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

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 b/sqrt2ZSK are also real numbers. We have,

\(\displaystyle{a}{<}{b}\Rightarrow\frac{{a}}{\sqrt{{2}}}{<}\frac{{b}}{\sqrt{{2}}}\)</span>

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 \(\displaystyle\frac{{b}}{\sqrt{{2}}}\Rightarrow\frac{{a}}{\sqrt{{2}}}{<}{c}{<}\frac{{b}}{\sqrt{{2}}}\)</span>

We got,

\(\displaystyle\frac{{a}}{\sqrt{{2}}}{<}{c}{<}\frac{{b}}{\sqrt{{2}}}\)</span>

Now , multiply by \(\displaystyle\sqrt{{2}}\) throughout, we get,

\(\displaystyle\sqrt{{2}}{\left(\frac{{a}}{\sqrt{{2}}}\right)}{<}\sqrt{{2}}\cdot{c}{<}\sqrt{{2}}{\left(\frac{{b}}{\sqrt{{2}}}\right)}\Rightarrow{a}{<}{c}\sqrt{{2}}{<}{b}\)</span>

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, csqrt2ZSK

Let, \(\displaystyle{c}\sqrt{{2}}={m}\Rightarrow{a}{<}{m}{<}{b}\)</span>

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 aTherefore , \(\displaystyle\frac{{a}}{\sqrt{{2}}}\) and b/sqrt2ZSK are also real numbers. We have,

\(\displaystyle{a}{<}{b}\Rightarrow\frac{{a}}{\sqrt{{2}}}{<}\frac{{b}}{\sqrt{{2}}}\)</span>

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 \(\displaystyle\frac{{b}}{\sqrt{{2}}}\Rightarrow\frac{{a}}{\sqrt{{2}}}{<}{c}{<}\frac{{b}}{\sqrt{{2}}}\)</span>

We got,

\(\displaystyle\frac{{a}}{\sqrt{{2}}}{<}{c}{<}\frac{{b}}{\sqrt{{2}}}\)</span>

Now , multiply by \(\displaystyle\sqrt{{2}}\) throughout, we get,

\(\displaystyle\sqrt{{2}}{\left(\frac{{a}}{\sqrt{{2}}}\right)}{<}\sqrt{{2}}\cdot{c}{<}\sqrt{{2}}{\left(\frac{{b}}{\sqrt{{2}}}\right)}\Rightarrow{a}{<}{c}\sqrt{{2}}{<}{b}\)</span>

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, csqrt2ZSK

Let, \(\displaystyle{c}\sqrt{{2}}={m}\Rightarrow{a}{<}{m}{<}{b}\)</span>

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 \(\displaystyle\mathbb{R}\) .Answer: Irrational numbers are dense in \(\displaystyle\mathbb{R}\) .