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?

  • 4
    $\begingroup$ Graham's number is much much bigger. As in, take 3^3^3^3^... for the rest of your life, and you aren't even remotely close in any sense. $\endgroup$
    – user233746
    Dec 18 '16 at 13:21
  • 1
    $\begingroup$ Better yet, have everyone alive from now until the universe ends, help you take 3^3^3^3^... as fast as possible. You still won't be able to reach Grahams number. $\endgroup$
    – user233746
    Dec 18 '16 at 13:27
  • 1
    $\begingroup$ Even iterating the number of $3$'s in the power tower $3\uparrow3 \uparrow\cdots \uparrow 3$ a googol times (starting with $3^{3^3}$) will not get even close to Graham's number $\endgroup$
    – Peter
    Dec 19 '16 at 11:40
  • $\begingroup$ This isn't really rigorous, but I use as a heuristic that pretty much any number you can reasonably define in a "normal" way (i.e., without invoking set theory or new operations or anything like that) is WAY smaller than Graham's number, and should never be compared to it. But by all means, let me know if there is some such number greater than Graham's number- that would be interesting. $\endgroup$ Jan 6 at 15:49

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.

  • $\begingroup$ After the first equality sign It should be $3^{3\times 10^{100}}$ instead of $3^{3\times 10^{10^{100}}}$. $\endgroup$
    – timon92
    Dec 18 '16 at 9:33
  • $\begingroup$ @timon92 Thank you for the correction. $\endgroup$
    – user371838
    Dec 18 '16 at 9:36
  • 2
    $\begingroup$ The inability to even conceive what $3\uparrow\uparrow\uparrow\uparrow3$ is should be enough. Or even drop an arrow. $\endgroup$ Dec 18 '16 at 13:43

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