Relation based on divisibility, on set $(\mathbb N \setminus \{0,1\})\times \mathbb N$

Question

So i have to see if $\prec$ is order relation where two elements $(a,b)$ and $(c,d)$ are in relation $\prec$ if $a|c$ and $2b^{2}+6b\leq2d^{2} + 6 d$. This relation is defined on set $(\mathbb N \setminus \{0,1\}) \times ( \mathbb N)$.
To prove that $\prec$ is order relation i need to show that it is refelexive, antisymetric and transitive.
To show that it is reflexive i need to show that $(a,b)\prec(a,b)$. $(a,b)\prec(a,b)$ iff $a|a$ and $2b^{2}+6b\leq2b^{2}+6b$. Form $a|a$ only if from $a=ka$ follows that $k=1$. And here is my question since $a$ is from $(\mathbb N \setminus \{0,1\})$ thas it mean that $k$ also has to be from $(\mathbb N \setminus \{0,1\})$? If so then i have that $a\neq ka$ since $k$ is from $(\mathbb N \setminus \{0,1\})$.So on $(\mathbb N \setminus \{0,1\})$ $a\nmid a$ and there fore relation fails to be reflexive and thus fails to be order on $(\mathbb N \setminus \{0,1\}) \times ( \mathbb N)$. Now on $ \mathbb N \times \mathbb N$ this would be order. Is this reasoning correct?

Answer 1

No, $k$ can be any integer and does not have to be an element of $\mathbb{N}\setminus\{0,1\}$. The relation $a\mid c$ is defined as a relation between two integers, and means that there exists an integer $k$ such that $c=ka$. The fact that in your context you are restricting your attention to $a$ and $c$ which are in $\mathbb{N}\setminus\{0,1\}$ does not change the meaning of $a\mid c$.