I have heard it stated before that Graham's number is so vast that it is completely beyond comprehension. It is way larger than the number of atoms in the universe, so cannot be related to real quantities, and unfathomably larger than even hypothetical quantities, such as the number of Planck unit-sided cubes that could fit in the observable universe.

People usually stop at this point, and state that it is simply beyond comprehension, which is fair enough. However, I want to know more about what numbers are within comprehension, and what "intutitional" devices can be used to comprehend them. For instance, I considered combinatorial analogies, such as the number of different ways there are of selecting half the atoms in the universe (which contains $10^{80}$ atoms):

$$10^{80} \choose 5 \times 10^{79}$$

or the number of different ways there are of visiting every single atom in the universe exactly once ("universal atom tours"?):

$$10^{80}! \over 2$$

or even, the number of ways of selecting half the universal atom tours!

$${10^{80}! \over 2} \choose {10^{80}! \over 4}$$

Clearly we can continue in this manner, sacrificing semantic succinctness and intuitive meaning for further gains in magnitude. Even if, as I imagine is the case, these quantities barely scratch the surface of Graham's number, I suppose I am asking two things:

  • How can the magnitudes of extremely large constructs like the ones I've given be evaluated and compared?

  • Even if Graham's number is out of reach, what are some particularly effective 'intuitional devices' for explaining the scale of extremely large quantities to laymen like myself?


  • 4
    $\begingroup$ "[Graham's number]," it says, "is big. Really big. You just won't believe how vastly, hugely, mindbogglingly big it is. I mean, you may think it's a long way down the road to [a googolplex], but that's just peanuts to [Graham's number], listen..." $\endgroup$ – Qiaochu Yuan Jul 14 '13 at 18:10
  • $\begingroup$ See also math.stackexchange.com/questions/2497/… $\endgroup$ – John Bentin Jul 14 '13 at 21:19
  • 1
    $\begingroup$ See if you can get a copy of Asimov's essay "Skewered!" about making sense of Skewes' number (also phenomenally large). He goes through a process similar to the one you begin above. Hint: he ultimately fails. $\endgroup$ – Ryan Reich Jul 21 '13 at 4:40
  • $\begingroup$ @ryan-reich I can't get hold of "Skewered!" sadly, but according to my evaluation the number of ways of selecting half the universal atom tours is larger than Skewes' number. It is at least in the same ballpark (I think), which suggests Skewes' number is within reach of intuition (or at least, the stretched form of intuition I speak of in the OP). $\endgroup$ – timxyz Jul 22 '13 at 10:42

I'm about to prove that Graham's number is a number that is so big there is no way of intuitively grasping the magnitude of it, but in order to do this we must start with the very small and work our way up.

We will start with the number 9 which can represented as 3+3+3. This isn't so bad when you only have three 3's to write down, but what happens if I pick a number that isn't practical to be represented by +3's, like 729, I would need the sum of 243 3's, so now let's use multiplication. 729 can now be represented as 3*3*3*3*3*3 now I can represent larger numbers than before, but I run into the same problem. I can pick a number that isn't practical to be represented by the repetition of the same operator of multiplication.

7,625,597,484,987, for example, is the multiplication of 27 3's. That isn't good way to represent this number so now let's use powers to represent this number which is $3^{3^3}$, another way to represent this is $3 \uparrow 3 \uparrow 3$ where each single arrow is "to the power of operator".

Mathematicians realized when dealing with numbers that aren't practical to write down in one operator, a new operator is required that is more powerful the previous one. Arrow notation is a way of going from one operator to another with ease. $\uparrow \uparrow$ is the next operator up from $\uparrow$, just as $\uparrow$ is the next operator up from multiplication, and just as multiplication is one operator up from addition. So increasing the number of consecutive arrows will increase the ability to work with larger and larger numbers.

