# Is there a number so large that we could never calculate it?

Note that I edited this post significantly to make it more clear (as clear as I think I could possibly make it). First, let me mention what I am NOT asking:

• I am NOT asking for the largest number we could calculate.
• I am NOT asking if there is a largest number (if there were, just add one and it would be bigger, so obviously this is false)
• I am NOT asking for something like "divide by zero" or for limits ("infinity" isn't an answer, that isn't a number)

I have been thinking a lot about numbers and I put a couple numbers into Wolfram Alpha just to see what happens. It can handle insanely huge numbers, but it fails as soon as you put in $10! ^ {10!}$. After thinking for a few days, I feel that there must be a largest number that we could possibly ever calculate or define. In fact, I think there are an infinite number of numbers greater than anything we could ever calculate (I am saying larger numbers exist, but we cannot do anything with them, except maybe prove there are numbers so large we can't calculate them).

I mean a number or an equation that results in an exact number (such as $x!^{x!^{x!^{x!^{x!^{...}}}}}$, where $x$ is an exact number that can be written out) I think there is a limit here. There is a largest number possible because there are only so many atoms in the observable universe. But how would one prove this (that there are numbers so large we couldn't define them)? Is there any information on this?

Interestingly, of the irrational numbers we use, they always have an alternative way to write them such that the answer is exact. One can simply write the infinitely long irrational number that is the square root of two as $\sqrt 2$. Two symbols and you define the infinitely long number that is the square root of $2$. One can use the formulas for finding $\pi$. But aren't there numbers that we couldn't even do that for? Aren't there numbers that we couldn't write in some form, such as how we write $\sqrt 2$? Such that the number is exact? I think these numbers must exist, but we can't do anything with them. But I have no idea how one could prove that or if there is any information regarding these numbers. So my question is: are there numbers that are just too large (or too precise) that we can't write them down in any way at all that expresses their exact value?

UPDATE:

I accidentally came across something that really helps with my question. I found this article here, and I saw point #6:

Unknowable Thing: There are numbers that can’t be computed.

This is another mind bender proved by Alan Turing.

That is exactly what I'm talking about! I am going to do some research on Alan Turing!

• This might be of interest: youtube.com/watch?v=GuigptwlVHo Also this link is good: scottaaronson.com/writings/bignumbers.html – Colm Bhandal Jul 20 '15 at 12:11
• What do you mean by prove? When are we not able to calculate? Again, check out the video on Graham's number and the page I provided in my other comment. To this day, it is not possible to write Graham's number down nor can you write down how many digits are in Graham's number (and if that's not big enough for you: you can use the ackermann function to get way bigger). – Hirshy Jul 20 '15 at 12:17
• I've always been fond of the pseudo-code which reads "Let N be the least positive integer too large to be defined by a program as short this one." The fact that this clearly can't define a number tends to suggest that our notion of what it means to define a number is at least a little hazy. – lulu Jul 20 '15 at 12:23
• All incomputable number (such as Busy Beaver numbers) are such that you can't compute them in any computable (algorithmic) sort of way. – orion Jul 20 '15 at 12:58
• It can be shown that in the context of ordinary mathematics (say ZFC) there are infinitely many well-specified positive integers whose decimal representations cannot be proved. E.g., for every $n \ge 10\uparrow\uparrow 10$, the Busy Beaver number $\Sigma(n)$ is well-defined and has some decimal representation $d_1d_2...d_k$, but there exists no proof that $\Sigma(n) = d_1d_2...d_k$. It isn't that the proof or the digit string is merely infeasible due to physical resource limitations; rather, such a proof is a logical impossibility. – r.e.s. Jul 24 '15 at 14:16

It can be shown that in the context of ordinary mathematics (say ZFC) there are infinitely many well-specified positive integers whose numerical representations cannot be proved. E.g., for every $n \ge 10\uparrow\uparrow 10$, the Busy Beaver number $\Sigma(n)$ is well-defined and has some decimal representation $d_1d_2...d_k$, but there exists no proof that $\Sigma(n) = d_1d_2...d_k$. It isn't that the proof or the digit string is merely infeasible due to physical resource limitations; rather, such a proof is a logical impossibility.

