Let's begin by rewriting this as a conditional statement. Let $z$ and $q(q≠0)$ be irrational and rational numbers respectively. This means, that $$∀a∈Z,∀b∈Z,b≠0∣\frac{a}{b}≠z$$$$∃c∈Z,∃d∈Z,d≠0∣\frac{c}{d}=q$$Let $P$ and $Q$ denote the top and bottom statements. $RTP$ : $(P∧Q)→z⋅q≠\frac{e}{f}$ where e and f are integers.
Proof : Let's assume that the $RHS$ of this implication is not true, and that the $LHS$ is true . This means$$z⋅q=\frac{e}{f}$$ $$z=\frac{e}{f}÷q$$$$z=\frac{eq}{fc}$$Also, since $LHS$ is true$$z≠\frac{a}{b}∣∀a∈Z,∀b∈Z,b≠0$$
This is a contradiction. Which means our implication is always true. Correct? I know it's similar to some previous questions, but I think the solutions are structured differently.
Best Answer
Correct... but can be much simplified.
Suppose that $r_1 x = r_2$ where $r_1,r_2 \in \mathbb Q$ with $r_1 \neq 0$ and $x \in \mathbb R \setminus \mathbb Q$. Then $x = r_2/r_1 \in \mathbb Q$ as $\mathbb Q$ is a field.
A contradiction.