# 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.

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