Density of the irrationals

Statement

Let $x$ and $y$ be real numbers with $x < y$. Then there is an irrational number $z$ with $x < z < y$.

Remarks

Proof: by the density of the rationals, there is a rational $q$ with $x / \sqrt{2} < q < y / \sqrt{2}$ and $q \neq 0$. Then $z = q \sqrt{2}$ is irrational and $x < z < y$. Thus both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$.

Explanations

L1 human

Between any two different real numbers, there is also a number that is not a fraction. So fractions and non-fractions are mixed together everywhere on the number line.