So we saw how the size of numbers that were created when using $\uparrow$. Now let's look at the $\uparrow \uparrow$. $3\uparrow \uparrow3\uparrow \uparrow3$ = $3\uparrow \uparrow(3\uparrow3\uparrow3)$=$3\uparrow \uparrow7,625,597,484,987$. Which means this is $\uparrow3$ is used on 3, 7,625,597,484,987 times!

${{{{{{{{{{3}^3}^3}^3}^3}^3}^3}^3}^3}^\cdot\Rightarrow}$ a stack of threes 7,625,597,484,987 high!

To calculate this value start with the top 3 and move downward.
$3^3 = 27$
$3^{7,625,597,484,987}$ is 3,638,334,640,024 digits long.
The fourth 3 is very large, but googolplex is 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 digits long. So still somewhat in our comprehension, but the fifth 3 is way out of our comprehension.
3 is raised to the power of a number that is 3,638,334,640,024 digits long.
Googolplex is 10 raised to the power of a number that is 101 digits long.
So we have used only 5 of the 3's on the 7,625,597,484,987 stack and it already becomes difficult to imagine, but this process is repeated over 7.6 trillion times. Doing this process 5 times took us from 3 to a number that makes googolplex look tiny. So the answer to the stack of 7,625,597,484,987 3's is a stupidly big number. This number is so big that if you memorized all the digits of this number your head would turn into a black hole. The maximum amount of entropy that can be stored in your head carries less information than the information of this stack of 3's.
[ http://m.youtube.com/watch?v=XTeJ64KD5cg ]

$$ \begin{array}{c|l|c|r} \text{operator representation} & \text{value} & \text{number of previous operator equivalent} \\ \hline 3*3*3 & 27 &\text{9 (+3)'s or 3+3+3+3+3+3+3+3+3}\\ 3\uparrow3\uparrow3& 7,625,597,484,987 &\text{27 (x3)'s or $3^{27}$} \\ 3\uparrow\uparrow3\uparrow\uparrow3 & \text {stupidly big} & \text{7,625,597,484,987 ($\uparrow$ 3)'s}\\ 3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3 & G_1 & \text {stupidly big ($\uparrow\uparrow3$)'s} \\ \end{array} $$

If you look at the table above you will notice that $3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3=G_1$.
Why is $G_1$ important? Well I'll get to that in a minute. But first I have to explain that each change from one operation to the next is unimaginably bigger than previous change so you would think that adding a few arrows would quickly get us to Graham's number, but it doesn't. It doesn't even come close.

So what if I were to put a subscript on the arrows to indicate how many arrows I have. So $3\uparrow_{1000000}3$ is one million arrows, so I am one million operations above multiplication and 999998 operations above "stupid big" and the value of each operation is unimaginably bigger than the previous one. So, surely I must have hit Graham's number by now. The answer is no.

Remember $G_1$? Well, I'm going to write this, $3\uparrow_{G_1}3$.

So let's see if I can break this down.

"stupidly big" made googolplex look tiny after 5 of 7,625,597,484,987 iterations. Applying $\uparrow\uparrow3$ to 3 "stupidly big" times gives me $G_1$ and $G_1$ is now going to be the number of times the operator is increased. Where each operation is unimaginably bigger in comparison to the previous one. So how well do I stand against Graham's number at this point? Not even close.


