Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating numbers should appear somewhere in Pi?
Since Pi is infinite and non-repeating, would that mean any finite sequence of non-repeating numbers should appear somewhere in Pi?
How about ANY FINITE SEQUENCE AT ALL?
You're viewing a single thread.
0.101001000100001000001 . . .
I'm infinite and non-repeating. Can you find a 2 in me?You can't prove that there isn't one somewhere
You can, it's literally the way the number is defined.
Defined where ?
It's implicitly defined here by its decimal form:
0.101001000100001000001 . . .
The definition of this number is that the number of 0s after each 1 is given by the total previous number of 1s in the sequence. That's why it can't contain 2 despite being infinite and non-repeating.
Pi is often defined as 3.141 592 653... Does that mean Pi does not contain any 7s or 8s?
That's a decimal approximation of Pi with an ellipsis at the end to indicate its an approximation, not a definition. The way the ellipsis is used above is different. It's being used to define a number via the decimal expansion by saying it's an infinite sum of negative powers of 10 defined by the pattern before the ellipsis.
So we have:
0.101001000100001000001 . . . = 10^-1 + 10^-2 + 10^-3 + 10^-4 +10^-5+ . . .
Pi, however, is not defined this way. Pi can be defined as twice the solution of the integral from -1 to 1 of the square root of (1-x^2), a function defining a unit semi-circle.
0.101001000100001000001 . . .
Might very well be :
0.101001000100001000001202002000200002000002 ...
Real life, is different from gamified questions asked in student exams.
Implicitly defining a number via it's decimal form typically relies on their being a pattern to follow after the ellipsis. You can define a different number with twos in it, but if you put an ellipsis at the end you're implying there's a different pattern to follow for the rest of the decimal expansion, hence your number is not the same number as the one without twos in it.
assumption ≠ definition
Math kind of relies on assumptions, you really can't get anywhere in math without an assumption at the beginning of your thought process.
Why couldn’t you?
Because you'd need to search through an infinite number of digits (unless you have access to the original formula)
Are you trying to say the answer to their question is no? Because if so, you're wrong, and if not I'm not sure what you're trying to say.
The conclusion does not follow from the premises, as evidenced by my counterexample. It could be the case that every finite string of digits appears in the decimal expansion of pi, but if that's the case, a proof would have to involve more properties than an infinite non-repeating decimal expansion. I would like to see your proof that every finite string of digits appears in the decimal expansion of pi.
Well that's just being pointlessly pedantic, obviously they fucking know that a repeating number of all zeros and ones doesn't have a two in it. This is pure reddit pedantry you're doing
It's a math proof. Chill.
It's not a valid one, though. See my response to your original comment.
It kind of does come across as pedantic -- the real question is just that "Does pi contain all sequences"
But because of the way that it is phrased, in mathematics you do a lot of problems/phrasing proofs where you would be expected to follow along exactly in this pedantic manner
You might want to stay away from higher maths and all discussions around it, like this one.
You might want to play in traffic
Infinite and non-repeating are one thing, but the number you gave is not transcendental. Transcendentalism is the nature of pi that causes it to contain every sequence. Also, fun fact, there are more transcendental numbers than there are non-transcendental numbers, but pi is one of the few we know about because proving transcendentalism is hard.
Are you Pi?
Shit, opsec fail.