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The domain of relation $D$ is the set of positive integers. For $x, y \in \mathbb Z^+$, $xDy$ if $x$ evenly divides $y.$

I do know this: that a positive integer $x$ evenly divides positive integer $y$ if and only if there is another positive integer $n$ such that $y = xn$.

I'm pretty sure I need to use this definition to prove whether or not relation D is symmetric, anti-symmetric, or neither?

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    $\begingroup$ What I've learned is that "A relation D on set A is symmetric if an arrow from x to y implies that there is an arrow from y to x." & "A relation D on set A is anti-symmetric if there are no pairs x and y (with x ≠≠ y) in which x and y point to each other." But I am not sure how to apply that idea to this problem. $\endgroup$ – Jigar Patel Jan 11 '18 at 5:27
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    $\begingroup$ If $x$ divides $y$, when does $y$ divide $x$? $\endgroup$ – Fabio Somenzi Jan 11 '18 at 5:29
  • $\begingroup$ @JigarPatel: Please edit your question to include what you provided in your comment, since it belongs in the question not a comment. Thanks! $\endgroup$ – user21820 Feb 25 '18 at 9:54
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Preliminaries: Just to be clear, when I use, e.g. $x \mid y$, that means $x$ divides $y$, (or, alternatively, that $y$ is divisible by $x$

Let's look at symmetry.

Let's pick even integers $x = 2, y=4$, just to test one case. Then $x\mid y,$ (because $2$ divides $4$, since $2\cdot 2 = 4$), but $y$ does not divide $x$ (because there is no positive integer $k$ such that $4k = 2$).

All we need is one counterexample to prove that the relation is not symmetric, because a symmetric relation requires that for all $x, y,\;\;$ if $\;x\mid y,\;\;$ then $\;\;y\mid x$.

We see that doesn't hold $x = 2, y = 4.$ Therefore, it doesn't hold for all $x, y$ such that $x\mid y$. Hence the relation, as noted, is not symmetric.


Let's look at antisymmetry.

Now there are some cases in which $x$ divides $y$, and also $y$ divides $x$. When does that happen?

This relation is antisymmetric if, for all $x, y$, whenever it happens that $x$ divides $y$ AND also $y$ divides $x$, then it must be the case that $x = y$.

So let's suppose it happens that $x$ divides $y$, and $y$ divides $x$.

Then by definition, $y=xn$ and $x = ym$, where $n, m$ are positive integers.

We can substitute $\;y = xn\;$ into the equation $x= ym = (xn)m = nmx.$ Clearly, if $x = xnm,$ then $nm = 1$. And the only way that two positive integers, when multiplied, can equal one is if they are both equal to $1$, i.e. $n=m=1$. In short, we have that $y=x$, and $x=y$. Hence, when $x\mid y$ and $y\mid x$, it follows that $x=y$.

Hence the relation D is, in fact, antisymmetric.

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