Never really thought this deeply about PI before

[MathJakob]

Was just watching an episode of person of interest and the student asks the teacher what pi is good for and he tells her that contained in pi lies every possible combination of words, every conversation that has ever taken place on Earth is located somewhere in pi. Every shakespear play, every song ever wrote, even word or setence you've ever strung together from your first word to the last you'll ever say. Your date of birth, your address, your social security number, everything about your life is somewhere in pi.

That got me thinking, if every possible combination of words or numbers that can exist does exist somewhere in the infinity of pi, then does this include the infinities themselves? Such as the number 1 occurring an infinite number of times without a break?

Or the number 1 occurring a billion times in a row, then the number 2 occurring once, then the number 1 occurring a billion times ect.

How true is this?

 

[atyy]

The teacher was making things up. What he said is not known to be true.

Take a look at Brian M. Scott's answer to http://math.stackexchange.com/questions/216343/does-pi-contain-all-possible-number-combinations

 

[1MileCrash]

That got me thinking, if every possible combination of words or numbers that can exist does exist somewhere in the infinity of pi, then does this include the infinities themselves? Such as the number 1 occurring an infinite number of times without a break?

This would make pi rational, which we know is false.

 

[phinds]

This would make pi rational, which we know is false.

Uh ... how did you reach that conclusion ?

 

[1MileCrash]

Uh ... how did you reach that conclusion ?

If the number one occurs an infinite number of times without a break, anywhere in PI, then clearly I can multiply pi by some (possibly very large, but it doesn't matter) 10^n and get a whole number plus 1/9.

Clearly if I multiply pi by a rational number and get a rational number, pi was rational by basic closure properties.

Of course, this is just a roundabout way of saying "he made it a repeating decimal." I wasn't expecting a follow-up question.

Unless you are suggesting that pi can return to being a non repeating decimal "after" the infinite string of 1s without a break, in which case we have a fundamental disagreement on what an infinite string of 1 means.

 

[phinds]

in which case we have a fundamental disagreement on what an infinite string of 1 means.

Yeah, I'd say that's where we differ. I think you can have an infinite string of 1's followed by an infinite string of 2's and still have the whole thing be the same order of infinity (aleph-null).

 

[1MileCrash]

Yeah, I'd say that's where we differ. I think you can have an infinite string of 1's followed by an infinite string of 2's and still have the whole thing be the same order of infinity (aleph-null).

You can if you want, but the value of those two such numbers in our decimal system isn't different.

.111...
.111...222..

Would you not agree that these are both 1/9? Otherwise, what is the second one?

This is often touched on in the .99..=1 equality discussion, in that ".0000...1" isn't different from 0.

https://www.physicsforums.com/showthread.php?t=507001

 

[MathJakob]

Yeah, I'd say that's where we differ. I think you can have an infinite string of 1's followed by an infinite string of 2's and still have the whole thing be the same order of infinity (aleph-null).

So it's true? That located within pi is every conversation that has and every will take place? And every possible combination of numbers that can every exist is located somewhere in pi, even the infinite strings of 1's, 2's, 3's 121's 101's ect? Even my mobile phone number repeats itself an infinite number of times consecutively... Which kinda makes me think that it isn't rational once it reaches this infinite repetitiveness?

 

[mfb]

Yeah, I'd say that's where we differ. I think you can have an infinite string of 1's followed by an infinite string of 2's and still have the whole thing be the same order of infinity (aleph-null).

You cannot, the decimal digits of pi are an ordered list.

So it's true? That located within pi is every conversation that has and every will take place?

Every finite conversation, yes.
This is not true for infinite strings. There are infinite strings, of course, but only a tiny subset of all possible infinite strings. And none of those infinite strings has pure repetitions of anything.

 

[dipole]

If it's true of pi, then it's true for almost every other number as well. I don't think that's a good reason to think pi is so great.

 

[mfb]

If it's true of pi, then it's true for almost every other number as well. I don't think that's a good reason to think pi is so great.

It is not true for rational numbers, and you can construct many irrational numbers where it is false as well.

 

[dipole]

It is not true for rational numbers, and you can construct many irrational numbers where it is false as well.

http://en.wikipedia.org/wiki/Almost_all

If it's true of pi, it's probably true of all or most transcendental numbers. Pi is special for better reasons than this, and even if the OP's original claim is true, it's misleading to tell people this is somehow a special property of pi.

 

[1MileCrash]

What's stopping me from constructing as many transcendentals as I want that never have "9" "4" or "2" in them, immediately making the property false?

 

[phinds]

You cannot, the decimal digits of pi are an ordered list.

Yes, I agree w/ that, and my discussion is NOT appropriate for PI, which is defined. I DO think you can have an irrational that has an infinite number of 1's in a row AND an infinite number of 2's in a row. If that's not true, then Hilbert's Hotel fills up.

 

[mfb]

http://en.wikipedia.org/wiki/Almost_all

Hmm, as almost everywhere, probably.

If it's true of pi, it's probably true of all or most transcendental numbers.

Certainly not for all.

I DO think you can have an irrational that has an infinite number of 1's in a row AND an infinite number of 2's in a row. If that's not true, then Hilbert's Hotel fills up.

No. To add guests to Hilbert's hotel, you have to relocate guests. If you take an arbitrary but fixed number, you cannot relocate its decimal digits. Every digit has to have a fixed position - I can ask you for the position of some "2" after your infinite string of 1, and you cannot give me a meaningful answer.

