Prove that the Archimedes property doesn't occur in the Polynomials

Question

I found this question and I can't solve it.

So by Definition of Archimedes property - Archimedes property doesn't occur if The $\mathbb{N}$ set is bounded.

Which means - this is what we need to prove - that $\mathbb{N}$ is bounded over Polynomials.

My try :

Prove by contradiction:

Lets assume that $\mathbb{N}$ is not bounded over $P(x)$ which means exist $n∈\mathbb{N}$:

such that : $∀P(x) : P(x)<n$

$n∈\mathbb{N}$ and $\mathbb{N}⊆\mathbb{R}$ $\Rightarrow$ $n∈\mathbb{R}$.

Lets take the polynomial $P(x)=x+1$ and do $P(n)=n+1$ and by assumption $n+1<n$ which is a contradiction.

Which means $\Rightarrow$ $\mathbb{N}$ is bounded above $P(x)$ $\Rightarrow$ The polynomials doesn't "have" the archimedes axiom.

is it ok to assume that ? or I did a mistake somewhere?

Answer 1

That problem only makes sense if we define an order relation on the polynomials. The usual one is this: $p(x)<q(x)$ if one of these conditions holds:

1. $\deg p(x)<\deg q(x)$
2. $\deg p(x)=\deg q(x)$ and, if $n$ is that common degree, then the coefficient of $x^n$ in $p(x)$ is smaller than the coefficient of $x^n$ in $q(x)$.

With this definition, it is clear that $(\forall n\in\Bbb N):n<x$. So, indeed, the Archimedes property doesn't hold in this context.