Explanation
$$ {\textbf{Step - 1: Formation}} $$
$$ \sqrt 2 {\text{ = }}\dfrac{{\text{p}}}{{\text{q}}} $$
$$ {{\text{Here p and q are coprime numbers and q}} \ne {\text{0}}} $$
$$ {\sqrt 2 {\text{ = }}\dfrac{{\text{p}}}{{\text{q}}}} $$
$$ {{\text{On squaring both the sides we get,}}} $$
$$ {{\Rightarrow \text{ 2 = }}{{\left( {\dfrac{{\text{p}}}{{\text{q}}}} \right)}^{\text{2}}}} $$
$$ {{\Rightarrow \text{ 2}}{{\text{q}}^{\text{2}}}{\text{ = }}{{\text{p}}^{\text{2}}}{{ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots }}..\left( {\text{1}} \right)} $$
$$ {\Rightarrow} \dfrac{{{p^2}}}{2} = q $$
$$ {\textbf{Step - 2: Calculation}} $$
$$ {\text{So 2 divides p and p is multiple of 2}} $$
$$ \Rightarrow {\text{p = 2m}} $$
$$ \Rightarrow {{\text{p}}^2} = 4{m^2} ..........................(2) $$
$$ {\text{From equation (1) and (2), we get}} $$
$$ \Rightarrow 2{q^2} = 4{m^2} $$
$$ \Rightarrow {q^2} = 2{m^2} $$
$$ {q^2}\;{\text{is amultiple of 2}} $$
$$ {\text{So, q is multiple of 2}} $$
$$ {\text{Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. }}$$
$${\text{Therefore, p/q is not a rational number}} $$
$$ \sqrt 2{\text{ is an irrational number.}}$$
$${\textbf{Hence, the correct answer is option B.}}$$
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