# Irrationality of $r\sqrt{5}$ for rational number $r$? [duplicate]

As $$5$$ is a prime number, thus $$\sqrt{5}$$ is an irrational number.

Now I am thinking about how to prove -

If $$r$$ is a rational number, then how do we prove $$r\sqrt{5}$$ is an irrational number?

I was thinking that since $$r$$ is a rational number, then $$r$$ can be expressed as the fraction in simplified form that is $$r = \frac{a}{b}$$ such that $$a,b \in \Bbb{Z}$$ and $$gcd(a,b)=1$$. So $$r\sqrt{5} = \frac{a\sqrt{5}}{b}$$, but How can this guarantee us the irrationality of $$r\sqrt{5}$$?

ALso let $$c =r\sqrt{5}$$, then $$c^2 = 5r^2$$, if we could prove it is a prime number, then its square-root $$c$$ must be irrational and ths proved but unfortunately we donot have $$c^2$$ prime as it has more than one factors like $$r$$ , $$5$$

## marked as duplicate by Eric Wofsey, Lord Shark the Unknown, Paul Frost, Don Thousand, Delta-uOct 15 '18 at 0:39

Now if $$r\ne 0$$ is rational and $$r\sqrt 5$$ is also rational, we have to have $$\frac {r\sqrt 5}{r} = \sqrt 5$$ to be rational and we know that it is not rational, so $$r\sqrt 5$$ must be irrational.
• Yes, you are right , $r$ must be non-zero. I fixed my answer. Thanks – Mohammad Riazi-Kermani Oct 14 '18 at 3:58