On Sun, 19 Jan 1997 12:20:45 -0600, Sir Robin of Locksley wrote: >Is it possible to prove that number 0.1234567891011121314151617181920... >iz irrational? Most definately. All you need to do is prove that the set of this number is uncountable, ergo is irrational. If you have friend who have done math in real analysis they can explain more. This makes no sense at all. Presumably "the set of this number" means the cardinality of the set {1, 2, 3, ...} which is quite obviously countable! The definition of countable is that it can be put in one to one correspondence with exactly this set! The number is, nevertheless, irrational. A rational number is one that can be written as a ratio of two integers (and infinity is _not_ an integer...or any kind of number, for that matter. Else there would be no irrational numbers). Any rational number will end with a repeating series of digits (which may be 0s, of course). This number clearly has no such repeating sequence. >Is it REALLY true that there are real numbers that cannot be generated >by any algorithm? Some guy said that since the set of algorithms is >countable, but the set of real numbers is more than countable, there >must be some numbers for which there is no algorithms that generate them. If you show me an algorythm that calculates the real number TT (=3.14.....) I'll give you a Nobel Prize personally! Some things are too easy. How about pi = 4(1 - 1/3 + 1/5 - 1/7 + ...) Evaluate. The prize should probably go to Liebniz, though... To answer the original question, yes, there are numbers (the vast majority of numbers, in fact) which cannot be calculated by an algorithm. The computable numbers (those which can be generated algorithmically) are countable (because a universal Turing machine can emulate any other Turing machine given the appropriate input (program), you can give each possible algorithm a number corresponding to the program that implements it. Such numbers are integers, hence countable, so the number of algorithms is countable. Since the real numbers are not countable, it's obvious that there are real numbers which are not computable.) >But I still do not believe him. > >Also, is it true that the sequence of digits in e is random because the ONLY way to get to the >p'th digit is to calculate the p-1'st digits? > Exactly! This is called recursive definition. To get the p-th set in the sequence you need to find p-1 first... The sequence of digits in e is not random! You mean the distribution of digits looks random (is it?). e is certainly computable, which implies there is an algorithm that generates successive digits of e given inputs of 0, 1, 2, 3, ... I have no idea whether every such algorithm must generate all preceding digits in doing so.

Also, is it the correct definition of a real number: > >``A real number is the class of numbers which can represent the length of >an arbitrary line.''

Well, that is not entirely true... The length of any arbitrary line can be any number, rational, natural, etc. Real numbers are the numbers that defy all other categorization (they are not rational, irrational, natural, etc.) They are complex and despite any instinctual perception, there are a lot of them! No, the original definition is perfectly good. Natural numbers are rational, and rational numbers are real numbers. Irrational numbers are also real numbers. The last sentence is correct, though -- real numbers are complex numbers (although that's not what you meant, I think) and there are a lot of them (an infinite number, of course, but a "bigger" infinity than the countable sets of naturals or rationals). -----BEGIN PGP SIGNATURE----- Version: 4.5 What's version 4.5? -- Paul Foley <mycroft@actrix.gen.nz> --- PGPmail preferred PGP key ID 0x1CA3386D available from keyservers fingerprint = 4A 76 83 D8 99 BC ED 33 C5 02 81 C9 BF 7A 91 E8 ---------------------------------------------------------------------- November, n.: The eleventh twelfth of a weariness. -- Ambrose Bierce, "The Devil's Dictionary"