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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-u Oct 15 '18 at 0:39

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  • $\begingroup$ Oh it's a duplicate, I also vote to close, the above link cleared my query!! $\endgroup$ – BAYMAX Oct 14 '18 at 3:57

Note that if you divide two non-zero rational numbers the result is a rational number.

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.

  • $\begingroup$ Yes, you are right , $r$ must be non-zero. I fixed my answer. Thanks $\endgroup$ – Mohammad Riazi-Kermani Oct 14 '18 at 3:58

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