Does it make sense to reason about the infinite?

[Humberto]

Consider the number π (pi) for example. π is an irrational number with an infinitely long decimal expansion. To my knowledge and understanding it is believed, but has not been proved, that there is no formula that outputs the n-th decimal of π given n. The first few decimals of π are

π = 3.1415926...

The Universe contains an estimated 10[SUP]80[/SUP] atoms. Hence an upper limit to the number of decimals of π that could ever actually be written down is 10[SUP]80[/SUP]. If indeed there is no formula that can describe what the n-th decimal of π looks like then it may not be possible to ever know what the decimal expansion of π looks like beyond its 10[SUP]80[/SUP]-th decimal. And even if there were a formula to compute the n-th decimal of π, one would still have the practical difficulty of needing at least 10[SUP]80[/SUP] atoms to be able to store the value of an integer n with n > 10[SUP]80[/SUP] that will be used to calculate the value of the n-th decimal of π.

It seems like π is off limits beyond its n-th decimal for some very large n. Any real number that has the same decimal expansion as π up to its n-th decimal, but differs in some decimal beyond that, and hence would be considered as a different real number to π, cannot ever be distinguished from π in our finite Universe.

Let n be the largest integer that can be represented using the finite number of atoms and particles in our Universe. Call n the observable finite constant.

My claim is that one cannot reason about the Universe beyond the observable finite constant. Reasoning about finite quantities beyond the observable finite constant is merely an intellectual abstraction.

 

[Keegan]

You lost me at "number π (pi)"

 

[guest2013-1]

I like pie

 

[broloks]

The Bailey–Borwein–Plouffe formula (http://en.wikipedia.org/wiki/Bailey–Borwein–Plouffe_formula) can be used to calculate any digit of Pi.

 

[Jings]

If you think pi is bad enough try Godel's theorem of incompleteness. Divide any number by zero and end up with an undefined value. Infinity cannot be defined by a numerical value.

 

[Humberto]

The Bailey–Borwein–Plouffe formula (http://en.wikipedia.org/wiki/Bailey–Borwein–Plouffe_formula) can be used to calculate any digit of Pi.

That is interesting. That would seem to imply that the digits of π are not randomly generated.

There is still a limit to how accurately π can be described though. Suppose there are K[SUB]0[/SUB] particles in the Universe and let n > K[SUB]0[/SUB]. It is impossible to compute the n-th digit of π since there aren't enough particles in the Universe with which to store the value of n when using the Bailey–Borwein–Plouffe formula.

Moreover, as time passes, entropy increases, so our potential computational ability diminishes.

Suppose m(t) is the largest integer n for which the n-th decimal of π can be computed at some time t. Then if t[SUB]1[/SUB] < t[SUB]2[/SUB] then m(t[SUB]1[/SUB]) ≥ m(t[SUB]2[/SUB]). Over time, our ability to compute the decimals of π grows less.

Let x be a real number defined as follows:

x = 3.x[SUB]1[/SUB]x[SUB]2[/SUB]x[SUB]3[/SUB]...

where x[SUB]i[/SUB] is the i-th decimal of π when i ≤ K[SUB]0[/SUB] and x[SUB]i[/SUB] = 0 otherwise.

In the real number system, π ≠ x. In fact, x is not even an irrational number. However, it is impossible to distinguish between π and x.

More generally, let t[SUB]1[/SUB] be some fixed time in the distant future and define the real number y as follows:

y = 3.y[SUB]1[/SUB]y[SUB]2[/SUB]y[SUB]3[/SUB]...

where y[SUB]i[/SUB] is the i-th decimal of π when i ≤ m(t[SUB]1[/SUB]), and y[SUB]i[/SUB] = 0 otherwise.

In the real number system, π ≠ y. In principle, we are able to distinguish between π and y up to and including the time t[SUB]1[/SUB]. After the time t[SUB]1[/SUB] we will not, in principle, be able to distinguish between π and y.

Over time, our ability to test the truth of statements diminishes.

 

[broloks]

It is impossible to compute the n-th digit of π since there aren't enough particles in the Universe with which to store the value of n when using the Bailey–Borwein–Plouffe formula.

As I understand it you can calculate any n-th digit of Pi with BBP from a storage point of view, since you do not require the previous n - 1 digits. The problem comes from a time perspective, in that BBP is a linearithmic function.

In the real number system, π ≠ x. In fact, x is not even an irrational number. However, it is impossible to distinguish between π and x.

Agreed

In principle, we are able to distinguish between π and y up to and including the time t[SUB]1[/SUB]. After the time t[SUB]1[/SUB] we will not, in principle, be able to distinguish between π and y.
Over time, our ability to test the truth of statements diminishes.

Disagree, contradicts previous statement.

