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**Failure** If $\displaystyle \sqrt{3}+\sqrt{5}$ were rational then $\displaystyle \sqrt{3}-\sqrt{5}$ would also be rational (for consider that $\displaystyle \sqrt{3}-\sqrt{5}=\frac{3-5}{\sqrt{3}+\sqrt{5}}$).

But from this it would follow that both $\displaystyle \sqrt{3}$ and $\displaystyle \sqrt{5}$ are rational, which is not the case.

Whether the square root of a natural number is rational or not is determined by wether or not its (essentially) unique prime factor decomposition consists only of *even* powers of prime numbers or not. In the former case its square root is a natural number, in the latter case its square root is irrational. So $\displaystyle \sqrt{3^1\cdot 5^1}, \sqrt{3^2\cdot 5^2}, \sqrt{3^1\cdot 5^2},\sqrt{3^{101}\cdot 5^{78}}$ are all irrational, whereas $\displaystyle \sqrt{3^2\cdot 5^2}, \sqrt{3^{102}\cdot 5^{78}}$ are natural numbers.