Meaning of Divisibility in the definition of a relation

Question

Consider {$ℝ+,≤$}, where "$≤$" defined as "a is divisible by b". Is {$ℝ+,≤$} a Partially Ordered Set (POSET) ?
While checking for antisymmetric property, a≤b : a is divisible by b

The axioms for a non-strict partial order state that the relation ≤ is reflexive, antisymmetric, and transitive. That is, for all a, b, and c in P, it must satisfy:

1. a ≤ a (reflexivity: every element is related to itself).
2. if a ≤ b and b ≤ a, then a = b (antisymmetry: two distinct elements cannot be related in both directions).
3. if a ≤ b and b ≤ c, then a ≤ c (transitivity: if a first element is related to a second element, and, in turn, that element is related to a third element, then the first element is related to the third element).

Does the above statement mean, we need to get the remainder as zero? My teacher said the definition of "divisibility" here is based on the concept of multiples and NOT on the concept of factors.

Can anyone please clarify this doubt? Please also prove that {$ℝ+,≤$} is NOT a POSET.

Answer 1

Taking your teacher's hint that "the definition of "divisibility" here is based on the concept of multiples" we can say that a is divisible by b means that $\fbox{$a=kb$ for some $k\in \Bbb N$}$

Then for reflexivity: Test $a=ka$; take $k=1\in \Bbb N$, $\checkmark$
For anti-symmetry: If $a=kb$ with $k\neq1$ ($a,b$ distinct); then $b=\frac 1k a$ but $\frac 1k\not\in \Bbb N$, $\checkmark$
For transitivity: $a=kb$ and $b=jc$; so $a=kj\cdot c$ and $kj\in\Bbb N$, $\checkmark$

Thus the defined set is a POSet.