# 10 enormous numbers

There's a story (of disputed veracity) that some primitive cultures have limited counting ability, only having words for "one", "two", and "many."

XKCD 764, Randall Munroe

But in a sense, this is just a matter of degree - it is surprisingly easy to find numbers of such inexpressible collossalness that our brains simply have to resort to bundling them into a single category: "many".

Who can name the bigger number?
You have fifteen seconds. Using standard math notation, English words, or both, name a single whole number—not an infinity—on a blank index card. Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named, by consulting only your card and, if necessary, the published literature.
- Scott Aaronson

There's a story (of disputed veracity) that some primitive cultures have limited counting ability, only having words for "one", "two", and "many."
I, too, question some of this linguistic analysis. One of these guys said that since our words "first" and "second" don't fit the format of "third, fourth," etc. (i.e., they are not formed by inflecting the word for the number itself), this implies that our ancestors could only count to two.

He's not much of a linguist if he didn't bother to look up the etymologies of the two words. "First" is the superlative inflection of "fore," meaning "the one that is in front." "Second" is taken from French and ultimately from Latin secundus, "the one that follows." These words were clearly invented as ordinals and were never meant to have any etymological relationship to the cardinal numbers one and two.

You have fifteen seconds. Using standard math notation, English words, or both, name a single whole number—not an infinity—on a blank index card. Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named, by consulting only your card and, if necessary, the published literature.
((((A googolplex to the googolplex power) to the googolplex power) to the googolplex power) to the googolplex power) to the googolplex power . . . . . until 12 of my 15 seconds run out, giving me time to make sure the parentheses are paired.

As far as I know, "googolplex" -- 10^(10^100), ten to the googol power -- is the largest number for which there is an English word. So it's reasonable to start with that.

If I'm not mistaken, we don't need a word for a larger number because this one is very much larger than the total number of subatomic particles in the universe.

((((A googolplex to the googolplex power) to the googolplex power) to the googolplex power) to the googolplex power) to the googolplex power . . . . . until 12 of my 15 seconds run out, giving me time to make sure the parentheses are paired.

As far as I know, "googolplex" -- 10^(10^100), ten to the googol power -- is the largest number for which there is an English word. So it's reasonable to start with that.

If I'm not mistaken, we don't need a word for a larger number because this one is very much larger than the total number of subatomic particles in the universe.
Agreed on the use of the word. My number was "googolplex(up-arrowed googolplex times)" but I didn't want to bother trying to figure out the tex to write it.

Because of the laws of exponents and they way you have your parentheses, you miss out on getting to truly large numbers.

((googolplex ^ googolplex) ^ googolplex) ^ googolplex
= googolplex^(googolplex^3)
= (10^(10^(10^2)))^( (10^(10^(10^2)))^3 )
= (10^(10^(10^2)))^(10^(3×10^(10^2)))
= 10^( 10^(10^2) × 10^(3×10^(10^2)) )
= 10^( 10^( 3×10^(10^2) + 10^2 ) )

And if you have N googolplexes in your expression, FR(N) then log log log log FR(N) = log log ( (N-1)×10^(10^2) + 10^2 ) ≈ log ( log (N-1) + 10^2 ) ≈ 2

While
$$\HUGE 10^{10^{10^{10^{10^{10}}}}}$$ is much larger (and easier to write). log log log log 10^(10^(10^(10^(10^10)))) = 10,000,000,000

Knuth's uparrow notation $$10 \uparrow \uparrow 6$$ is the same number.
So we can write $$10 = 10\uparrow \uparrow 1 \lt 10 \uparrow \uparrow 2 \lt \textrm{googol} \lt 10 \uparrow \uparrow 3 \lt \textrm{googolplex} \lt 10 \uparrow \uparrow 4 \lt FR(2) \lt FR(10^9) \lt 10 \uparrow \uparrow 5 \lt 10 \uparrow \uparrow 6 \lt 10 \uparrow \uparrow 10 = 10 \uparrow \uparrow \uparrow 2$$

http://mathworld.wolfram.com/PowerTower.html

Agreed on the use of the word. My number was "googolplex(up-arrowed googolplex times)" but I didn't want to bother trying to figure out the tex to write it.
Did you mean $$\left( 10^{10^{10^2}} \right) \uparrow \uparrow \left( 10^{10^{10^2}} \right) = \left( 10^{10^{10^2}} \right) \uparrow \uparrow \uparrow 2$$ ?

Did you mean $$\left( 10^{10^{10^2}} \right) \uparrow \uparrow \left( 10^{10^{10^2}} \right) = \left( 10^{10^{10^2}} \right) \uparrow \uparrow \uparrow 2$$ ?
...actually I was thinking something like $$\left( 10^{10^{10^2}} \right) \uparrow^{\left( 10^{10^{10^2}} \right)}\left( 10^{10^{10^2}} \right)$$ but thanks for laying out the starting tex!

