Is this proof concerning field axioms valid?

Question

If $r$ is a rational number and $x$ irrational, prove $r + x$ and $rx$ are irrational.

First, suppose

$$\exists p, q \in \Bbb N : {p \over q} = r + x$$

This is equivalent to $${p \over q} - {p_{0} \over q_{0}} = x$$

Also equivalent to $${pq_{0} - p_{0}q \over qq_{0}} = x$$

Since the numerator is an element of $\Bbb N$, then $p \over q$ is rational. But $x$ is irrational, a contradiction. Thus $r + x$ is irrational.

To solve the latter exercise, suppose

$$\exists p, q \in F : {p \over q} = {p_{0}\over q_{0}}x$$

Then we can say

$${p \over qp_{0}} = {1\over q_{0}}x = 1$$

because of the multiplicative inverse axiom. Hence $x = q_{0}$. But an irrational number cannot equal a natural number, and this is a contradiction. Hence $rx$ is irrational.

Answer 1

Lemma: If $s, t$ are rational, then $-t$, $s+t$, and $st$ are rational. Consequently, $s+(-t)$ is rational.

Proof: If $t=\frac{c}{d}$, where $c$ and $d$ are integers, then $-t=\frac{-c}{d}$.

If furthermore $s=\frac{a}{b}$ then $s+t=\frac{ad+bc}{bd}$ and $st=\frac{ac}{bd}$.

Answer 2

The first part is correct, but in the second part your conclusion $$\frac x{q_0}=1$$ is complete nonsense (comes from nothing). If $r,x$ are rationals then $rx$ being a rational should certainly not imply $x$ is an integer. Up to $${p \over qp_{0}} = {1\over q_{0}}x$$ you are fine so multiply by $q_0$ we get $${{pq_0} \over qp_{0}}=x$$