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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.

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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
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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
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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
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(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
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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.

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