Answer:

Step-by-step explanation:

LET us ASSUME on the contrary that \sqrt {3}  

3

​  

 is a RATIONAL number. Then, there exist positive integers aa and bb such that

\sqrt { 3 } =\dfrac { a }{ b }  

3

​  

=  

b

a

​  

 where, aa and bb, are co-prime i.e. their HCFHCF is 11

Now,

\sqrt { 3 } =\dfrac { a }{ b }  

3

​  

=  

b

a

​  

\Rightarrow \quad 3=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } }⇒3=  

b  

2

a  

2

​  

\Rightarrow \quad 3{ b }^{ 2 }={ a }^{ 2 }⇒3b  

2

=a  

2

\Rightarrow \quad 3|{ a }^{ 2 }\quad \quad \left[ \because 3|3{ b }^{ 2 } \right]⇒3∣a  

2

[∵3∣3b  

2

]  

\Rightarrow \quad 3|a\quad \quad ...\left( i \right)⇒3∣a...(i)  

\Rightarrow \quad a=3c⇒a=3c for some integer cc

\Rightarrow \quad { a }^{ 2 }=9{ c }^{ 2 }⇒a  

2

=9c  

2

\Rightarrow \quad 3{ b }^{ 2 }={ 9c }^{ 2 }\quad \quad \left[ \because { a }^{ 2 }=3{ b }^{ 2 } \right]⇒3b  

2

=9c  

2

[∵a  

2

=3b  

2

]  

\Rightarrow \quad { b }^{ 2 }={ 3c }^{ 2 }⇒b  

2

=3c  

2

\Rightarrow \quad 3|{ b }^{ 2 }\quad \quad \left[ \because 3|3{ c }^{ 2 } \right]⇒3∣b  

2

[∵3∣3c  

2

]  

\Rightarrow \quad 3|b\quad \quad ...\left( ii \right)⇒3∣b...(ii)

From (i)(i) and (ii),(ii), we observe that aa and bb have at least 33 as a common factor. But, this contradicts the FACT that aa and bb are co-prime. This MEANS that our assumption is not correct.

Hence, \sqrt {3}  

3

​  

 is an irrational number.