Explanation
Assume that $$\sqrt p + \sqrt q $$ is rational. So,
$$\sqrt p + \sqrt q = \dfrac{a}{b}$$
$$p + q + 2\sqrt {pq} = \dfrac{{{a^2}}}{{{b^2}}}$$
$$\sqrt {pq} = \dfrac{1}{2}\left( {\dfrac{{{a^2}}}{{{b^2}}} – p - q} \right)$$
Since the RHS of the above equation is rational but $$\sqrt {pq} $$ is an irrational number, so the assumption is wrong.
Therefore, it is true that $$\sqrt p + \sqrt q $$ is irrational.
$${\textbf{Step - 1: Solving A}}$$
$${\text{Prime factorizing 8 and 40,}}$$
$$ \Rightarrow {\text{ 8 = 2 }} \times {\text{ 2 }} \times {\text{ 2}}$$
$$ \Rightarrow {\text{ 40 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 5}}$$
$${\text{Multiplying common prime factors to get HCF, }}$$
$${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 }} \times {\text{ 2 = 8}}$$
$${\textbf{Step - 2: Solving B}}$$
$${\text{Prime factorizing 15 and 25,}}$$
$$ \Rightarrow {\text{ 15 = 3 }} \times {\text{ 5}}$$
$$ \Rightarrow {\text{ 25 = 5 }} \times {\text{ 5}}$$
$${\text{H}}{\text{.C}}{\text{.F = 5}}$$
$${\textbf{Step - 3: Solving C}}$$
$${\text{Prime factorizing 16 and 60,}}$$
$$ \Rightarrow {\text{ 16 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2}}$$
$$ \Rightarrow {\text{ 60 = 2 }} \times {\text{ 2 }} \times {\text{ 3 }} \times {\text{ 5}}$$
$${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 = 4}}$$
$${\textbf{Step - 4: Solving D}}$$
$${\text{Prime factorizing 14 and 35,}}$$
$$ \Rightarrow {\text{ 14 = 2 }} \times {\text{ 7}}$$
$$ \Rightarrow {\text{ 35 = 5 }} \times {\text{ 7}}$$
$${\text{H}}{\text{.C}}{\text{.F = 7}}$$
$${\textbf{Step - 5: Solving E}}$$
$${\text{Prime factorizing 15, 25 and 60,}}$$
$$ \Rightarrow {\text{ 25 = }}5{\text{ }} \times {\text{ 5}}$$
$${\text{H}}{\text{.C}}{\text{.F = }}5$$
$${\textbf{Step - 6: Solving F}}$$
$${\text{Prime factorizing 16, 48 and 80,}}$$
$$ \Rightarrow {\text{ 48 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 3}}$$
$$ \Rightarrow {\text{ 80 = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 5}}$$
$${\text{H}}{\text{.C}}{\text{.F = 2 }} \times {\text{ 2 }} \times {\text{ 2 }} \times {\text{ 2 = 16}}$$
$${\textbf{Hence, we have found H.C.F of different pair of numbers using prime factorization method}}$$
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