*that's supposed to be an up-arrow raised to the googolplex power...

Because of the laws of exponents and they way you have your parentheses, you miss out on getting to truly large numbers.

((googolplex ^ googolplex) ^ googolplex) ^ googolplex
= googolplex^(googolplex^3)
= (10^(10^(10^2)))^( (10^(10^(10^2)))^3 )
= (10^(10^(10^2)))^(10^(3×10^(10^2)))
= 10^( 10^(10^2) × 10^(3×10^(10^2)) )
= 10^( 10^( 3×10^(10^2) + 10^2 ) )

And if you have N googolplexes in your expression, FR(N) then log log log log FR(N) = log log ( (N-1)×10^(10^2) + 10^2 ) ≈ log ( log (N-1) + 10^2 ) ≈ 2

While
$$\HUGE 10^{10^{10^{10^{10^{10}}}}}$$ is much larger (and easier to write). log log log log 10^(10^(10^(10^(10^10)))) = 10,000,000,000

Knuth's uparrow notation $$10 \uparrow \uparrow 6$$ is the same number.
So we can write $$10 = 10\uparrow \uparrow 1 \lt 10 \uparrow \uparrow 2 \lt \textrm{googol} \lt 10 \uparrow \uparrow 3 \lt \textrm{googolplex} \lt 10 \uparrow \uparrow 4 \lt FR(2) \lt FR(10^9) \lt 10 \uparrow \uparrow 5 \lt 10 \uparrow \uparrow 6 \lt 10 \uparrow \uparrow 10 = 10 \uparrow \uparrow \uparrow 2$$

http://mathworld.wolfram.com/PowerTower.html
What you're saying here is that if Fraggle Rocker omitted the parentheses completely he would produce a much greater result..?

What you're saying here is that if Fraggle Rocker omitted the parentheses completely he would produce a much greater result..?
Maybe this number would be bigger:

I can't see any way to write this in notation without the parentheses. And I don't see any way at all to write it in strict prose.

But this way was much harder. I'm still not sure I've got everything paired correctly!

Maybe this number would be bigger:

I can't see any way to write this in notation without the parentheses. And I don't see any way at all to write it in strict prose.

But this way was much harder. I'm still not sure I've got everything paired correctly!
Hmm, what about $$G^{G^{G^{G^{G^{G^{G^{G^{G^{G^{G}}}}}}}}}}...$$?
(granted, you still have to pair up the damn brackets in TEX

Maybe this number would be bigger:

$$\left({10^{10^{10^2}}} \right)^{\left( {10^{10^{10^2}}}\right)^{\left({10^{10^{10^2}} }\right)^{\left({10^{10^{10^2}} }\right)^{\left( {10^{10^{10^2}} }\right)^{10^{10^{10^2}}} }}}}$$ where exponentiation is evaluated right-to-left.

$$FR_2(1) = 10^{10^{100}} \\ FR_2(n+1) = { FR_2(1) }^{ FR_2(n) }$$
$$\log \, FR_2(1) = 10^{100} \\ \log \, FR_2(n+1) = 10^{100} \times FR_2(n)$$
$$\log \, \log \, FR_2(1) = 100 \\ \log \, \log \, FR_2(2) = 10^{100} + 100 \\ \log \, \log \, FR_2(n+2) = 100 + 10^{100} FR_2(n)$$
$$\log \, \log \, \log \, FR(1) = 2 \\ \log \, \log \, \log \, FR_2(n+1) \approx \log \, \log \, FR_2(n)$$
$$\log \, \log \, \log \, \log \, FR_2(1) \approx 0.3 \\ \log \, \log \, \log \, \log \, FR_2(2) = 2 \\ \log \, \log \, \log \, \log \, FR_2(n+2) \approx \log \, \log \, FR_2(n) \\ \log \, \log \, \log \, \log \, FR_2(n+4) \approx 100 + 10^{100} FR_2(n)$$
And $$\log\, \log \, \log \, \log \, 10 \uparrow \uparrow 4 = 1, \quad \quad \log\, \log \, \log \, \log \, 10 \uparrow \uparrow (n+5) = 10 \uparrow \uparrow (n+1)$$

So $$\log \, \log \, \log \, \log \, FR_2(6) \approx 100 + 10^{100} \times { \left( {10^{10^{10^2}}}\right)}^{10^{10^{10^2}}} \approx 10^{\left(100 + 10^{\left(10^{100} + 100 \right)} \right)}$$

So $$10 \uparrow \uparrow 8 \quad \lt \quad FR_2(6) \quad \lt \quad 10 \uparrow \uparrow 9$$ (Corrected)

