Topics Mathematics

Question Is this for philosophers, mathematicians, or logicians? But here goes: Given that the decimal places of pi continue to infinity, does this imply that somewhere in the sequence of numbers of pi there must be, for instance, a huge (and possibly infinite) number of the same number repeated? 77777777777777777777777777... , say? If Pi goes on forever, you might think it must be. After all, if you checked pi to the first googol decimal places you obviously would't find an infinite number of anything. Try a googlplex! Still nothing. But we haven't scratched the surface, even though the universe would have fizzled out by now. If pi's decimal places go on forever, there may be, (not just 77777777777777... or 1515151515151) but all of them, in all combinations, forever. After all, you only have to say "You've only checked a googolplex. There's still an infinite number to to check. The universe is long gone, but pi goes on and on." Philosophers, mathematicians, logicians, any ideas? Mark G.

Accepted:
November 17, 2009

Comments

It doesn't imply that, because that's false. If pi had an infinite number of the same digits repeated, it would be a rational number. But it's an irrational number. See Wikipedia's article on pi.

There are two questions here that need to be distinguished: (1) Does the decimal expansion of pi contain a large but finite string of consecutive 7's--say, 1000 consecutive 7's? (2) Does it contain an infinite string of consecutive 7's?

The second question is the easier one. The only way that pi could contain a string of infinitely many consecutive 7's is if all the digits are 7's from some point on. And, as William has pointed out, that can't happen because pi is irrational.

But the first question is harder. The digits of pi "look random." Imagine a number whose digits are generated by some random process--for example, we might roll a 10-sided die repeatedly to generate the digits. One could compute the probability that a string of 1000 consecutive 7's appears in the first n digits of this number. For n =1000, this number would be extremely small--you'd have to roll 1000 consecutive 7's on your first 1000 rolls, and that's very unlikely. But as n increases, the probability increases, and in fact as n approaches infinity the probability approaches 1. Thus, if you generate an infinite string of digits this way, then the probability that you will eventually get 1000 consecutive 7's is 1, although that doesn't mean that you're guaranteed to get 1000 consecutive 7's. (For more on the difference between "probability 1" and "guaranteed to happen," see this question. Well, actually the discussion there is about the difference between "probability 0" and "guaranteed not to happen," but the idea is the same.)

This heuristic argument makes it seem very likely that the decimal expansion of pi contains 1000 consecutive 7's. But it doesn't actually prove anything--after all, the digits of pi are not generated by a random process, they just "look random" to us. If pi is a normal number, then it must contain 1000 consecutive 7's. But proving that pi is normal is an unsolved problem.