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I don't think so. You can create a number that has only 1s and 0s (in the decimal system) that is still transcendental. Thus, not all transcendental numbers ha
by TomasSedovic 17y ago
I don't think so. You can create a number that has only 1s and 0s (in the decimal system) that is still transcendental.
Thus, not all transcendental numbers have all possible numeric subsets in them.
So, while it can be the case, pi being transcendental doesn't automatically mean that it contains every finite sequence of digits.
Sorry, I'm too tired for a formal proof now. Please tell me if this is wrong.
- oomkiller 17y agoYou are correct here, but what i meant is that specifically since pi by definition goes on forever and never has a pattern, it is possible that any sequence of numbers will appear in pi at some point in its calculation.
- pg 17y agoPossible seems the wrong word to use here. Either every sequence appears, or some don't, and though the former seems likely, as far as I know it hasn't been proven yet.
- pc 17y agoRight—it hasn't even been proved that the decimal expansion of pi doesn't eventually revert to a digit sequence containing only 1s and 0s. (Or any other two digits.)
- pc 17y agoIt depends what you mean by "pattern", but it hasn't been shown that pi is a normal number. (I.e., that every sequence of digits is equally likely to occur.)