Correct option is (C) positive irrational number

Assume contrary that \(\sqrt3 +\sqrt5 \) is a rational number.

\(\therefore\) \(\sqrt3 +\sqrt5 \) \(=\frac ab\)  (Every rational number can be written as \(\frac pq\) form)

\(\Rightarrow\) \((\sqrt3 +\sqrt5 )^2\) \(=\frac{a^2}{b^2}\)   (By squaring both sides)

\(\Rightarrow\) \(3+5+2\sqrt{15}\) \(=\frac{a^2}{b^2}\)

\(\Rightarrow\) \(2\sqrt{15}\) \(=\frac{a^2}{b^2}-8\) \(=\frac{a^2-8b^2}{b^2}\)

\(\Rightarrow\) \(\sqrt{15}\) \(=\frac{a^2-8b^2}{2b^2}\)   ___________(1)

\(\because\) a & b both are integers & \(b\neq0\)

\(\therefore\) \(a^2-8b^2\) is an integer.

\(\Rightarrow\) \(\frac{a^2-8b^2}{2b^2}\) is a rational number.

But \(\sqrt{15}\) is an irrational number.

This is a contradiction.   (From (1))

\(\therefore\) Our assumption is wrong.

\(\Rightarrow\) \(\sqrt3 +\sqrt5 \) is a positive irrational number.

\((\because\) Sum of two positive number is always positive)

Correct option is C) positive irrational number