Prove/disprove, that the relation is reflexive, symmetric, antisymmetric and transitive

Question

I want to prove or disprove, if the relation $R$ with \begin{align} iRj:\Longleftrightarrow (\forall k \in \mathbb{N} \text{ with } k \text{ is a prime number}:k \mid i \Longrightarrow k \mid j) \end{align} is reflexive, symetric, asymetrical, antisymetric and transitive.

Reflexive: $\forall x\in R: xRx$ \begin{align} \Longleftrightarrow k \mid x \Longrightarrow k \mid x \text{ (trivial, if k devides x then k devides x.)} \end{align}

Symetric: $\forall x,y \in R: xRy \Longrightarrow yRx$

EDIT:

Actually your relation is not symmetric: For example you have $3R6$ because the only prime dividing $3$ is $3$, and $3$ also divides $6$. On the other hand you do not have $6R3$ because for $k=2$ you get $2\mid6$ but $2\nmid3$. (contributed by araomis)

THIS IS WRONG! \begin{align} \Longleftrightarrow (k \mid x \Rightarrow k \mid y) \Longrightarrow (k \mid y \Rightarrow k \mid x) \end{align} I'm not really sure how I should prove this relation. I know that I should probably use the fact that \begin{align} a\mid b: \text{ ('}a\text{'} \text{ devides } \text{'}b\text{'}) \Longleftrightarrow c\cdot a=b \end{align} but I can't really figure it out:

$ \big((\underbrace{k\cdot a=x \Rightarrow k\cdot b =y}_{\rightarrow a=\dfrac{x}{k}; b=\dfrac{y}{k}}) \Longrightarrow (\underbrace{k\cdot b =y \Rightarrow k\cdot a =x}_{*})\big) $
$a,b$ in $(*)$: \begin{align} (k\mid y \Rightarrow k\mid x)\Longleftrightarrow \left(k\cdot \dfrac{y}{k}=y\Rightarrow k\cdot \dfrac{x}{k}=x\right) \Longleftrightarrow (y=y \Rightarrow x=x) \Longleftrightarrow \left(x=x\Rightarrow y=y\right) \Longleftrightarrow \left(k\cdot \dfrac{x}{k}=x \Rightarrow k\cdot \dfrac{y}{k}=y\right)\Longleftrightarrow \left(k\cdot a= x \Rightarrow k\cdot b=y\right) \Longleftrightarrow (k\mid x \Rightarrow k\mid y) \rightarrow (k \mid x \Rightarrow k \mid y) \Longrightarrow (k\mid y \Rightarrow k \mid x) \end{align} Is this enough to prove symmetry?

I'm not really sure about the rest, because that is probably similar to the last one.

Answer 1

The relation is reflexive (indeed evident).

The relation is not symmetric: $1R3$ and not $3R1$

Also it is not asymmetric: $3R9$ and $9R3$

Also it is not antisymmetric: $3R9$ and $9R3$ but $3\neq9$

The relation is transitive. If $iRj$ and $jRm$ and $k\in\mathbb N$ is prime with $k\mid i$ then $k\mid j$ (because $iRj$) and then also $k\mid m$ (because $jRm$).

Answer 2

Actually your relation is not symmetric: For example you have $3 R 6$ because the only prime dividing 3 is 3, and 3 also divides 6. On the other hand you do not have $6 R 3$ because for $k = 2$ you get $2 | 6$ but $2 \nmid 3$.