Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This might sounds stupid, but I really don't know can I show Irrational numbers in proves? And if so, how to show it?

For example, when I want to show Rational numbers, I do this:

$\frac{m}{n} $ , $m, n $ are integer $,$ $n\ne0$

Can I do something like that with Irrational numbers?

share|cite|improve this question
That depends on the number you want to prove the irrationality of... Say, $\sqrt{n}$ where $n$ is not a perfect square is easily proven to be irrational whereas $e$ and $\pi$ have more complicated proofs. – M.B. Sep 5 '12 at 12:09
@M.B. - I think the OP wants some kind of a "closed form", not how to prove a number is irrational – Belgi Sep 5 '12 at 12:14
There's no way of describing irrationals that's as simple as the way you've described a rational. It's easy enough, though, to simply say that a number is not rational. This, as you likely know, is a common way to show that a number is irrational: assume it were (i.e., equal to $n/m$ for $n, m$ integers) and then argue to a contradiction. This is seen in many proofs that $\sqrt{2}$ is irrational. – Rick Decker Sep 5 '12 at 12:16
One can easily construct numbers that are irrational if and only if the Goldbach conjecture is true, for example. Thus an (ir)rationality proof can be arbitrarily complicated. – Hagen von Eitzen Sep 5 '12 at 12:24
up vote 5 down vote accepted

A real number $a$ is irrational iff its decimal (or binary) expansion doesn't become ultimately periodic. You can formulate this in the following way: $$a=a_0.a_1a_2a_3\ldots\qquad\bigl(a_0\in{\mathbb Z}, \ a_k\in\{0,1,\ldots,9\}\ (k\geq1)\bigr)$$ is irrational iff $\forall p\in{\mathbb N}_{\geq1}$, $\forall n\in{\mathbb N}_{\geq1}$ there is a $k>n$ with $a_k\ne a_{k+p}$.

share|cite|improve this answer
Your definition is right. But I couldn't undrestand how can we sure that the number won't become repeating? – Mostafa Farzán Sep 5 '12 at 13:08
@MostafaFarz'an: That's exactly what is assured by the condition "$\ \forall p\in{\mathbb N}_{\geq1}$, $\forall n\in{\mathbb N}_{\geq1}$ there is a $k>n$ with $a_k\ne a_{k+p}\ $". Think about it! – Christian Blatter Sep 5 '12 at 14:19

You can say it is an element of $\mathbb{R}$ that is not in $\mathbb{Q}$.

There is no "general representation" of irrational numbers

share|cite|improve this answer

In principle, as you point out, showing that a number $r$ is rational is easy. All we need to do is to exhibit integers $a$ and $b$, with $b\ne 0$, such that $a=rb$.

Proving that a number $x$ is irrational is in principle, and often in practice, much harder. We have to show that there do not exist integers $a$ and $b$, with $b\ne 0$, such that $a=xb$. So in principle we have to examine all ordered pairs $(a,b)$ of integers, with $b\ne 0$, and show that none of them can possibly "work."

This in principle involves examining an infinite set. That cannot be done by simply exhibiting a pair of integers, like in a proof of rationality. Sometimes, as in the case $x=\sqrt{2}$, there is a relatively simple proof of irrationality. But all too often, like in the case of the Euler-Mascheroni constant $\gamma$, no proof of irrationality is known, despite the fact that considerable effort has been expended trying to prove that $\gamma$ is irrational.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.