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“But it is most assuredly not the case that all irrational numbers have every sequence of digits, for instance the number with decimal expansion”

Yes, I stand corrected, you are right. However, since there are an infinite number of irrational numbers between any two whole numbers, or between any two numbers of any kind, for that matter, then it’s obvious that there are still an infinite amount of irrational numbers that DO contain any possible sequence of numbers. By their very nature, it wouldn’t be anything extraordinary at all to find such a thing to be true about any number you random picked.

Whether you can mathematically prove that or not is another matter, but I think it’s a foregone conclusion logically, which is why mathematicians bother to try to prove it in the first place. It’s intuitively true, so someone must find the proof someday, or so it would seem.

44 posted on 04/02/2013 10:41:59 PM PDT by Boogieman

To: Boogieman

Actually, it’s fairly trivial to prove that between any two distinct real numbers there is a number whose decimal expansion contains every finite sequence of decimal digits: first construct one such number (necessarily between 0 and 1) as follows:

Concatenate all one digit strings, followed by the concatenation of all 2 digit strings, followed by the concatenation of all 3 digit strings, etc. with the strings of given length listed in lexicographic order with respect to the usual order on the digits

(the number’s decimal expansion begins

.0123456789000102030405060708091011121314151617181920212223... )

Now it’s just a matter of multiplying this by a small enough power of 10 and adding some digits at the front to get it in between the two given numbers. I leave the details as an exercise.

I think, however, that what you are intuitively grasping for is the assertion that
almost every real number has a decimal expansion in which every finite sequence of decimal digits occurs — where “almost every” had the technical meaning of the complementary set being of measure zero (can be contained in a union of intervals the sum of whose lengths may be chosen to be less than any given positive number).

I think I can see a route to proving this: since the number of finite decimal strings is countable and countable unions of sets of measure zero are of measure zero, it actually suffices to show that for every finite decimal string, the set of real numbers whose decimal expansion does not contain it is of measure zero. And I think the set you get by removing the places where the given string occurs in the decimal expansion will end up being a (subset of a) Cantor set, and thus of measure zero (but it’s getting late here and I’m not going to try to set down a rigorous proof tonight).

By the way — if it’s a foregone conclusion logically, then there’s a mathematical proof, and conversely. The only things which are foregone conclusions logically are mathematical theorems.

73 posted on 04/03/2013 9:19:43 PM PDT by The_Reader_David (And when they behead your own people in the wars which are to come, then you will know...)

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