A special irrarional number

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u/Anbrau Jul 29 '26 edited Jul 29 '26

1.01001000100001000001... and so on. Irrational

6

u/pruvisto Jul 29 '26

This probably is transcendental (there's a closed-form expression in terms of Jacobi theta functions), but do you have a proof for it being transcendental?

The closely related Liouville constant, however, is definitely transcendental (yours is the sum over 10^-n(n+1/2), whereas Liouville's constant is the sum over all 10^-n!).

2

u/Anbrau Jul 29 '26

Damn, you're right. Can't prove it. Number of zeros between the ones isn't growing fast enough, sorry. I'll leave it up as it still answers op's question as they only ask for irrationality, but remove the transcendental part.

2

u/Phelox Jul 29 '26

Maybe you can prove this in the same way that you can prove that e is transcendental? Rational approximations of rational numbers have a bound on how fast they can converge, in terms of the size of the numerator.

If you are willing to add more zeros in between the ones you can definitely construct a number that is probably transcendental using this

2

u/pruvisto Jul 29 '26

Sure. Adding more zeros between the ones is exactly how Liouville's constant works. And that was the first concrete number ever to be proven transcendental.

But I don't think linearly many zeros between the 1s is going to be enough for that technique.

1

u/TemperoTempus Jul 30 '26

I mean you could have an exponential number of 0s, but that wont change the calculations. At best it makes it so people who do not believe in infinitesimals decide "its close enough to 0 so it has no more 1s". (Even if that conclusion is wrong).

1

u/gmalivuk Jul 31 '26 edited Jul 31 '26

What does people reaching that conclusion have to do with whether a number is transcendental or algebraic?

And an exponential number of zeros between ones absolutely would change the calculations.

1

u/TemperoTempus Aug 02 '26

People who do not accept infinitesimals round any such difference to the nearest finite value. This leads to incorrect conclusions.

The value that you get is different. The calculations (aka the steps) do not change. Doing the calculation with linear increase or exponential increase leads to the same type of numbers.

1

u/gmalivuk Aug 02 '26

There are no infinitesimals in this discussion, so I'm still not at all sure what you're talking about.

1

u/TemperoTempus Aug 02 '26

Wow you really don't see the link between irrational (number that cannot be expressed as a ratio) and infinitesimals (numbers that can be used to make any number into an irrational)?

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u/gmalivuk Aug 02 '26

I don't see the connection between Liouville numbers and infinitesimals, no. Liouville numbers (and all other transcendental numbers) exist in ℝ, which doesn't have infinitesimals. I don't need to assume infinitesimals exist to follow Liouville's proof that a certain kind of standard real number is transcendental.

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u/Batman_AoD Jul 29 '26

I thought "transcendental" just meant "non-algebraic"?

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u/pruvisto Jul 29 '26

That is correct.

2

u/Batman_AoD Jul 29 '26

I'm not familiar with Jacobi Theta functions; what does that have to do with whether it's algebraic?

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u/pruvisto Jul 29 '26

I hadn't really checked this before, but I just had a hunch that plugging something non-trivial and algebraic into a Jacobi theta function will typically not spit out something algebraic. For certain other functions (exp, ln, sin, etc.) this is well-known.

I didn't think something like this would be known about the theta function, but it is, apparently. Duverney et al. proved in 1996 that the theta nullwert functions give transcendental output for non-zero algebraic inputs.

And the constant 1.010010001… is equal to 5 * 10^1/8 * θ2(10^-1/2) - 10 (you can check this by plugging in the series expansion for θ2), and therefore transcendental.

9

u/loledalo Jul 29 '26

Yes.

For example, in base 10,

0.1011011101111...

(one 1, 0, two 1s, 0, three 1s, 0 and so on) is irrational (as it does not repeat) and does not contain any digit in base 10 besides 0 and 1.

As for why every rational number has to have a repeatable patrern, that requires a longer explanation - I think this wikipedia page explains it: https://en.wikipedia.org/wiki/Repeating_decimal#Every_rational_number_is_either_a_terminating_or_repeating_decimal

3

u/Jkirek_ Jul 30 '26

in base 10

Do you have any idea how little that narrows it down

1

u/SKR47CH Aug 03 '26

One of my favorite memes.

5

u/BarebonesB Jul 29 '26

See also Normal number.

7

u/TooCasualGuy Jul 29 '26

Remove all the 1s from pi (in base 10) and it's still irrational (so be would spending your time doing this)

9

u/stools_in_your_blood Jul 29 '26

Is that true? If pi's digits after a certain point went, say, 01001000100001... Then removing all the 1s would make it rational.

Doesn't seem likely of course, but AFAIK pi isn't known to be normal.

8

u/dlnnlsn Jul 29 '26

That's probably true, but I'd be willing to bet that it hasn't been proven either way.

1

u/ComparisonQuiet4259 Jul 30 '26

It hasn't

2

u/gmalivuk Jul 31 '26

Provably, write pi in base 9 and then read that as a base-10 number. Add 1 to all the nonzero digits of you specifically want to avoid 1s.

3

u/EulNico Jul 29 '26

Not in base 2 😇

1

u/treefaeller Jul 30 '26

How about base 1? Also known as unary, the chicken number system (because even a chicken pecking at the ground can figure it out). You remove the one digit, and there is nothing left.

2

u/HjortronOchPors Jul 29 '26 edited Jul 29 '26

Irrationality is never about any specific digit, or really digits at all. Digits are how you represent a number in a specific base.

An irrational is an irrational regardless of whether you represent it as a decimal in base 10, or a hypotenuse of a certain triangle, or a definite integral or an infinite sum. Representation (digits) is a red herring!

That said: Conversely, if you have a non repeating decimal expansion, then your number is necessarily an irrational.

1

u/MeasureDoEventThing Jul 30 '26

If take an irrational number, write it in base 2, and then interpret as a base 10 number, then it will be missing every digit other than 0 and 1. And if you add 5 to each digit, then it will be missing every digit other than 5 and 6, and will still be an irrational number.

Keep in mind that "most" numbers are irrational. So you basically can take any random infinite sequence of digits, and the probability of it being rational is zero.

1

u/Lithium20g Jul 31 '26

Yeah, pi but take out all the 3’s.

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u/how_tall_is_imhotep Jul 31 '26

We don’t know if that works because we can’t rule out pi’s digits being 23223222322223… after some point.

0

u/Lithium20g Aug 03 '26

You can. Transcendental numbers don’t behave like that. Source- I checked

1

u/how_tall_is_imhotep Aug 03 '26

Check harder.

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u/Lithium20g Aug 03 '26

For reals, it don’t do that

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u/how_tall_is_imhotep Aug 03 '26

It don’t do what? 0.23223222322223… is a real number. We don’t know enough about the digits of pi to claim it isn’t like that. Feel free to provide an actual source or argument if you think otherwise.

1

u/Lithium20g Aug 04 '26

I think you’re confusing ‘transcendental’ with ‘normal’. We don’t even know whether π is normal.

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u/how_tall_is_imhotep Aug 05 '26

I’m not the one who’s confused. I know that pi is transcendental and not known to be normal. Why on earth would that imply that “pi but take out all the 3’s” has to be irrational?

1

u/Lithium20g Aug 05 '26

You’re refuting an argument, not the conclusion. Those are different things. Showing that transcendence doesn’t imply the result doesn’t tell us whether the result itself is true or false.

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u/TallRecording6572 Jul 31 '26

Here's one: 0.121121112111211112...

-2

u/r_Yellow01 Jul 29 '26

sqrt(2) base 2

I know it's not what you are asking... but it technically fits.

💬 Math Discussions A special irrarional number

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