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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?

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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

3 Answers 3

up vote 3 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 $a_k\ne a_{k+p}$.

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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
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@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

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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.

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