2007-09-09 20:04:07 · 4 answers · asked by Aken 3 in Science & Mathematics ➔ Mathematics
Yes, there are well-defined but noncomputable numbers. Chaitin's constant Ω is perhaps the best known.
The Wikipedia article on definable numbers notes:
"The numeric representations of the Halting problem, Chaitin's constant, the truth set of first order arithmetic, and 0# are examples of numbers that are definable but not computable. Many other such numbers are known."
2007-09-09 20:55:25 · answer #1 · answered by Scarlet Manuka 7 · 2⤊ 0⤋
Yes I think that such numbers exist.
There would be mathematical expressions so complex that there is no possibility that they will ever be computed. For example a google is a number represented by 1 with 100 zeros after it. A google is so big that there are less than 1 google atoms in the known universe. A google plex is a number represented by 1 with a google zeros behind it.
Question - what is the google plex digit of pi? No one will ever know that - only god knows that (if there is a god).
In theory, if you had a big enough computer with enough time you could solve the above problem (although there is aparently a limited amount of time and space in the universe).
However there is the halting problem mentioned by another poster. That's a classic example of a non computable number that even an infinitely large computer with an infinite amount of time could not solve.
2007-09-10 05:34:01 · answer #2 · answered by Ben O 6 · 0⤊ 0⤋
â2 comes to mind. Like Ï it is an 'irrational' number (one which cannot be exactly expressed as the quotient of 2 integers). Any of the irrationals have the interesting property that their decimal expansions are not only infinite, but are also random. Unlike, say, 1/6 which had a decimal expansion of .1666666666 -> â which has a never ending expansion but is --not-- random.
Ï also has another characteristic (shared with a few other numbers, such as e, the base of the 'natural' logarithm system) in that it is also 'transcendental'. In simplest terms, this means that it cannot be the root of any polynomial expression which has rational coefficients.
HTH
Doug
2007-09-10 03:42:42 · answer #3 · answered by doug_donaghue 7 · 0⤊ 1⤋
pi
It is not calculable because it has an infinite number of decimal places. But it is well defined and is a crucial element in mathematics.
2007-09-10 03:20:33 · answer #4 · answered by rohak1212 7 · 1⤊ 1⤋