 

[1MileCrash]

In the context of the decimal system, .1111...222... and .111... are the same number. That's why we never talk about numbers like the first.

 

[dipole]

What's stopping me from constructing as many transcendentals as I want that never have "9" "4" or "2" in them, immediately making the property false?

Ok - all is obviously not the case, but can you prove to me that there is an uncountable set of transcendentals that don't contain every digit? I'm inclined to think that most transcendentals contain every digit in their decimal expansion, which would make pi ordinary in this context.

 

[someGorilla]

Every finite conversation, yes.

Although this is probably true, it's also true for the majority of irrational numbers, so why concentrate on pi when sqrt(12345) is equally likely to contain Hamlet in ascii code?

 

[1MileCrash]

Ok - all is obviously not the case, but can you prove to me that there is an uncountable set of transcendentals that don't contain every digit?

Probably not, but I can try to explain why I think there is and you can let me know if you agree.

Let T be the set of all transcendentals.

Let S be a set of transcendentals such that every element of S is different in more than one way at more than one decimal place. What I mean by that is, let x and y be elements of S, I can find an nth decimal place on x, and call its value l, such that the nth decimal place on y is not l. I can then find an mth decimal place on x, whose value is k (and k=/=l), such that the mth decimal place on y is not k. I think that this set is uncountable.

Now, let P be the set created by replacing every 9 in every element of S with a 0.

This change shouldn't effect the cardinality of S, because this change won't cause any members of S to become equivalent since they differ in more than one way at more than one place (thus if they are as close to being the same as they can possibly be while still being members of S, this change could not possibly make them the same.)

If S is uncountable and P has the same cardinality, P is uncountable.

I'm inclined to think that most transcendentals contain every digit in their decimal expansion, which would make pi ordinary in this context.

I agree. I think the way I would put it, is that to find transcendentals that didn't contain every digit, we would almost have to construct them with that goal in mind.

 

[mfb]

Ok - all is obviously not the case, but can you prove to me that there is an uncountable set of transcendentals that don't contain every digit? I'm inclined to think that most transcendentals contain every digit in their decimal expansion, which would make pi ordinary in this context.

If you express every irrational number between 0 and 1 in binary and then read it as decimal number (with just 0 and 1 in the decimal expansion), they are all irrational. An uncountable set of irrational numbers has to contain an uncountable set of transcendental numbers, as the set of algebraic numbers in this set is countable.

@someGorilla: Sure, many irrational numbers share that property.

 

[phinds]

No. To add guests to Hilbert's hotel, you have to relocate guests. If you take an arbitrary but fixed number, you cannot relocate its decimal digits. Every digit has to have a fixed position - I can ask you for the position of some "2" after your infinite string of 1, and you cannot give me a meaningful answer.

OK, I'm willing to believe that I've got this wrong, but explain why my logic, below, is wrong.

We have an irrational number

A = 1.2929802902980<POSITION A>9230983209820382039<POSITION B>8203948023894092380238... infinite string of random digits

There is nothing in POSITION A or POSITION B

Now, we create a new number by putting a 1 in POSITION B, then another different new number, by adding another 1 in POSITION B, and we do that an infinite number of times.

Then we create a new number from that by putting a 2 in POSITION A, then another, and so forth.

Why does this not work?

 

[mfb]

Now, we create a new number by putting a 1 in POSITION B, then another different new number, by adding another 1 in POSITION B, and we do that an infinite number of times.

That is not a well-defined process to define a real number. It is as meaningless as the following question:
Take 1, double it (2), double it again (4), ..., double it an infinite number of times. Which natural number do you get?

 

[atyy]

So it's true? That located within pi is every conversation that has and every will take place? And every possible combination of numbers that can every exist is located somewhere in pi, even the infinite strings of 1's, 2's, 3's 121's 101's ect? Even my mobile phone number repeats itself an infinite number of times consecutively... Which kinda makes me think that it isn't rational once it reaches this infinite repetitiveness?

Every finite conversation, yes.
This is not true for infinite strings. There are infinite strings, of course, but only a tiny subset of all possible infinite strings. And none of those infinite strings has pure repetitions of anything.

Is that known to be true?

Knowing that pi is irrational is not enough to conclude this. http://math.stackexchange.com/questions/216343/does-pi-contain-all-possible-number-combinations

 

[mfb]

Oh... looks like it is still an open problem, and it is just expected to be true.

 

[phinds]

That is not a well-defined process to define a real number. It is as meaningless as the following question:
Take 1, double it (2), double it again (4), ..., double it an infinite number of times. Which natural number do you get?

OK, I don't get how the two are the same but I'm not enough into math to know better, so I'll take your word for it. Thanks.

 

[someGorilla]

Ok - all is obviously not the case, but can you prove to me that there is an uncountable set of transcendentals that don't contain every digit?

Yes, it's uncountable. And yet you're right on this:

I'm inclined to think that most transcendentals contain every digit in their decimal expansion, which would make pi ordinary in this context.

Look at this. The set of non normal irrationals is uncountable but has Lebesgue measure zero. If you choose a random irrational (or a random real, for that matter) you have 0% chance of landing on a non normal number, that is you have 100% chance (though not certainty ) of landing on a number that contains every finite string of digits.

 

[someGorilla]

Have fun

 

[krash661]

pi = -4*i*ln((-.5)^.5+.5^.5)