Question

# Prove that $$\sqrt {3}$$ is an irrational number.

Solution

## Let us assume on the contrary that $$\sqrt {3}$$ is a rational number. Then, there exist positive integers $$a$$ and $$b$$ such that$$\sqrt { 3 } =\dfrac { a }{ b }$$ where, $$a$$ and $$b$$, are co-prime i.e. their $$HCF$$ is $$1$$Now,$$\sqrt { 3 } =\dfrac { a }{ b }$$$$\Rightarrow \quad 3=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } }$$ $$\Rightarrow \quad 3{ b }^{ 2 }={ a }^{ 2 }$$ $$\Rightarrow \quad 3$$ divides $${ a }^{ 2 }\quad \quad \left[ \because 3 \ \text{divides}\ 3{ b }^{ 2 } \right]$$ $$\Rightarrow \quad 3$$ divides $$a\quad \quad ...\left( i \right)$$ $$\Rightarrow \quad a=3c$$ for some integer $$c$$$$\Rightarrow \quad { a }^{ 2 }=9{ c }^{ 2 }$$ $$\Rightarrow \quad 3{ b }^{ 2 }={ 9c }^{ 2 }\quad \quad \left[ \because { a }^{ 2 }=3{ b }^{ 2 } \right]$$ $$\Rightarrow \quad { b }^{ 2 }={ 3c }^{ 2 }$$ $$\Rightarrow \quad 3$$ divides $${ b }^{ 2 }\quad \quad \left[ \because 3 \ \text{divides}\ 3{ c }^{ 2 } \right]$$ $$\Rightarrow \quad 3$$ divides $$b\quad \quad ...\left( ii \right)$$ From $$(i)$$ and $$(ii),$$ we observe that $$a$$ and $$b$$ have at least $$3$$ as a common factor. But, this contradicts the fact that $$a$$ and $$b$$ are co-prime. This means that our assumption is not correct.Hence, $$\sqrt {3}$$ is an irrational number.Mathematics

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