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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.

Eric Wofsey
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Chris
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    Your title does not match the body. – Moo Feb 17 '16 at 05:07
  • @ForeverMozart It is enough to show $sqrt(15)$ is irrational, but you really should be careful what you say. For instance $9$ is odd but $sqrt(9) = 3$ is rational. – cpiegore Feb 17 '16 at 05:12
  • "I can't immediately say the 15b^2 is even"... Then don't. Even has nothing to do with anything now. So 15b^2 is divisible by 3 or 15b^2 is divisible by 5. There's nothing significant about 2 other than it is a prime. But there are other primes and they are more relevant. 15b^2 is divisible by 3 so b is divisible by3 and and so c is divisible by 3 so... – fleablood Feb 17 '16 at 05:47
  • The same proof as linked works here if you use the prime $3$ isntead of $2$. You get $,3\mid c^2,\Rightarrow,3\mid c,,$ so $, 3(c/3)^2 = 5b^2,\Rightarrow, 3\mid b,,$ so $,3\mid b,c,,$ contra hypothesis that $,b,c,$ are coprime (wlog). – Bill Dubuque Nov 29 '16 at 23:37

3 Answers3

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$\sqrt{60}=2\sqrt{15}$, 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.

Forever Mozart
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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.

Ng Chung Tak
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  • Implicit in these style proofs is a descent on denominators: if $,n,$ is a denom for $, r = \sqrt{60},$ then so too is the numerator $,m = nr.,$ Thus a smaller denom is $, m\bmod n, $ because denoms are closed under subtraction (so under mod too). See the 3rd proof here and the following remark for more on the innate ideal-theoretic structure (Dedkeind's conductor ideal). – Bill Dubuque Nov 29 '16 at 23:25
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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 $x^2 - 60=0$.

Maffred
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