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transcendental number

created by Thomas Miconi

(idea) by ariels (17.4 hr) (print)   ?   (I like it!) Tue Jun 06 2000 at 7:12:35

(Real or complex) numbers which are not the root of any polynomial equation anxn+...+a1x+a0=0 with integer coefficients aiZ (the set of all such roots is the set of algebraic numbers). Note that this is the same as requiring that aiQ be rational numbers.

Any proportion which can be constructed with straightedge and compass is algebraic, hence not transcendental. But many algebraic numbers cannot be constructed in this manner.

The set of algebraic numbers is countable, hence has measure zero (and is of first category). Thus, almost every real number is transcendental. Proving a number transcendental is a hard problem of number theory. The first numbers proved transcendental were specially constructed. For instance, if we define Liouville's number 0<w<1 by requiring that the n'th digit after the decimal point be 1 iff n=k! for some k, and 0 otherwise, then it is fairly easy to show w is transcendental (by Liouville's lemma on algebraic numbers; see the number's node for a proof). Unfortunately, such contrived examples are of little interest; proving an interesting real number is transcendental is much harder.

Of the three problems of antiquity, one relates to transcendentality. Squaring the circle is impossible due to pi being transcendental. The same is not true of the other 2 problems. Doubling the cube is possible iff you can construct a line segment cbrt(2) times as long as a given line segment (i.e. given a segment of length 1, construct a segment of length x, where x3=2); this too is impossible. And by trisecting an angle, you could construct some other cube root (e.g. by constructing a 20o angle), which too is impossible. Both these problems involve algebraic numbers which are not constructible.


(idea) by Kenan (5.1 y) (print)   ?   (I like it!) Tue Nov 27 2001 at 21:46:06

Since the set of algebraic numbers is countable, it follows that the set of transcendental numbers is uncountable. In other words, almost all of the real numbers are transcendental. Nonetheless, it is extremely difficult to find a transcendental number, or to prove that a number is transcendental.

It was not until 1873 that e was proved to be transcendental (by Hermite), and not until 1882 that pi was proved to be transcendental (by Lindemann)


(idea) by JpB (5.4 y) (print)   ?   (I like it!) Fri Dec 06 2002 at 7:17:06

To make things more complex, the same function can have transcendental/algebraic number answers depending on the domain. For example, sin (1 radian) is transcendental, while sin (1 degree) is algebraic.
Since e is transcendental, ln a (where a is not equal to 0 or 1 and is algebraic) is also transcendental.
A fundamental problem about transcendental numbers that still needs to be proved is whether a^b, where a is not equal to 0 or 1 and is algebraic and b is algebraic but not rational, is always transcendental.

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