Question:
The relation R on {1,2,3,....} where aRb means a|b.
Is it reflexive, is it symmetric, is it antisymmetric, is it transitive?
obviously, for all $\displaystyle a \in \mathbb{N}$ we have $\displaystyle a|a$, thus the relation is reflexive
for $\displaystyle a,b \in \mathbb{N}$, if $\displaystyle a|b$ does it mean that $\displaystyle b|a$?is it symmetric
for $\displaystyle a,b \in \mathbb{N}$, if $\displaystyle a|b$ and $\displaystyle b|a$ does it mean that $\displaystyle a = b$?is it antisymmetric
for $\displaystyle a,b,c \in \mathbb{N}$. if $\displaystyle a|b$ and $\displaystyle b|c$, does it mean that $\displaystyle a|c$?is it transitive?
Hello, oldguy!
The relation $\displaystyle R$ on $\displaystyle \{1,2,3, \cdots\}$ where $\displaystyle a\text{R}b$ means $\displaystyle a|b.$
Is it reflexive? .Symmetric? .Antisymmetric? .Transitive?
For any natural numnber $\displaystyle a,\;a \div a \:=\:1$
. . That is: .$\displaystyle a|a$
Hence, $\displaystyle \text{R}$ is reflexive.
If $\displaystyle a$ divides $\displaystyle b$, it does not follow that $\displaystyle b$ divides $\displaystyle a.$
Hence, it is not symmetric.
If $\displaystyle a|b$ and $\displaystyle b|a$, then $\displaystyle a = b.$
Hence, $\displaystyle \text{R}$ is antisymmetric.
If $\displaystyle a|b$ and $\displaystyle b|c$, then $\displaystyle a|c.$ . **
Hence, $\displaystyle \text{r}$ is transitive.
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** Proof of transitivity
If $\displaystyle a|b$, then: .$\displaystyle b \:=\:ma$ for some integer $\displaystyle m.$
If $\displaystyle b|c$, then: .$\displaystyle c \:=\:nb$ for some integer $\displaystyle n.$
Then: .$\displaystyle c \:=\:nb \:=\:n(ma) \:=\:(mn)a$
Therefore: .$\displaystyle a|c.$