 

[Humberto]

As I understand it you can calculate any n-th digit of Pi with BBP from a storage point of view, since you do not require the previous n - 1 digits.

My claim is that there are not enough particles in the Universe to store very large numbers (for example integers n with n > K[SUB]0[/SUB]) so that any formula or algorithm can only be used to compute decimals up to a certain point.

Disagree, contradicts previous statement.

The number y is not the same as x.

 

[Jehosefat]

My claim is that one cannot reason about the Universe beyond the observable finite constant. Reasoning about finite quantities beyond the observable finite constant is merely an intellectual abstraction.

Reasoning is, pretty much by definition, an abstraction. Just because something cannot be practically measured does not mean it is not useful. I would venture that all mathematics is an "intellectual abstraction" but that does not mean mathematics is not useful or practical. To illustrate this point, numbers are an abstraction. You cannot show me 5. You can show me the symbol for it or you can show me 5 objects and you can demonstrate or explain the concept to me but 5 itself is not a tangible thing that can be seen or held. Therefore, the observable finite constant is in no way a limit on numbers because they are a concept and not something tangible that is made up of matter.

You could also have made your exact same point about 0 as follows: because there are atoms in the universe, one cannot reason about 0 because there is always "something" and 0 can only truly be defined where nothing tangible/physical exists.

Further on your point of using atoms to limit the observable value, why not use sub-atomic particles like protons neutrons and electrons? Instantly higher observable constant. But why stop there? Why not go even further to quarks, muons etc? Sure, eventually you will hit the ceiling but that in no way invalidates the concept of integers higher than this.

 

[SaiyanZ]

^ What he said.

 

[Humberto]

Reasoning is, pretty much by definition, an abstraction. Just because something cannot be practically measured does not mean it is not useful. I would venture that all mathematics is an "intellectual abstraction" but that does not mean mathematics is not useful or practical. To illustrate this point, numbers are an abstraction. You cannot show me 5. You can show me the symbol for it or you can show me 5 objects and you can demonstrate or explain the concept to me but 5 itself is not a tangible thing that can be seen or held. Therefore, the observable finite constant is in no way a limit on numbers because they are a concept and not something tangible that is made up of matter.

You could also have made your exact same point about 0 as follows: because there are atoms in the universe, one cannot reason about 0 because there is always "something" and 0 can only truly be defined where nothing tangible/physical exists.

Further on your point of using atoms to limit the observable value, why not use sub-atomic particles like protons neutrons and electrons? Instantly higher observable constant. But why stop there? Why not go even further to quarks, muons etc? Sure, eventually you will hit the ceiling but that in no way invalidates the concept of integers higher than this.

The observable finite constant was defined to be the largest integer that can be represented using the finite number of atoms and particles in our Universe.

---

Consider the following statement I made:

The Universe contains an estimated 10[SUP]80[/SUP] atoms. Hence an upper limit to the number of decimals of π that could ever actually be written down is 10[SUP]80[/SUP].

I made a mistake with this statement for at least two reasons:

1. There are particles other than atoms available too with which to represent the decimals of π.
2. Given n particles, there may be more than n integers that can be represented using them. For example, given two hydrogen atoms, one can represent at least the following numbers: 0, 1, 2, 3. A system may contain zero, one or two hydrogen atoms, and if containing two hydrogen atoms, they may be bonded or not. Hence there are at least four numbers that can be represented with just two hydrogen atoms.
   
   Hence given 10[SUP]80[/SUP] atoms, it may actually be possible, using a suitable syntax, to write down more than the first 10[SUP]80[/SUP] decimals of π.

---

I now also realise that the notion of an observable finite constant, as I defined it, is not well defined. To explain this, note that the observably finite constant should be unique, and hence satisfy the following property:

If F is the observable finite constant and n is an integer with F < n then there exists at least one digit of n that cannot be computed.

Yet there exists a sufficiently large integer k such that 10[SUP]k[/SUP] > F, and every digit of 10[SUP]k[/SUP] can be easily computed: the i-th digit of 10[SUP]k[/SUP] is 1 when i = 1, and 0 for all other i.

---

Let n be a very large integer such that there is at least one digit of n, say the j-th digit, that cannot be computed due to insufficient material or computational resources in the Universe. Hence we cannot uniquely define n because if one asks, "what is the j-th digit of n," then it is impossible to compute that digit. If one person claims that the j-th digit of n is 3, and another person claims that the j-th digit of n is 8, it is impossible to settle their argument.

 

[Keegan]

E = MC^2

enough said.

 

[Jehosefat]

The bit I wrote about using sub-atomic particles was not meant to be taken seriously :erm:

Let n be a very large integer such that there is at least one digit of n, say the j-th digit, that cannot be computed due to insufficient material or computational resources in the Universe. Hence we cannot uniquely define n because if one asks, "what is the j-th digit of n," then it is impossible to compute that digit. If one person claims that the j-th digit of n is 3, and another person claims that the j-th digit of n is 8, it is impossible to settle their argument.

