Is a dividing relation on the natural numbers an symmetric/antisymmetric relation?

Question

Let $R = \{(x,y) \in (\mathbb{N}\times\mathbb{N}): x \mid 2y\}$. I.e. $R$ is the set of all pairs $(x,y)$ of natural numbers (excluding $0$) such that $x$ divides $2y$.

Is such a relation antisymmetric? Is such a relation symmetric? I can't even find $x$, $y$ such that $(x,y)\in R\land (y,x)\in R$.

If there really were no cases of $x \not = y$ such that $(x,y)\in R\land (y,x)\in R$, then the relation $R$ would be antisymmetric — Henry, Nov 18 '16 at 15:01

score 2 · Accepted Answer · edited Dec 14 '19 at 08:38

2

Hint: $1 \mid (2 \cdot 2)$ and $2 \mid (2 \cdot 1)$

edited Dec 14 '19 at 08:38

Martin Sleziak

53,687

answered Nov 18 '16 at 14:36

Dominik

19,963

score 0 · Answer 2 · edited Dec 14 '19 at 08:38

0

Consider $x=2$ and $y=4.$ We have that $x\mid 2y$ and $y\mid 2x.$ But $x\ne y.$

edited Dec 14 '19 at 08:38

Martin Sleziak

53,687

answered Nov 18 '16 at 14:37

mfl

29,399

Is a dividing relation on the natural numbers an symmetric/antisymmetric relation?

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