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A useful thing to note is that $Y \\subseteq X$. We can prove this by taking a generic element of $Y$ and proving that it also belongs to $X$.

If $y \\in Y$ then $y$ has the shape $y = \\var{c}k + \\var{d}$ for some $k \\in \\mathbb Z$. We need to show that it also has the shape $\\var{a}n + \\var{b}$ for some integer $n$ and so belongs to $X$ as well.

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

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

(a)

So for this part of this question you can choose any element of $Y$. All of them will automatically be included in $X$. There are infinitely many possible answers!

For example we can choose $k=1 \\in \\mathbb Z$. Then $y = \\var{c} \\times 1 + \\var{d} = \\var{c+d} \\in Y$. Since $X \\subseteq Y$ then $\\var{c+d} \\in X$ as well.

If we wanted to check that $\\var{c+d} \\in X$, we simply have to show that $\\var{c+d} = \\var{a}n + \\var{b}$, where $n$ is an integer.

Well we can solve this to give $n= \\frac{\\var{c+d}-\\var{b}}{\\var{a}} =\\var{(c+d-b)/a}$ which is an integer as required.

(b)

For the second question you just need to find an element of $X$ which which is not an element of $Y$. The easiest way to procede is to just keep trying different values of $n \\in \\mathbb Z$, calculate $\\var{a}n + \\var{b}$, and then check whether this value $\\in Y$.

There are infinitely many possible answers!

e.g. We could try $n=0$ which gives $x=\\var{a}n + \\var{b}=\\var{a} \\times 0 + \\var{b} = \\var{b} \\in X $

Now to check whether $\\var{b} \\in Y $, we write $\\var{b}=\\var{c}k + \\var{d}$, which we can solve to give $k = \\frac{\\var{b}-\\var{d}}{\\var{c}}=\\var{(b-d)/c}$ which is not an integer.

Hence $\\var{b} \\notin Y $ as required.

(c)

By definition the sets $X$ and $Y$ are equal when both $Y\\subseteq X$ and $X \\subseteq Y$ are true.

We showed in the first question that the relation $Y\\subseteq X$ is true.

However we showed in the second question that $X \\nsubseteq Y$, which means that the sets cannot be equal.

Since the sets are not equal, instead we could say that $X$ 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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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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The sets $X$ and $Y$ are not equal because

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$X \\nsubseteq Y$

", "

$Y \\subseteq X$

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