(Real or complex) numbers which are not the root of any polynomial equation a_{n}x^{n}+...+a_{1}x+a_{0}=0 with integer coefficients a_{i}∈**Z** (the set of all such roots is the set of algebraic numbers). Note that this is the same as requiring that a_{i}∈**Q** 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 *x*^{3}=2); this too is impossible. And by trisecting an angle, you could construct some other cube root (e.g. by constructing a 20^{o} angle), which too is impossible. Both these problems involve algebraic numbers which are not constructible.