// Numbas version: exam_results_page_options {"name": "Basic Set Theory: element not in a set", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"extensions": [], "metadata": {"description": "

Introductory exercise about subsets using custom grading code.

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Find an element of $Y$ which is also an element of $X$.

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Find an element of $X$ which is not an element of $Y$.

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$X \\nsubseteq Y$

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$Y \\subseteq X$

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The sets $X$ and $Y$ are not equal because

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The first thing to note is that $Y \\subseteq X$. To prove this, show that every $y \\in Y$ is also in $X$. Essentially you have to prove an infinite amount of stuff here (because $Y$ is infinitely large), and we can do this by considering any generic element of $Y$ and checking that it is also in $X$.

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Suppose $y$ is any generic element of $Y$. Then it has the shape $y = \\var{c}k + \\var{d}$ for some $k \\in \\mathbb Z$. But elements of $X$ have the shape $\\var{a}n + \\var{b}$ for some integer $n$. We can $\\var{c}k + \\var{d}$ into $\\var{a}n + \\var{b}$ as follows:

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

\n

where $n = (\\var{u}k+1) \\in \\mathbb Z$. So $y=\\var{a}n + \\var{b}$ for some integer $n$, which means that any element of $Y$ is also an element of $X$, i.e. $Y \\subseteq  X$.

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So for this part of this question you can choose any element of $Y$. All of them will automatically be included in $X$.

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For the second question 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 $X$.

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By definition the sets $X$ and $Y$ are equal when both $Y\\subseteq X$ and $X \\subseteq Y$ are true. We showed in part b that $X \\nsubseteq Y$. So the sets are not equal. We can be more precise and see from part a that $Y\\subseteq X$ and so $X$ is a proper subset of $Y$: $X \\subset Y$.

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

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