this post was submitted on 26 Dec 2024
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How about ANY FINITE SEQUENCE AT ALL?

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[โ€“] ped_xing@hexbear.net 21 points 2 weeks ago (4 children)

0.101001000100001000001 . . .

speech-r I'm infinite and non-repeating. Can you find a 2 in me?

[โ€“] flashgnash@lemm.ee 5 points 2 weeks ago (2 children)

You can't prove that there isn't one somewhere

[โ€“] Danitos@reddthat.com 11 points 2 weeks ago (1 children)

You can, it's literally the way the number is defined.

[โ€“] mukt@lemmy.ml 3 points 1 week ago (1 children)
[โ€“] cosecantphi@hexbear.net 7 points 1 week ago* (last edited 1 week ago) (2 children)

It's implicitly defined here.

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.

[โ€“] mukt@lemmy.ml 1 points 1 week ago (1 children)

0.101001000100001000001 . . .

Might very well be :

0.101001000100001000001202002000200002000002 ...

Real life, is different from gamified questions asked in student exams.

[โ€“] cosecantphi@hexbear.net 1 points 1 week ago* (last edited 1 week ago) (1 children)

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.

[โ€“] mukt@lemmy.ml 1 points 1 week ago (1 children)

assumption โ‰  definition

[โ€“] cosecantphi@hexbear.net 1 points 1 week ago* (last edited 1 week ago) (1 children)

Math kind of relies on assumptions, you really can't get anywhere in math without an assumption at the beginning of your thought process.

[โ€“] mukt@lemmy.ml 1 points 1 week ago (1 children)

Obviously. But still maths avoids stuff like "I assume the answer is X. QED."

[โ€“] cosecantphi@hexbear.net 1 points 1 week ago* (last edited 1 week ago) (1 children)

Right and the point of defining this number as a non-repeating infinite sequence of 0s and 1s is just to show that non-repetition of digits alone is not sufficient to say a number contains all finite sequences.

[โ€“] mukt@lemmy.ml 1 points 1 week ago

That trivial point is not the one we (you and me) are contending.

The issue is that OP hasn't actually defined the sequence, just given some properties (which does not lead to any definition or determination of the location of the number/s on the number line, by itself). Assuming that he has defined it, doesn't change anything as any other commentator can assume something different, which consistent with OP's post.

[โ€“] Hawk@lemmy.dbzer0.com 1 points 1 week ago (1 children)

Pi is often defined as 3.141 592 653... Does that mean Pi does not contain any 7s or 8s?

[โ€“] cosecantphi@hexbear.net 1 points 1 week ago* (last edited 1 week ago)

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.

[โ€“] xigoi@lemmy.sdf.org 4 points 2 weeks ago (1 children)
[โ€“] flashgnash@lemm.ee 2 points 1 week ago

Because you'd need to search through an infinite number of digits (unless you have access to the original formula)

[โ€“] BrainInABox@lemmy.ml 2 points 1 week ago (1 children)

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.

[โ€“] ped_xing@hexbear.net 8 points 1 week ago (1 children)

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.

[โ€“] BrainInABox@lemmy.ml -3 points 1 week ago (3 children)

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

[โ€“] ExFed@lemm.ee 5 points 1 week ago (1 children)
[โ€“] spireghost@lemmy.zip 2 points 1 week ago

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

[โ€“] mukt@lemmy.ml -1 points 1 week ago (1 children)

You might want to stay away from higher maths and all discussions around it, like this one.

[โ€“] BrainInABox@lemmy.ml 0 points 1 week ago

You might want to play in traffic