I know there are lots of post about this but I wanted to know this proof would work also.
Proposition.
Let $x,y ∈ \mathbb{R}$ with $x < y$. There exists an irrational number $z$ such that $x < z < y$.
Proof.
Let $x,y ∈ \mathbb{R}$ with $x < y$
We know $$0 <\frac1{\sqrt2}<\frac{\sqrt2}{\sqrt2}=1$$
Since $y – x > 0$, we can multiply the inequality by $(y – x)$, getting
$$0<(y-x)\frac1{\sqrt2}<y-x$$Also adding $x$ to the inequality, we have $$x<x+(y-x)\frac1{\sqrt2}<y$$
Note that multiplying an irrational number by a non-zero rational number yields an irrational number. so $$(y-x)\frac1{\sqrt2}\in \mathbb{R}-\mathbb{Q}$$ Also $$\text{since}\;x \in \mathbb{R}, \;\;x+(y-x)\frac1{\sqrt2}\in \mathbb{R}-\mathbb{Q}$$
Hence we prove there is an irrational number between real number x and y.
Best Answer
Your proof is fine if $x$ and $y$ are rational, but your claim is that it works for $x$ and $y$ real. You assume that $y-x$ is rational. If not, $(y-x)\frac 1 {\sqrt 2}$ might be rational and the proof fails.