Just because n cannot be calculated in practice does not mean that it does not exist nor does it mean that it is not unique. But then again, a really, really, really large n has absolutely nothing to do with the concept of infinity nor its usefulness. Practically, defined/calculable numbers above a certain threshold are not useful however the concept of infinity is a necessary foundation for a large chunk of mathematics that does serve a practical purpose.

For example, the fact that we can approximate π to an arbitrary level of precision is largely due to the fact that we can express it as a series that we can prove is exactly equal to π when the number of terms is ∞. Without the concept of infinity, we could not prove that the series converges to exactly π and then any approximation made using a portion of the series would have, at best, questionable accuracy.

 

[RiaX]

lol when humberto stops asking why he will get a nobel prize or something.

 

[ponder]

E = MC^2

enough said.

Apparently there might be a problem with that...

 

[bullzeye.za]

You're working under the assumption that using a base 10 number system is the correct way to go about things.

Consider instead using base pi, or base e. Then we would be left with one less irrational number.

It'd take a whole rewrite of our number system, for sure - But there is considerable value in it for mathematics, and it's actually why we avoid expressing numbers where pi is involved in any other way than nπ

 

[battletoad]

You're working under the assumption that using a base 10 number system is the correct way to go about things.

Consider instead using base pi, or base e. Then we would be left with one less irrational number.

It'd take a whole rewrite of our number system, for sure - But there is considerable value in it for mathematics, and it's actually why we avoid expressing numbers where pi is involved in any other way than nπ

Taking this idea further (although I think Jehosefat provided a solid explanation)

Assuming base 10.

Let n be the least incalculable integer; its first incalculable digit k in say the j-th position, is at the least digital position among all incalculable integers, i.e. every other incalculable number has its first incalculable digit at a later progression in its decimal expansion when compared to n.

Now, if we represent n in say base 17 (keeping it prime and all), k now becomes "part" of the i-th digit where i<<j. We can calculate digits in the (i-1)-th, i-th and (i+1)-th positions by your argument for the fundamental limit on the universe, hence we can calculate k's value by investigating these i's. This means that k is calculable, a contradiction.

E.g. 13568 base 10 = 2CG2 base 17 where C=12 base 10 and G=16 base 10. Point is, you need an equal or less amount of digits to represent numbers in base 17 as opposed to base 10.

You could take your argument further by considering all number bases less than this least incalculable integer, but then you can always make squiggles or whatever to represent numbers, colours even which to my knowledge there are infinitely many of. Because ultimately, numbers are used to make sense of things.

Side note: I always wondered what the most efficient, simple number base was. I guess it depends on what one is working on.

 

[Humberto]

Definition: Call an integer n incalculable in base k if there is a digit of n that cannot be computed.

Conjecture: If an integer is incalculable in any base, then it is incalculable in every base.

 

[Humberto]

You're working under the assumption that using a base 10 number system is the correct way to go about things.

Consider instead using base pi, or base e. Then we would be left with one less irrational number.

It'd take a whole rewrite of our number system, for sure - But there is considerable value in it for mathematics, and it's actually why we avoid expressing numbers where pi is involved in any other way than nπ

In base pi though, I think we'd introduce other irrational numbers that are not irrational in base 10. For example, I think the number 1 would become irrational in base pi, but I'm not sure of this.

 

[Humberto]

Let n be the least incalculable integer; its first incalculable digit k in say the j-th position, is at the least digital position among all incalculable integers, i.e. every other incalculable number has its first incalculable digit at a later progression in its decimal expansion when compared to n.

Now, if we represent n in say base 17 (keeping it prime and all), k now becomes "part" of the i-th digit where i<<j. We can calculate digits in the (i-1)-th, i-th and (i+1)-th positions by your argument for the fundamental limit on the universe, hence we can calculate k's value by investigating these i's. This means that k is calculable, a contradiction.

I suspect though that if the j-th digit of n written in base 10 is incalculable, then so is the i-th digit of n written in base 17. In that case, there isn't yet a contradiction.

By way of example, suppose we had a binary number N = 1101k1 with k not known. N written as a decimal number is

1 x 2[SUP]5[/SUP] + 1 x 2[SUP]4[/SUP] + 0 x 2[SUP]3[/SUP] + 1 x 2[SUP]2[/SUP] + k x 2[SUP]1[/SUP] + 1 x 2[SUP]0[/SUP] = 53 + 2k.

Hence the second digit of N in decimal notation is either 3 or 5. The unknown 5-th digit of N in binary notation corresponds to an unknown 2-nd digit of N in decimal notation. Without knowing the 5-th digit of N in binary notation, we cannot compute the 2-nd digit of N in decimal notation.