# There exists an irrational number z such that x<z<y

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.

• y-x might not be a rational number so your proof is incorrect. – user247608 Apr 29 '16 at 3:26
• Consider to work with the different part in the infinite or finite decimal-digit-representations of x and y and to construct a number between them based on the first different digit... – Gottfried Helms Apr 29 '16 at 10:19

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.