eli5, is it true that irrational numbers like pi contains every single number combination in it?
In: 0
It goes on forever with no overall repeating pattern. As such, most any reasonable pattern you can think of likely exists. Looking it up, it seems like around 60 trillion digits have been found; meaning that a pattern of 70 trillion 5s in a row could exit but we haven’t found where that pattern in pi exists.
The probability that a certain number occurs, is different from zero, since pi is infinite.
If you roll n dice, you will get at some point a random or guessed number. And pi it’s like that, with infinite dice throws.
This is only true is the number is also something called ‘normal’. A normal number’s decimal expansion contains every possible finite string of digits with a frequency inversly proportional to the length of that string. There is no test for whether or not a given number is normal, but we do know that there are more normal numbers than i.e., the set of normal number is uncountable while the set of non-normal is countable.)
Pi is probably normal, given that it is just the most likely possibility, but we cant prove it.
For some irrational numbers, no, the digits don’t repeat but also are not in a random uniform distribution. For example,
1.0110111011110111110…
is irrational but has only 1s and 0s.
The digits of pi are thought to be randomly uniformly distributed, and so you would expect to find any given finite sequence somewhere in it, but it hasn’t been proven. And that’s because if you randomly select N digits an infinite number of times (which pi is thought to do), you will eventually get every sequence of N digits.
Edit: I rescind this argument. It’s already down voted but won’t delete for the record.
Pi follows no pattern. We can’t predict (neither affirmatively nor negatively) next number based on previous.
Assume there exists a number combo (1234) that definitely does not exist in Pi.
Finding all the numbers in the combo except the last (123), you would know the next number couldn’t be a 4.
But we said Pi does not allow for such “next number” prediction from the outset and have a contradiction somewhere.
By contradiction, our assumption (1234) does not exist in Pi was wrong.
Not necessarily. A number could be irrational, but not contain every digit, never mind every combination.
An irrational number is just a number that you can’t make as a fraction of whole numbers. It’s basically one that has infinitely many digits, but doesn’t settle into a repeating loop. For example, 0.33333… is a rational number because it’s just ⅓. Or, you could have 0.142897142897142897… where those 6 digits repeat. That’s just ⅐.
If an irrational number has roughly equal amounts of all digits, and there’s never a point where it just stops using a particular digit, then we call it a *normal* number. In this case, going on forever and not settling into a repeating patter and using all the digits, any finite sequence will eventually appear.
Okay, so how could you have an irrational number that *isn’t* normal? Well, let’s take the number 0.1101001000100001000001… where each group of 0s just has one more than the last. It’s irrational because it doesn’t fall into a loop (although there is a pattern there). But it doesn’t contain any sequences involving any of the digits 2-9.
We *think* that pi is a normal number. But we don’t know that for sure. It could be that after some point, it just stops using the digit 8. It’s an open question in maths, but most people agree that it’s probably normal.
[removed]
First thing: I suspect what you mean is: “do the *decimal expansions* of irrational numbers like pi contain every *finite* number combination?”
It’s important to distinguish between the number itself and its decimal representation. Also, the decimal expansion of pi clearly doesn’t contain every infinite number combination (since, for example, an infinite sequence of 0’s never occurs there).
That said: there are clearly irrational numbers that *don’t* have this property. For example, consider this number:
0.01101010001010001…
where the n’th decimal place is a 1 if *n* is prime, and 0 otherwise. This number is irrational, but its decimal representation clearly doesn’t have every finite number combination since it doesn’t contain any digit other than 0 or 1.
Another question: does pi have this property? **Pi is conjectured to have this property, but it hasn’t been proven**. In fact, pi is conjectured to have the stronger property of being “normal.” A normal number’s decimal representation is contains every finite digit sequence of a given length roughly the same number of times (in other words, the digits are approximately uniformly distributed).
Interestingly, there are a large number of irrational numbers that are conjectured to have this property, but it is very difficult to prove, and proofs are only known for a small number of numbers.
Pi does not contain the number 2*Pi, or Pi – .07, or one third. At best it contains every finite integer combination within its decimal expansion.
Latest Answers