Here are a few relevant online sources:

NB: In connection with the computability of numbers, note that an uncomputable number cannot be an integer (because each integer has a purely finite representation, unlike the situation for real numbers). That's why the "computable-but-unprovable" results mentioned above seem especially poignant, since they apply specifically to positive integers, without complicating the situation with infinite objects such as the digital representations of uncomputable real numbers.

In a completely different (and much more mundane) sense, a digital representation of a positive integer can be "too big to calculate" for reasons of physical infeasibility implied by the assumed laws of physics:

• An absolute upper bound on any computer's operational speed is $1/t_{Planck} = \sqrt{\frac{c^5}{Gh}}\ \lessapprox\ 2\cdot 10^{43}\ \tt{bits}\ \tt{per}\ \tt{second}.$
• An absolute upper bound on any computer's storage capacity is
$Volume_{observable\ universe} /l^3_{Planck}\ \lessapprox\ 9 \cdot 10^{184} \ \tt{bits}.$

See the Wikipedia article on Physical limits to computation, and also the absolute bounds mentioned in the external weblink provided in the article on Bremermann's limit.

• I feel this answer is kind of misleading. Not being able to prove the decimal expansion of a number from some definition doesn't really have anything to do with size. For instance, define a number $x$ which is $1$ if ZFC is consistent and $0$ if it is inconsistent. Then this is a uniquely defined natural number, but ZFC (assuming it is consistent) cannot determine what it is. – Eric Wofsey Jul 28 '15 at 15:18
• @EricWofsey - I would rather say "Not being able to prove the decimal expansion of a number from some definition doesn't necessarily have anything to do with size." For your definition of $x$ it may not, but for the definition of $\Sigma(10\uparrow\uparrow 10)$ it evidently does (as per the sources cited above). – r.e.s. Jul 28 '15 at 19:50
• @r.e.s. I'm a little late to the game but just wanted to point out that you can now use just the 8000th busy beaver per scottaaronson.com/blog/?p=2725. (You already commented on a post there saying this recently so I know you already know but I wanted to make this more visible since I find it pretty amazing). – exfret Sep 5 '17 at 5:30

You need to define what you mean by calculating a number. I remember finding a website with a program to "calculate a googleplex". What the program really did was output a $1$ followed by $10^{100}$ zeros. Is that calculating a googleplex? Do you really have to multiply $10$ by itself that many times? If you make a rigorous definition, you fall afoul of the Berry paradox. If you don't, the question does not have an answer. Given a way to calculate $N$, I can calculate $N+1$ in the obvious way. When you move from integers to reals, you have the additional result that almost all reals cannot be defined, simply because there are only countably many definitions and there are uncountably many reals.

• I realize how my question isn't the most clear (the barrage of responses clued me in on that). It is difficult to formulate what I am thinking. I know the $N$ and $N + 1$ issue, but I'm trying to think of a way to ask.. Are there numbers that are just so large that we can't calculate them? To put down a formula or number for $N$ would require ever bit of the universe, then you wouldn't have enough for $N + 1$ (assuming you can't change come numeral, maybe 7 to 8, somewhere, and assume there is not a more compact formula). – zagadka314 Jul 25 '15 at 19:07
• I think you are getting the idea of my question with the reals. There are only countable many definitions, but there are more reals than that. But how does one prove that? And does that mean what I am indirectly asking? There are numbers between 0 and 1 that we just can't "access." They can't be defined (nor calculated or any other way we migh access them), it just isn't possible for us to do that. My indirect question is: are there numbers we can't access? It is difficult to ask my question. – zagadka314 Jul 25 '15 at 19:09
• You prove that there are only countably many definitions by noting that a definition is a finite string of symbols from a finite alphabet. There are a countably infinite set of these. Some (most) do not define a real, but once you prove the reals are uncountable you know almost all reals are not definable. To make a mathematical argument about numbers we cannot access, you need to define what it means to access them. This is hard. – Ross Millikan Jul 26 '15 at 3:37
• To do it right, you get deep into model theory and need to pay attention to the difference between first order and second order logic, which sounds beyond you now. If you understand the Löwenheim–Skolem theorem you are on the way. – Ross Millikan Jul 26 '15 at 3:38
• I haven't gotten into second order logic, but I am a good way through my first order logic class and book. I only have a couple chapters left. I want to continue studying logic and mathematics, then hopefully, later this will make more sense to me. – zagadka314 Jul 26 '15 at 7:59

A necessary condition for being able to calculate a number is that we can specify a computation whose result is that number.

