Elements in sets, subsets

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Lois Rollings

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Lois Rollings 7 years, 6 months ago

Created this as a copy of Basic Set Theory: element not in a set.

Name	Status	Author	Last Modified
Basic Set Theory: element not in a set	Ready to use	Daniel Mansfield	23/09/2019 01:18
Elements in sets, subsets	draft	Lois Rollings	02/10/2017 12:11
Simon's copy of Basic Set Theory: element not in a set	draft	Simon Thomas	07/03/2019 14:20
Simon's copy of Simon's copy of Basic Set Theory: element not in a set	draft	Simon Thomas	07/03/2019 14:55

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Question statement

Consider the sets $X = \left\{ \var{a}n + \var{b} | n \in \mathbb Z\right\}$ and $Y = \left\{ \var{c}k + \var{d} | k \in \mathbb Z\right\}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
b	integer	8
c	integer	12
d	integer	11
u	integer	4

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a) In fact $Y \subseteq X$ , so that any element of $Y$ will also be an element of $X$ .

b) Here you need to find an element of $X$ which which is not an element of $Y$ . This does not require an infinite amount of work - all you need is one element of $X$ which is not in $Y$ . The easiest way to proceede is to just try a few different values of $X$ .

c) $Y \subseteq X$ is true. The question does not ask you to prove this, but this is how you would proceed if it did.

You need to prove that for every $y \in Y$ , $y \in X$ .

$Y$ is an infinite set so you cannot check each element individually but you can take a general element of $Y$ and prove that it also belongs to $X$ .

If $y \in Y$ then $y$ has the shape $y = \var{c}k + \var{d}$ for some $k \in \mathbb Z$ . So

$\begin{align*} y & = \var{c} k + \var{d} \\ & = \var{a}\times\var{u} k + \var{a} + \var{b} \\ & = \var{a}(\var{u} k + 1) + \var{b} \\ & = \var{a}n + \var{b}. \end{align*}$

where $n = (\var{u}k+1) \in \mathbb Z$ . Because $y = \var{a}n + \var{b}$ for some integer $n$ , it is also an element of $X$ .

$X \nsubseteq Y$ is also true. You have already shown this in part b, by finding an element of $X$ that is not in $Y$ .

Note that these results mean that the sets are not equal and that $X$ is a proper subset of $Y$ : $X \subset Y$ .

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This question is used in the following exam:

Set Notation Practice by Lois Rollings in Algebra 1 - Possibilities.

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Lois Rollings 7 years, 6 months ago

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Elements in sets, subsets

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Lois Rollings 7 years, 6 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Generated value: `integer`