Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

It is known that constants $\pi$ and $e$ are irrational numbers but also transcedental. Where consist difference between irrationality and transcedentality. How we know that given irrational number is not tanscedental.

share|cite|improve this question
In general determining whether an irrational number is transcendental is very hard. For example, it is not known whether $.1010010001\cdots$ is transcendental. – Alex Becker May 24 '12 at 17:39
Even proving a number to be irrational, let alone trascendental, is extremely difficult. For example, it is still not known whether Euler-Mascheroni constant is irrational. – Holdsworth88 May 24 '12 at 17:45
For example $\sqrt{2}$ is irrational but not transcendental ... Maybe an interesting version of this question would ask for examples of numbers that are algebraic irrationals, but not obviously algebraic. – GEdgar May 24 '12 at 17:57
Somewhat trivia: it is not known whether $e+\pi$ and $e\pi$ are irrational or not. However, using that they are both transcendental it is easy to see that at most one of them is rational. – M.B. May 24 '12 at 18:00
(Since it took me a moment to see M.B.'s 'easy to see', a quick fleshing-out: consider the polynomial $(x-e)(x-\pi) = x^2-(e+\pi)x+e\pi$; if both the coefficients $e+\pi$ and $e\pi$ were rational (or even algebraic) then the roots of this polynomial (namely $e$ and $\pi$) would both be algebraic.) – Steven Stadnicki May 24 '12 at 18:38
up vote 6 down vote accepted

A number $x$ is irrational if there are no integers $a_0, a_1$ such that $a_1x + a_0 = 0$. That is, if there is no integer polynomial $P$ of degree 1 with $P(x)=0$.

A number $x$ is transcendental if there is no positive integer $n$ and no integers $a_0, \ldots a_n$ such that $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0$. That is, if there is no integer polynomial $P$ of any degree $n$ with $P(x)=0$.

All transcendental numbers are irrational, because we can take $n=1$. Not all irrational numbers are transcendental. Non-transcendental numbers are called algebraic. $\sqrt2$ is irrational, but not transcendental, because $(\sqrt2)^2 - 2 = 0$. (That is, $n=2, a_2 = 1, a_1 = 0, a_0=-2$.)

Nobody knows methods that work in general to show that a particular number is rational, irrational, or transcendental. (Many methods are known that work in particular cases.) $\pi$ and $e$ are known to be transcendental, but nobody knows the answer even for simple combinations of $\pi$ and $e$ such as $\pi+e$ or $\pi e$. The important constant $\gamma$ has been studied for hundreds of years, but nobody has yet proved that it is not rational.

Historically the first example of a specific number known to be transcendental was Liouville's number, which is:

$$ \sum_{i=1}^\infty {1\over 10^{i!}} = \frac1{10^{\vphantom1}} + \frac1{10^2} + \frac1{10^6} + \frac1{10^{24}} +\cdots = 0.1100010000000000000000010\ldots $$

The proof that Liouville's number is transcendental is particularly simple. If you want to see a proof that a number is transcendental, that is a good place to start.

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.