If we accept r.e.s.'s figure that the universe cannot contain more than $10^{185}$ bits, this should mean that $$\text{the first number whose Kolmogorov complexity is}\ge 10^{185}+1$$ cannot be produced in any deterministic way, in any effective notation.

(It's not just that if we have that number in our hands, we cannot prove that it answers to the above description -- though that is certainly true too. It's that it's impossible to "have that number" in any meaningful way, even without knowing it satisfies that definition).

For calculations, it is the computer memory and how numbers are stored that is the limit.

Simple programs use 32 bit or 64 bit integers, while complex program use string-integers. But even string-integers depend on how much memory is available.

The number $10!^{10!}$ would require about 52 Mb memory to store the number, not to mention the auxiliary numbers that could appear in a recursion step.

The number $10!^{10!}$ has about $54810892$ digits in the number, so yes - such numbers require much memory.

A number like $15!^{15!}$ would require about $33 Tb$ - think about it 33 TERABYTE! Most computers could not even store such a number on the HD!!!

• You are assuming that we represent numbers as a binary bitstring. One can define other representations that are much more compact for some large numbers (while being less compact for others). $10!^{10!}$ is a fine representation of a number, which does not use very many bytes at all. This is the point of my claim that we have to define what it means to compute a number. $15!^{15!}$ is not so big. You are right that my hard drive could not store it in binary, but I could easily buy enough hard drives to do so. $15^{15!^{15!}}$ on the other hand.... – Ross Millikan Jul 25 '15 at 5:26
• @RossMillikan, true, but compact representations require more processing time. And yes, when we use compact number we save some space. If we use bits, then $15!^{15!}$ would be about 5 Tb - still large. – johannesvalks Jul 25 '15 at 5:31
• OP didn't talk about processing time. The problem with more compact representations is that they can only represent a few of the numbers with substantially fewer bits than straight binary-you can get that by counting how many shorter strings there are. If I use $15!^{15!}$, that represents a large number in just a few bits, but the bit pattern it represents will be stolen from representing another number. – Ross Millikan Jul 25 '15 at 13:46

Graham's number.

Graham's number is much larger than many other large numbers such as a googol, googolplex, Skewes' number and Moser's number. Indeed, like the last two of those numbers, the observable universe is far too small to contain an ordinary digital representation of Graham's number, assuming that each digit occupies one Planck volume. Even power towers of the form $a ^{ b ^{ c ^{ \cdot ^{ \cdot ^{ \cdot}}}}}$ are insufficient for this purpose, although it can be described by recursive formulas using Knuth's up-arrow notation or equivalent, as was done by Graham. The last 12 digits of Graham's number are ...262464195387.

• I already saw references to this in the comments, though this does not answer the actual question, as the number may indeed be written in arrow notation. – zagadka314 Jul 25 '15 at 19:01
• en.wikipedia.org/wiki/Computable_number "Countable but not computably enumerableedit While the set of real numbers is uncountable, the set of computable numbers is only countable and thus almost all real numbers are not computable." So yes, there are many numbers that fit your criteria. For obvious reasons, we cannot give examples of them. – Race Bannon Jul 25 '15 at 19:06
• I think you have given me one of the best links so far. This is what I'm trying to get at! I just apparently fail at asking questions! – zagadka314 Jul 25 '15 at 19:11
• What part of mathematics studies the idea of definable and countable numbers? – zagadka314 Jul 25 '15 at 19:18
• My guess would be computer science and mathematical logic. Specifically, the branch of logic dealing with computability. – Race Bannon Jul 25 '15 at 19:24