I`m trying to prove that this relation is partially ordered set:
$R$ relation on $\mathbb {N}$ iff $n\in \{1,5,5^{2},\dots,5^{k}\}$ and $x=n \cdot y$
the condition for partially ordered set are:
- if for all $a \in A$(the functions set) implies $(a,a)\in R \rightarrow$ Reflexivity
- for all $(a,b) \in R , (b,a) \in R \rightarrow a=b $ i.e. Anti - Symmetry
- if for all $(a,b) \in R $ and $(b,c) \in R \rightarrow (a,c)\in R $ Transitivity
so for (1.) what I did is $xRx \rightarrow x=5^{0}x$ and $x\in \mathbb{N}$
for (2.) $(x,y)\in R \wedge (y,x)\in R \rightarrow x=n\cdot y$ and $ y=n\cdot x$
we get $y=n\cdot n\cdot y$ because $n\in \mathbb{N} \rightarrow n=1$ then we get the anti symmetry, what I did is right? or there is another way to prove it?
what about transitivity?
I would like to get some advice
Thanks!
Thanks!
Your proof of reflexivity is fine.
Your proof of antisymmetry is not, though the basic idea is there. If $\langle x,y\rangle\in R$ and $\langle y,x\rangle\in R$, then there are $k,\ell\ge 0$ such that $x=5^ky$ and $y=5^\ell x$, but $k$ and $\ell$ certainly don’t have to be the same. (If you prefer, there are $m,n\in\{1,5,5^2,\ldots\}$ such that $x=my$ and $y=nx$, but you can’t assume that $m=n$.) It follows that $x=5^ky=5^k(5^\ell x)=5^{k+\ell}x$. Clearly this implies that $5^{k+\ell}=1$, and since $k+\ell\ge 0$, we must have $k=\ell=0$ and $x=y$.
For transitivity you can use the same kind of calculation that I used in the previous paragraph. If $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$, then there are integers $k,\ell\ge 0$ such that $x=5^ky$ and $y=5^\ell z$. Now just combine these two equations to express $x$ in terms of $z$.