[robxorb]

Yes and that contradiction is already present in my premise which is the point. Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance. Unless, not random...

This applies to every normal, "irrational" number, the name with which I massively agree, because the only way they can be not purely random suggests they are compressible further and so they have to be purely random, and thus... can't be.

It is a completely irrational concept, thinking rationally.

[linearrust]

> Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance.

What you are essentially saying is that pi = 3.14....pi...........

If that was the case, wouldn't it mean that the digits of pi are not countably infinite but instead is a continuum. So you wouldn't be able to put the digits of pi in one to one correspondence with natural numbers. But obviously we can so shouldn't our default be to assume our premise was wrong?

> It is a completely irrational concept, thinking rationally.

It is definitely interesting to think about.

[traes]

The belief that a normal number must eventually contain itself arises from extremely flawed thinking about probability. Like djkorchi mentioned above, if we knew pi = 3.14....pi..., that would mean pi = 3.14... + 10^n pi for some n, meaning (1 - 10^n) pi = 3.14... and pi = (3.14...) / (1 - 10^n), aka a rational number.

[linearrust]

> The belief that a normal number must eventually contain itself arises from extremely flawed thinking about probability.

Yes. There is an issue with the premise as it leads to a contradiction.

> Like djkorchi mentioned above, if we knew pi = 3.14....pi..., that would mean pi = 3.14... + 10^n pi for some n, meaning (1 - 10^n) pi = 3.14... and pi = (3.14...) / (1 - 10^n), aka a rational number.

Yes. If pi = 3.14...pi ( pi repeats at the end ), then it is rational as the ending pi itself would contain an ending pi and it would repeat forever ( hence a rational number ). I thought the guy was talking about pi contain pi somewhere within itself.

pi = 3.14...pi... ( where the second ... represents an infinite series of numbers ). Then we would never reach the second set of ... and the digits of pi would not be enumerable.

So if pi cannot be contained within ( anywhere in the middle of pi ) and pi cannot be contained at the end, then pi must not contain pi.

[thaumasiotes]

> If that was the case, wouldn't it mean that the digits of pi are not countably infinite but instead is a continuum.

No; combining two countably infinite sets doesn't increase the cardinality of the result (because two is finite). Combining one finite set with one countably infinite set won't give you an uncountable result either. The digits would still be countably infinite.

Looking at this from another direction, it is literally true that, when x = 1/7, x = 0.142....x.... , but it is obviously not true that the decimal expansion of 1/7 contains uncountably many digits.

[linearrust]

> No; combining two countably infinite sets doesn't increase the cardinality of the result (because two is finite).

Agreed. But pi = 3.14...pi... isn't combing 2 infinite sets. It 'combining' infinite amounts of infinite sets and not in a linear fashion either.

You have to keep in mind the 2nd pi in the equation can be expanded to 3.14...pi...

pi = 3.14...pi... when expanded is pi = 3.14...(3.14...pi...)...

and you can keep expanding the inner pi forever.

> The digits would still be countably infinite.

How can you ever reach the first number after the inner pi in (pi = 3.14...pi...). Or put another way how do you get to the 4th '.'? You can't.

This is a classical example of countably infinite and a continuum.

[jcranmer]

> Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance.

A normal number would mean that every finite sequence of digits is contained within the number. It does not follow that the number contains every infinite sequence of digits.

In general, something that holds for all finite x does not necessarily hold for infinite x as well.

[robxorb]

Exactly - and when you remove the assumptions, what's left?

Pi is assumed to be infinite, random, and normal. The point here is not these assumptions may be wrong. Underneath them may sit a greater point; that irrationality is defined in a contradictory way - which may be correct, or not, or, both.

Given proof Pi is infinite lay on irrationality, it is rather an important issue. Pi may not be infinite, and a great place to observe that may be Planck.

[orbillius]

> A normal number would mean that every finite sequence of digits is contained within the number.

Is that true? I don't see how that could be true. The sequence 0-9 repeated infinitely is, by definition, a normal number (in that the distribution of digits is uniform)

...and yet nowhere in that sequence does "321" appear ...or "654" ...or "99"

There are an infinite number of combinations of digits that do not appear in that normal number I've just described. So, I don't think your statement is true.

