Correct option is (D) None of these

We are disproving all given facts in options by giving a counter example against the fact.

(A) Let distinct irrational numbers are \(\sqrt8\;and\;\sqrt2.\)

Then, their product \(=\sqrt8\times\sqrt2=\sqrt{8\times2}=\sqrt{16}=4\) which is a rational number.

\(\therefore\) The product of two distinct irrational numbers is not always irrational.

(B) Let \(1+\sqrt2\) is irrational number.

Then, \(1-\sqrt2\) is one rationalising factor of \((1+\sqrt2).\)

\((\because\) \((1-\sqrt2)(1+\sqrt2)=1^2-(\sqrt2)^2\) = 1 - 2 = -1 (rational number)).

Also, any multiple of \((1-\sqrt2)\) is another rationalising factor of \((1+\sqrt2).\)

For example \(2(1-\sqrt2)(1+\sqrt2)\) = 2 (1-2) = -2 (rational number)

Thus, \(2(1-\sqrt2)\) is another rationalising factor of \((1+\sqrt2).\)

\(\therefore\) The rationalising factor of a number is not always unique.

(C) Let both distinct irrational numbers are \(\sqrt2\;and\;\sqrt3.\)

Their sum \(=\sqrt2+\sqrt3\) which is an irrational number.

Thus, The sum of two distinct irrational numbers is not always rational.

Correct option is (D) None of these