The rational numbers are dense in RR. This means that between any two real numbers a and b with a < b, there exists a rational number q such that a < q < b. Using this fact, establish that the irrational numbers are dense in RR as well.

We need to prove that Irrational numbers are dense in ${\displaystyle \mathbb{R}}$ .We are given that Rational numbers are dense in ${\displaystyle \mathbb{R}}$ , which means between any two real numbers there exist a rational number. Now, ${\displaystyle \sqrt{2}}$ is an irrational number .
Let, a and b be two real numbers, such that, a Since ${\displaystyle a,b,\sqrt{2}}$ are real numbers therefore there division is also a real number
Therefore , ${\displaystyle \frac{a}{\sqrt{2}}}$ and $\frac{b}{\sqrt{2}}$ are also real numbers. We have,
${\displaystyle a<b\Rightarrow \frac{a}{\sqrt{2}}<\frac{b}{\sqrt{2}}}$
And we are given that between any two real numbers there exist a rational number .
Therefore , let c in ${\displaystyle \mathbb{Q}}$ be a rational number between ${\displaystyle \frac{a}{\sqrt{2}}}$ and ${\displaystyle \frac{b}{\sqrt{2}}\Rightarrow \frac{a}{\sqrt{2}}<c<\frac{b}{\sqrt{2}}}$
We got,
${\displaystyle \frac{a}{\sqrt{2}}<c<\frac{b}{\sqrt{2}}}$
Now , multiply by ${\displaystyle \sqrt{2}}$ throughout, we get,
${\displaystyle \sqrt{2}\left(\frac{a}{\sqrt{2}}\right)<\sqrt{2}\cdot c<\sqrt{2}\left(\frac{b}{\sqrt{2}}\right)\Rightarrow a<c\sqrt{2}<b}$
Now, product of a rational and an irrational number is an irrational number , therefore , since c in ${\displaystyle \mathbb{Q}}$ and ${\displaystyle \sqrt{2}\in \frac{\mathbb{R}}{\mathbb{Q}}}$, therefore, $c\sqrt{2}$
Let, ${\displaystyle c\sqrt{2}=m\Rightarrow a<m<b}$
and a and b are real numbers and m is irrational number .
Therefore, we get that between any two real numbers a and b, with a Therefore, Irrational numbers are dense in real numbers .
Answer: Irrational numbers are dense in ${\displaystyle \mathbb{R}}$ .