

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Question statement

For two sets $A,B$ the Cartesian product is

$A \times B = \left\{(a,b) | a \in A, b \in B\right\}.$

A function $f$ from $A$ to $B$ , written $f : A \mapsto B$ , is a subset of $A\times B$ subject to the condition that for all $a\in A$ there is exactly one $b\in B$ such that $(a,b) \in f$ .

In other words, for every $a$ the function must determine a unique output $b$ . Since $a$ determines $b$ we usually indicate this with the function notation $f(a) = b$ .

For example, consider the sets $A = \left\{\var{a0},\var{a1},\var{b1},\var{a2}\right\}$ and $B=\left\{\var{b0},\var{b1},\var{b2}\right\}$ and $f : A \mapsto B$ defined by

$f = \left\{ (\var{a0},\var{b0}),(\var{a1},\var{b1}),(\var{a2},\var{b0}),(\var{a0},\var{b1})\right\}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a0	integer	2
a1	integer	7
a2	integer	4
b0	integer	1
b1	integer	6
b2	integer	9

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Generated value: `integer`

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$(\var{a0}, \var{b0}) \in f$

Marks: Distractor Message:
$(\var{b0}, \var{a0}) \in f$

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$f(\var{a2}) = \var{b0}$

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$f(\var{b0}) = \var{a2}$

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$(\var{a1}, \var{b1}) \in f$

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$f(\var{a2}) = \var{b1}$

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Advice

This question exhibits the different ways to show that $f$ is not a function.

Part a introduces the notion of a function as a subset of the Cartesian product, and is designed to get you used to the syntax.

For part b, there are two values of $\var{b0},\var{b1} \in B$ which are equal to $f(\var{a0})$ . This means that $f(\var{a0})$ is not uniquely determined, and so $f$ can not be a function.

For part c, we see that $\var{b1} \in A$ but there is no function value for $f(\var{b1})$ . This means that $f(a)$ is not uniquely determined for every $a \in A$ and so $f$ can not be a function.

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This question is used in the following exams:

Week 3: functions by Daniel Mansfield in Discrete Mathematics.
Week 1: Basic Set Theory by Daniel Mansfield in Discrete Mathematics.
Week 1: Basic Set Theory - V2 by Sean Gardiner in Discrete Mathematics.
Shivendra's copy of Week 1: Basic Set Theory - V2 by Deactivated user in Deactivated user's workspace.
Shaheen's copy of Week 1: Basic Set Theory - V2 by Shaheen Charlwood in Shaheen's workspace.

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Daniel Mansfield 7 years, 9 months ago

Daniel Mansfield 7 years, 9 months ago

Daniel Mansfield 7 years, 9 months ago

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Functions: definition of a function

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Taxonomy: mathcentre

Taxonomy: Kind of activity

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Contributors

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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

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Generated value: integer

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Generated value: `integer`