[thaumasiotes]

> I don't see how that could be true. The sequence 0-9 repeated infinitely is, by definition, a normal number (in that the distribution of digits is uniform)

Well, your first problem is that you don't know the definition of a normal number. Your second problem is that this statement is clearly false.

Here's Wolfram Alpha:

> A normal number is an irrational number for which any finite pattern of numbers occurs with the expected limiting frequency in the expansion in a given base (or all bases). For example, for a normal decimal number, each digit 0-9 would be expected to occur 1/10 of the time, each pair of digits 00-99 would be expected to occur 1/100 of the time, etc. A number that is normal in base-b is often called b-normal.

Your "counterexample" is not a normal number in any sense, most obviously because it isn't irrational, but only slightly less obviously because, as you note yourself, the sequences "321", "654", and "99" do not ever appear.

[orbillius]

> Your "counterexample" is not a normal number in any sense, most obviously because it isn't irrational, but only slightly less obviously because, as you note yourself, the sequences "321", "654", and "99" do not ever appear.

lol. Your counterargument is a tautology because it contains "the sequences "321", "654", and "99" do not ever appear."

It's like if you claim, "A has the property B" then I say, "based on this definition, I don't think A has property B"

Then you say, "if it doesn't have property B, then it's not A"

...okay, but my point is, the definition that I had (from wikipedia) doesn't imply B. So for you to say, "if it doesn't have B, then it's not A" is just circular.

Now, you can point out that the definition I got from wikipedia is different from the one you got from wolfram. That's fine. That's also true. And you can argue that the definition you used does indeed imply B.

But what you cannot do is use B as part of the definition, when that's the thing I'm asking you to demonstrate.

You: all christians are pro-life

Me: I don't see how that's true. Here's the definition of christianity. I don't see how it necessarily implies being against abortion.

You: your """"counterexample"""" (sarcastic quotes to show how smart I am) is obviously wrong because, as you note yourself, that person is pro-choice, therefore, not a christian.

^^^^^ do you see how this exchange inappropriately uses the thing you're being asked to prove, which is that christians are pro-life, as a component of the argument?

Again, it's totally cool if you fine a different definition of christian that explicitly requires they be pro-life. But given that I didn't use that definition, that doesn't make it the slam dunk you imagine.

[thaumasiotes]

> But given that I didn't use that definition, that doesn't make it the slam dunk you imagine.

You might have a better argument if there were more than one relevant definition of a normal number. As you should have read in the other responses to your comment, the definition given on wikipedia does not differ from the one given on Wolfram Alpha.

> And you can argue that the definition you used does indeed imply B.

Given that the implication of "B" is stated directly within the definition ("For example, ..."), this seemed unnecessary.

> but my point is, the definition that I had (from wikipedia) doesn't imply B. So for you to say, "if it doesn't have B, then it's not A" is just circular.

Look at it this way:

1. You provided a completely spurious definition, which you obviously did not get from wikipedia.

2. You provided a number satisfying your spurious definition, which - not being normal - didn't have the properties of a normal number.

3. I responded that you weren't using the definition of a normal number.

4. And I also responded that it's easy to see that the number you provided is not normal, because it doesn't have the properties that a normal number must have.

Try to identify the circular part of the argument.

And, consider whether it's cause for concern that you believe you got a definition of "normal number" from wikipedia when that definition of "normal number" is not available on wikipedia.

[orbillius]

> Try to identify the circular part of the argument.

I did. Should I repeat it?

[moefh]

It depends on your definition of "normal number". You seem to be using what wikipedia[1] calls "simply normal", which is that every digit appears with equal probability.

What people usually call "normal number" is much stronger: a number is normal if, when you write it in any base b, every n-digit sequence appears with the same probability 1/b^n.

[1] https://en.wikipedia.org/wiki/Normal_number

[jumhyn]

IIRC the property ‘each single digit has the same density’ is the definition for a ‘simply normal number’ (in a given base), while ‘each finite string of a particular length has the same density as all other strings of that length’ is the definition for a ‘normal number’ (in a given base). And then ‘normal in all bases’ is sometimes called ‘absolutely normal’, or just ‘normal’ without reference to a base.