Do irrational numbers contain everything infinitely times?

I have a question about irrational numbers.
If I start writing a number of n digits does this sequence of numbers exist on every irrational number infinitely many times for all nΕΝ?
So for example if myself is nothing more than a really really large number (total information of myself) and every 10^10^10 numbers of pi is another universe (total information of the universe) do I have infinite replicas of myself ?
I guess that this happens for a completely random number, so is my question equivalent to 'Are irrational numbers completely random?'?
Is there any proof for this?
I'm a high school student ,don't expect any further knowledge from me.

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Reply 1

username1560589

No. A counterexample is the irrational number

\displaystyle\sum_{k=1}^{\infty} 10^{-k!}

, which only contains the digits 1 and 0, so doesn't contain the digit sequence 2.

You could construct a number that satisfies your property. List every one digit sequence, then every two digit sequence, then three digit, then one digit, then two digit, three, four, one, two, three, four, five, one, ....

It's not yet known whether pi has this property.

Reply 2

Zacken

Another (more general) counter example is constructing an irrational number

0.\mathrm{dedeedeeedeeeedeeeeed}\ldots

where

d

and

e

are any two digits you like.

Reply 3

Gregorius

Original post by Quantum boy

In addition to the counter-examples other folk have already posted, you might like to have a look at the notion of a normal number, the requirements for which are a bit stronger than those you advance. But you might find it amusing!

Reply 4

Quantum boy

Original post by Zacken

Another (more general) counter example is constructing an irrational number

0.\mathrm{dedeedeeedeeeedeeeeed}\ldots

where

d

and

e

are any two digits you like.

I was talking about the 'natural' irrational numbers: pi, e, nth roots of integers. Do they all have these properties?

Original post by morgan8002

You could construct a number that satisfies your property. List every one digit sequence, then every two digit sequence, then three digit, then one digit, then two digit, three, four, one, two, three, four, five, one, ....

What if you just list every 1 digit sequence, then 2,3,4,5,....
The n digit sequences will contain every sequence of 1,2,...,n-1 in them.
Doesn't this mean every sequence will be repeated infinitely?

So irrational numbers don't necessarily carry infinite information, but do pi for example satisfy this?

Original post by Gregorius

So normal numbers are completely random (No digit or sequence is "favored"). I got the general idea ,but what is natural density anyway?

Reply 5

Zacken

Original post by Quantum boy

I was talking about the 'natural' irrational numbers: pi, e, nth roots of integers. Do they all have these properties?

Literally no clue what you're going on about...

What makes those examples you've given any more "natural" than my counter-example.

Also, nth root of integers aren't always irrational, e.g:

(a^n)^{1/n} = a

for some

a \in \mathbb{N}

Reply 6

Gregorius

Original post by Quantum boy

I was talking about the 'natural' irrational numbers: pi, e, nth roots of integers. Do they all have these properties?

If you take a number like

\pi

the answer is that we simply do not know if every finite sequence of digits occurs somewhere. This sort of question is, in general, very hard.

So normal numbers are completely random (No digit or sequence is "favored" :wink:

. I got the general idea ,but what is natural density anyway?

Calculate the proportion of digits having some property (for example "equal to seven" or "even" :wink:

up to some large number of digits N. The natural density is x if, as N gets larger and larger, this proportion gets closer and closer to x.

Reply 7

Johann von Gauss

Original post by Quantum boy

I was talking about the 'natural' irrational numbers: pi, e, nth roots of integers. Do they all have these properties?

\pi

and e (transcendentals) are a different class of irrationals to irrational roots

No idea what you mean by natural irrationals...

Reply 8

Gregorius

Original post by Johann von Gauss

No idea what you mean by natural irrationals...

It is a term used informally in analytic number theory to describe those irrational numbers that just keep popping up everywhere!

Reply 9

Johann von Gauss

Original post by Gregorius

It is a term used informally in analytic number theory to describe those irrational numbers that just keep popping up everywhere!

Is there a rigorous definition?

Reply 10

Quantum boy

Original post by Johann von Gauss

\pi

and e (transcendentals) are a different class of irrationals to irrational roots

No idea what you mean by natural irrationals...

Original post by Zacken

Literally no clue what you're going on about...

What makes those examples you've given any more "natural" than my counter-example.

Also, nth root of integers aren't always irrational, e.g:

(a^n)^{1/n} = a

for some

a \in \mathbb{N}

Sorry guys for this expression, I made it up.
I meant the ones which have occured naturally : they were found in an application (ratio of length of circle to diameter, a number e so that (e^x)'=e^x,... ). The 0.deedeeeeeddddededde example is a random number generator ,it has little or no meaning.
Anyway, sorry for the misunderstanding.

Reply 11

Radical Don Juan

Actually a very interesting question.

Certainly not every irrational number does, but it would be interesting if there were a proof that a particular one has this property.

Reply 12

Johann von Gauss

Original post by Quantum boy

But you could say 0.dedeedeeedeeeed... is the simplest irrational where only 2 digits occur in the decimal expansion...

And what about the physical constants like the speed of light in a vacuum, etc.

You can't really prove anything about something unless it is rigorously defined, so its tough to say if what you are asking about is true or not.

(edited 8 years ago)

Reply 13

Gregorius

Original post by Johann von Gauss

Is there a rigorous definition?

Nope.

Reply 14

Quantum boy

Original post by Johann von Gauss

But you could say 0.dedeedeeedeeeed... is the simplest irrational where only 2 digits occur in the decimal expansion...

You can't really prove anything about something unless it is rigorously defined, so its tough to say if what you are asking about is true or not.

Ok let's make it more specific. They might be many natural irrational numbers, but the ones I know are pi,e,nth roots (ok not all of them).

Reply 15

Forum User

What about Sum(x=1 -> inf): 10^(-x^2)? isn't that a pretty 'simple' and 'natural-looking' irrational number that obviously doesn't contain every sequence of n digits?

[I have no idea how to insert equations!]

Reply 16

Lord of the Flies

Yes

Spoiler

Reply 17

Quantum boy

Original post by Forum User

Original post by Lord of the Flies

Yes

Spoiler

Since there is no strict definition there is really no point in discussing this I guess.

Reply 18

Craig1998

There must be an infinite amount of irrational numbers that have this property, but determining whether a number, like pi or e, does have it must be impossible.

Reply 19

Quantum boy

Original post by Johann von Gauss

I think the physical constants satisfy my conditions ,because they are completely random .
They are in the form 'write any number from 0 to 9 randomly' .
But perhaps they are not even irrational numbers.