Prove $\sqrt{60}$ is irrational? Prove $\sqrt{60}$ is irrational. I'm trying to base this off a proof that $\sqrt{6}$ is irrational, which I found here.
I followed the exact same steps and ended up with $c^2$ = $15b^2$. This was at the point where the example had $2c^2$ = $3b^2$, but the example was able to deduce that $3b^2$ is even, as it is equal to a number times $2$ which is always even, meaning $b^2$ itself has to be even because its coefficient is odd. And because $b^2$ is even, $b$ must be as well, which leads to the contradiction I'm trying to get to.
With my example, I can't immediately say that $15b^2$ is even because there is not an even coefficient in front of $c^2$. I've been stumped from there.
 A: $\sqrt{60}=2\sqrt{15}$, so we just need to prove $\sqrt 15$ is irrational.  Suppose not; $\sqrt{15}=a/b$. Then $15b^2=a^2$. But the factor $3$ appears an odd number of times on the left and must appear an even number of times on the right (the same is true of $5$), so this equality is impossible.
A: Tribute to G. H. Hardy
Suppose $\sqrt{60}=\frac{m}{n}$ where $m,n\in \mathbb{N}$.
Let $\frac{m}{n}$ be in lowest terms.
See also that $\sqrt{60}=60\times \frac{1}{\sqrt{60}}=\frac{60n}{m}$.
Hence $\sqrt{60}=\frac{60n}{m}=\frac{7m}{7n}=\frac{60n-7m}{m-7n}$.
Note that $7 < \sqrt{60} < 8$.
Thus $7 < \frac{m}{n} < 8$ or $0<m-7n<n$.
As a result, $\frac{60n-7m}{m-7n}$ is lower than $\frac{m}{n}$, this contradicts with the assumption.
Therefore $\sqrt{60}$ must be irrational.
A: Just a more tricky strategy using Continued Fractions:
it is not hard to show that a real number is rational iff it has a continued fraction representation which has all $0$'s starting from a certain point like this: $[a_0;a_1,\cdots a_n,0,0,0,\cdots]$ with $a_0 \geq 0$ integer and $a_i > 0$ integers. In fact if you have this kind of CF it is trivially in $\mathbb Q$, the converse is given by the euclidean division in $\mathbb Z$ which eventually ends.
Now compute $\sqrt {60}=[7;\overline{1,2,1,14}]$ for example with a calculator or by hands, this is periodic so it is not $0$ definitively.
Morover, since it is periodic, it is a 
quadratic irrational number, which is not a big surprise since it solves $x^2 - 60=0$.
