If possible, let ${\displaystyle \frac{1}{\sqrt{2},}}$ be rational. Then, there exist positive co-primes $a$ and $b$ such that

$$\begin{gathered} \Rightarrow 2a^2=b^2[\because 2\mathrm{\mid }2a^2] \\ <\mathrm{/}div><div> \end{gathered}$$\Rightarrow 2|{ b }^{ 2 }\\$<\mathrm{/}div><div>$\Rightarrow 2|{ b }\\$<\mathrm{/}div><div>$\Rightarrow b=2c\\$forsomepositiveinteger$c$<\mathrm{/}div><div>$ \therefore 2{ a }^{ 2 }={ b }^{ 2 }\Rightarrow 2{ a }^{ 2 }=4{ c }^{ 2 }\Rightarrow { a }^{ 2 }=2{ c }^{ 2 }\Rightarrow 2|{ a }^{ 2 }\qquad \left[ \because 2|2{ c }^{ 2 } \right] \\ $<\mathrm{/}div><div>$\Rightarrow 2|{ a }$<\mathrm{/}div><div>Thisisacontradictiontothefactthat$a,\ b$areco-primes\mathrm{.}<\mathrm{/}div><div>Hence,$\dfrac { 1 }{ \sqrt { 2 }  } $$ is irrational.