big numbersnumber-comparison
See YouTube or wikipedia for the defination of Graham's number.
A Googol is defined as $10^{100}$.
A Googolplex is defined as $10^{\text{Googol}}$.
A Googolplexian is defined as $10^{\text{Googolplex}}$.
Intuitively, it seems to me that Graham's number is larger (maybe because of it's complex definition).
Can anybody prove this?
OK, let's attempt a sorting of the names having at most three operands. I'll make several observations, and then use them to assemble the order section by section, beginning with the part below googol stack.
googol bang bang bang $\lt$ googol stack. It seems clear that we shall be able to iterated bangs many times before exceeding googol stack. Since googol bang bang bang is the largest three-operand name using only plex and bang, this means that all such names will interact only with each below googol stack.
plex $\lt$ bang. This was established in the question.
plex bang $\lt$ bang plex. This was established in the question, and it allows us to make many comparisons in terms involving only plex and bang, but not quite all of them.
googol bang bang $\lt$ googol plex plex plex. This is because $g!!\lt (g^g)^{g^g}=g^{gg^g}=10^{100\cdot gg^g}$, which is less than $10^{10^{10^g}}$, since $100\cdot gg^g=10^{102\cdot 10^{100}}\lt 10^{10^g}$. Since googol bang bang is the largest two-operand name using only plex and bang and googol plex plex plex is the smallest three-operand name, this means that the two-operand names using only plex and bang will all come before all the three-operand names.
googol plex bang bang $\lt$ googol bang plex plex. This is because $(10^g)!!\lt ((10^g)^{10^g})!=(10^{g10^g})!=(10^{10^{g+100}})!\lt (10^{10^{g+100}})^{10^{10^{g+100}}}=10^{10^{g+100}10^{10^{g+100}}}= 10^{10^{(g+100)10^{g+100}}}\lt 10^{10^{g!}}$.
Combining the previous observations leads to the following order of the three-operand names below googol stack:
googol
googol plex
googol bang
googol plex plex
googol plex bang
googol bang plex
googol bang bang
googol bang bang
googol plex plex plex
googol plex plex bang
googol plex bang plex
googol plex bang bang
googol bang plex plex
googol bang plex bang
googol bang bang plex
googol bang bang bang
googol stack
Perhaps someone can generalize the methods into a general comparison algorithm for larger smallish terms using only plex and bang? This is related to the topic of the Velleman article linked to by J. M. in the comments.
Meanwhile, let us now turn to the interaction with stack. Using the observations of the two-operand case in the question, we may continue as follows:
googol stack plex
googol stack bang
googol stack plex plex
googol stack plex bang
googol stack bang plex
googol stack bang bang
Now we use the following fact:
The order therefore continues with:
googol plex stack
googol plex stack plex
googol plex stack bang
The order therefore continues with:
googol bang stack
googol bang stack plex
googol bang stack bang
Thus, the order continues with:
googol plex plex stack
googol plex bang stack
googol bang plex stack
googol bang bang stack
This last item is clearly less than googol stack stack, and so, using all the pairwise operations we already know, we continue with:
googol stack stack
googol stack stack plex
googol stack stack bang
googol stack plex stack
googol stack bang stack
googol plex stack stack
googol bang stack stack
googol stack stack stack
Which seems to complete the list for three-operand names. If I have made any mistakes, please comment below.
Meanwhile, this answer is just partial progress, since we have the four-operand names, which will fit into the hierarchy, and I don't think the observations above are fully sufficient for the four-operand comparisons, although many of them will now be settled by these criteria. And of course, I am nowhere near a general comparison algorithm.
Sorry for the length of this answer. Please post comments if I've made any errors.
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:
The Wikipedia article on Σ, complexity and unprovability.
This answer to the math.SE question How can Busy beaver($10 \uparrow \uparrow 10$) have no provable upper bound?.
Pascal Michel's webpage on Busy beavers and unprovability.
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:
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.
Best Answer
Googolplex can be bounded from above like a tower of exponents, so $$10^{10^{100}} < (3^3)^{10^{100}} = 3^{3\times 10^{100}} <3^{10^{101}} < 3^{(3^3)^{101}} = 3^{3^{303}} < 3^{3^{3^{3^3}}}$$ In the last step, we have used the fact that $303$ is much,much smaller than $3^{27}$. Now, take the Googolplexian.We can thus, easily check that, $$10^{10^{10^{100}}} < (3^3)^{10^{10^{100}}} < 3^{3^{3^{3^{3^3}}}}$$ So, Googolplexian is much smaller than a tower of exponents of $3$'s of length $6$, or in other words Googolplexian is less than $3\uparrow \uparrow6$.(using Knuth's up-arrow notation.)
Now, compare this with just the first layer of Graham's number,i.e., $3\uparrow \uparrow \uparrow \uparrow 3$. Hope it helps.