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So we know that $\pi$ is irrational, that's fact! So we can't write it as $\frac{p}{q}$ where $p$ and $q$ are integers.

We also know that the square root of a prime number is irrational/

But what if $\pi$ can be written as the square root of $\frac{p}{q}$ where $p$ and $q$ are integers? Since$\frac{p}{q}$ would surely be some non integer number and its square root would surely be irrational?

So, is my intuition wrong?

Jack M
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user134384
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    Firstly, $\pi$ cannot be written as the square root of a rational number, because it is not only irrational but also transcendental. Secondly, even if it were possible (for some other number in place of $\pi$, say), there is no contradiction in what you say. – ShreevatsaR Mar 10 '14 at 12:02
  • which numbers will you choose as p and q? – Vikram Mar 10 '14 at 12:03
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    What is "your intuition"? Is it the belief that $\pi$ is the square root of some rational number? – Jack M Apr 26 '14 at 10:36

2 Answers2

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Two things are wrong here. First of all, the square root of $p/q$ is not necessarily irrational. For example, $\sqrt{4/9} = 2/3$.

Second, it is known that $\pi$ can't be the square root (or even the $nth$ root) of any rational number. $\pi$, on top of being irrational, is in fact transcendental. That is, $\pi$ it is not a root of any non-zero polynomial equation with rational coefficients.

Ben Grossmann
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If the following is true: $$\pi=\sqrt{\color{white}{\overline{\color{black}{\dfrac pq\,}}}}\tag{1}$$ then it would mean that $\pi^2$ is rational, by squaring both sides of $(1)$. But we know that $\pi^2$ is irrational.

In fact, $\pi$ can never be written down as: $$\pi=\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\,\cdots \sqrt{\color{white}{\overline{\color{black}{\dfrac pq\,}}}}}}}}}$$ since $\pi^n$ is irrational for every $n\in\mathbb N$. Here's the proof:

If $\pi^{n}$ was rational, then $\pi$ would not be transcendental, as it would be the root of $ax^n−b=0$ for some integers $a$,$b$.

Hakim
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  • Yes. The other answers, which rely on $\pi$ being transcendental, are overkill. It is much easier to prove that $\pi^2$ is irrational. – Gerry Myerson Mar 10 '14 at 12:23
  • @GerryMyerson: What is the easy proof that that $\pi^2$ is irrational? (Anyway this answer also invokes $\pi$ being transcendental.) (I understand that in principle it may be easier to prove $\pi^2$ irrational than to prove $\pi$ transcendental, as it is a much weaker statement. But I don't know if the actual proof is much easier.) – ShreevatsaR Mar 10 '14 at 12:42
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    @ShreevatsaR, the original version of the answer stopped at $\pi^2$ being irrational, and didn't invoke transcendence which, as I say, is not needed here. I didn't say the proof for $\pi^2$ is easy; I said it was easier. It uses nothing beyond integration by parts of functions of the form, polynomial times sine or cosine. It's due to Ivan Niven, and can be found in many intro Number Theory textbooks, and probably all over the internet, as well. – Gerry Myerson Mar 10 '14 at 12:47
  • @GerryMyerson: Thanks! – ShreevatsaR Mar 10 '14 at 12:51
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    http://faculty.washington.edu/rvanderp/Teaching/Archives/308Fall_10/Example2.pdf gives a proof. – Gerry Myerson Mar 10 '14 at 12:52