prove that $\sqrt{\frac{3}{5}}$ is irrational number

Question

I need to prove that $\sqrt{\frac{3}{5}}$ is irrational number.

How can I do it?

I know $\frac{3}{5}$ is non-shortened fraction because 3 and 5 are prime numbers. The only ways of writing it as a product are 1 × 3 or 3 × 1, involve 3 itself and 1×5 or 5×1.

I assume that $\frac{3}{5}$ = ${(\frac{m}{n}})^2$

so $\frac{3}{5}$ $\neq$ ${\frac{m^2}{n^2}}$ because $1^{2}$<3<$2^{2}$ and $2^{2}$<5<$3^{2}$

I don't know if that's good way of thinking. Can somebody help?

In the penultimate line, are you claiming that $m^2 = 3$ and $n^2 = 5$? — anomaly, Mar 08 '20 at 13:01

score 3 · Answer 1 · answered Mar 08 '20 at 13:02

3

One way to go about it is to use the fact that $\sqrt{3/5}$ is a root of $5x^2-3$ and apply the Rational Roots Theorem

answered Mar 08 '20 at 13:02

WaveX

5,440

score 3 · Accepted Answer · answered Mar 08 '20 at 13:02

3

We may assume that $gcd(m, n)=1$ and $3n^2=5m^2$. Since $3\mid 5m^2$ we conclude that $3\mid m$ since $3$ is prime. Hence $9\mid 5m^2=3n^2$, so that $3\mid n^2$ and hence $3\mid n$. This is a contradiction to $gcd(m, n)=1$.

answered Mar 08 '20 at 13:02

Dietrich Burde

130,978

prove that $\sqrt{\frac{3}{5}}$ is irrational number

2 Answers2