> [!NOTE] Lemma > For any two [[Real numbers|real numbers]] $a<b,$ there is a [[Rational Number|irrational number]]. **Proof**: By [[Existence of a rational in any closed real interval]], there exists $x \in \mathbb{Q}$ such that $\frac{a}{\sqrt{ 2 }} < x < \frac{b}{\sqrt{ 2 }}$ Since [[The sum and product of a rational and irrational is irrational]], $x \sqrt{ 2 } \not \in \mathbb{Q}$ and so $a < x \sqrt{ 2 } < b$ so we're done.