$G_{64}$=$3\uparrow_{G_{63}}3=\text{Graham's number}$

  • 1
    $\begingroup$ I think that you just turned me into an ultra-finitist. $\endgroup$ – Baby Dragon Jul 21 '13 at 4:57
  • 1
    $\begingroup$ AMAZING!!!! Too good a description. This answer deserves a $G_{64}$ number of upvotes. (obviously exaggerating!) $\endgroup$ – Parth Thakkar Jul 21 '13 at 10:56
  • $\begingroup$ Great answer. Seems silly now to have ever considered Graham's number as accessible to intuition! I still wonder what its limits are however, and what functions particularly effective intuitional devices would take advantage of. $\endgroup$ – timxyz Jul 22 '13 at 10:48
  • $\begingroup$ So why stop at $G_{64}$? What about $G_{G_{64}}$? or $G_{G_{G_{64}}}$? And what if we increase the supscript-stack up to $G_{64}$ layers? You will get to even unbelievable bigger numbers $\endgroup$ – Hubert Schölnast Jul 3 '15 at 9:56
  • $\begingroup$ @HubertSchölnast in terms of fast growing hierarchy (my answer below) you will find that regardless of how you layer it, you'll never beat, say, $f_{\omega^\omega}(n)$. Indeed, this is so much larger than layers of layers of layers of $G$... Or even if we stack the layers $G64$ times... You'll never get there. Indeed, if you have heard that, say, TREE(3) is beyond comprehensive power...the fast growing hierarchy will prove you wrong :-) $\endgroup$ – Simply Beautiful Art Mar 8 '17 at 11:59

Imagine all of the atoms in the universe, we dont know how many their are, but lets say it's x. Now if each atom in the universe contained another universe inside, and each atom inside that contained yet another universe, and that pattern continued on x amount of times. If you added up the total number of atoms in all of the universes then you would still be less than the googleplexth root of grahams number. :)


I'm going to take this one step further than quantus14. Once you reach Graham's number, the question then becomes

How to effectively compare and understand even larger numbers?

Well, the first thought should be to see if you can find a similar number to yours on the Googology Wiki, which is dedicated entirely to large numbers and their study.

My second recommendation is to learn fast growing hierarchy (FGH), which has a similar construction to Knuth's up-arrow notation, but it extends much further. Here is a good YouTube series.

We define the first two rules of FGH as follows:

  1. $f_0(n)=n+1$

  2. $f_{\alpha+1}(n)=\underbrace{f_\alpha(f_\alpha(\dots f_\alpha(n)\dots))}_{n\text{ amount of }f_\alpha's}$

A quick demonstration will show you how fast this grows:



$$\begin{align}f_2(160)&=\underbrace{f_1(f_1(\dots f_1(160)\dots))}_{160}\\&=5\times2^{165}\\f_3(5)&=f_2(f_2(f_2(f_2(f_2(5))))\\&=f_2(f_2(f_2(f_2(160)))\\&=f_2(f_2(f_2(f_2(f_2(5))))\\&=f_2(f_2(f_2(5\times2^{165}))\\&=f_2(f_2(\underbrace{f_1(f_1(\dots f_1(5\times2^{165})\dots))}_{5\times2^{165}})\end{align}$$

Indeed, you will find these numbers quickly exceed anything you could hope to write on a sheet of paper.

We also have approximations to your numbers, and by a quick check, they are around the range of $f_4(n)$ for some small values of $n$.

...and then we have rule 3:

  1. $f_\alpha(n)=f_{\alpha[n]}(n)$

This rule is a little bit confusing, so let me explain:

$\alpha$ is a limit ordinal, which basically translates into the limit of some sequence of numbers (a list of numbers (or a list of lists) if you are a programmer):


We first define $\omega$ to be the limit of the following sequence:


We then have $\omega[5]=5$ as the fifth term of this sequence, so


And from there we know this is very large...much larger than the previous numbers.

To deal with $\omega+1$, we apply the second rule, followed by the third:


So you can see how large this gets, and it gets pretty big. Graham 's number has a nice approximation:

$$f_{\omega+1}(64)\approx\text{Graham 's number}$$

So this is rather simple and

$$f_{\omega+1}(64)=\underbrace{f_\omega(\dots f_\omega(64)\dots)}_{64}$$

Note that we can even go further:


For example:


And we would then expand further and further and further...

And we have some other larger ordinals:


And beyond!


While it's not very... visual, it does become clear that you can make some very very big numbers while also being able to compare large numbers. One of the great advantages of FGH is the clearness of how to compare Graham's number to another number.

And the only thing holding you back from producing even larger numbers is how you want to define even larger ordinals.

And one last example:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.