prove there exists $x$ in ${\mathbb R}$ using the completeness axioms

Question

Let $a, b \in {\mathbb R}$ with $a < b$. Prove that there exists $x\in {\mathbb R}$ which is NOT a rational number such that $a < x < b$.

This is what i have at the moment.. It does not have a max nor a min, but it does have a sup and inf.

Given the that fact that $(a,b)=\{x\in {\mathbb R}: a<x<b\}$.

If I let $X$ be a set that is bounded above and below then there exist real numbers $x_1$ and $x_n$ such that $X\subset(x_1,x_n)$

Answer 1

We assume that it is already known that $\sqrt{2}$ is irrational Let $N$ be an integer such that $N\gt \frac{4}{b-a}$. Then $\frac{\sqrt{2}}{N}\lt \frac{b-a}{2}$.

There are integers $n$ such that $n\frac{\sqrt{2}}{N}\ge b$. Let $s$ be the smallest such integer.

Then $(s-1) \frac{\sqrt{2}}{N}\lt b$. Since $\frac{\sqrt{2}}{N}\lt \frac{b-a}{2}$, we have $(s-1) \frac{\sqrt{2}}{N}\gt a$. Also, $(s-2) \frac{\sqrt{2}}{N}\gt a$.

One at least of $s-1$ or $s-2$ is non-zero. Call it $k$. It follows that $k \frac{\sqrt{2}}{N}$ is in our interval. It is irrational, else $\sqrt{2}$ would be rational.

Answer 2

Hint: The rationals are countable, and $(a,b)$ is not.