But RJBeery's number is $$\left(10^{10^{100}} \right) \overbrace{\uparrow\uparrow\uparrow \dots \uparrow\uparrow\uparrow}^{10^{10^{100}} \; \textrm{times}} \left(10^{10^{100}} \right)$$ which is more simply written in Conway's chained arrow notation as $$\left(10^{10^{100}} \right) \rightarrow \left(10^{10^{100}} \right) \rightarrow \left(10^{10^{100}} \right)$$.

$$\quad 10 \uparrow \uparrow 6 \quad \lt \quad 10 \uparrow \uparrow 10 = 10 \uparrow \uparrow \uparrow 2 \quad \lt \quad 10 \uparrow \uparrow \uparrow 10 \quad \lt \quad 10 \overbrace{\uparrow\uparrow\uparrow \dots \uparrow\uparrow\uparrow}^{10 \; \textrm{times}} 10 = 10 \rightarrow 10 \rightarrow 10 \quad \lt \quad \left(10^{10^{100}} \right) \rightarrow \left(10^{10^{100}} \right) \rightarrow \left(10^{10^{100}} \right) \quad \lt \quad \left(10^{10^{100}} \right) \rightarrow \left(10^{10^{100}} \right) \rightarrow 2 \rightarrow 2$$

But I think both Graham's number and $$3 \rightarrow 3 \rightarrow 3 \rightarrow 3$$ are larger still.

But I think both Graham's number and $$3 \rightarrow 3 \rightarrow 3 \rightarrow 3$$ are larger still.
Ahh, there ya go. Since we only have 15 seconds, say Graham's number = G

Then

LARGE NUMBER = $$G \rightarrow G \rightarrow G...\rightarrow (G times)...\rightarrow G$$

That's pretty big

Another interesting challenge would be to make the largest definite number possible using any operators but only, say, 5 unique digits (once each) and no variables.

How do you guys do that with the numbers and such? Does it work for any character? Let's see...

$$B^{A^{L^{E^{R^{I^{O^{N^{!^{!^{!}}}}}}}}}}$$

Okay, I think I just won the internet.

Another interesting challenge would be to make the largest definite number possible using any operators but only, say, 5 unique digits (once each) and no variables.
But you invite the series
$$5 \rightarrow 6 \rightarrow 7 \rightarrow 8 \rightarrow 9 \\ (5!) \rightarrow (6!) \rightarrow (7!) \rightarrow (8!) \rightarrow (9!) \\ ((5!)!) \rightarrow ((6!)!) \rightarrow ((7!)!) \rightarrow ((8!)!) \rightarrow ((9!)!)$$
etc.

All of which are (finite!) magnitudes too large to easily place in terms of mere power towers.

But you invite the series
$$5 \rightarrow 6 \rightarrow 7 \rightarrow 8 \rightarrow 9 \\ (5!) \rightarrow (6!) \rightarrow (7!) \rightarrow (8!) \rightarrow (9!) \\ ((5!)!) \rightarrow ((6!)!) \rightarrow ((7!)!) \rightarrow ((8!)!) \rightarrow ((9!)!)$$
etc.

All of which are (finite!) magnitudes too large to easily place in terms of mere power towers.
Ah yes, unary operators break things.

Largest number using 5 unique symbols? Only previously well defined symbology, please.

(Berry's paradox is around here somewhere.)

Maybe :
$$\nolimits 7\uparrow^{9!}8$$

Maybe :
$$\nolimits 7\uparrow^{9!}8$$

$${\nolimits 7\, \uparrow^{\tiny9!} \, 8} = 7 \, \overbrace{\uparrow\uparrow\uparrow \dots \uparrow\uparrow\uparrow}^{362880 \; \textrm{times}} \, 8 \quad = 7 \rightarrow 8 \rightarrow 362880$$

This number is so ridiculously large Pete. I think you have something like $$1814400$$ nested parentheses before your have the first non-trivial exponentiation. But the power towers grow and grow and grow to the point where just keeping track of height of the power tower is ridiculous number.

$$3 \rightarrow 2 \rightarrow 2 = 3 \uparrow\uparrow 2 = 3^3 = 27 4 \rightarrow 2 \rightarrow 2 = 4 \uparrow\uparrow 2 = 4^4 = 256 3 \rightarrow 3 \rightarrow 2 = 3 \uparrow\uparrow 3 = 3^{3^3} = 7625597484987 3 \rightarrow 2 \rightarrow 3 = 3 \uparrow \uparrow \uparrow 2 = 3 \uparrow\uparrow 3 = 3^{3^3} 4 \rightarrow 3 \rightarrow 2 = 4 \uparrow \uparrow 3 = 4^{4^4} = 4^{256} \approx 10^{154} 4 \rightarrow 2 \rightarrow 3 = 4 \uparrow \uparrow \uparrow 2 = 4 \uparrow\uparrow 4 = 4^{4^{4^4}} \approx 10^{8 \times 10^{153}} 3 \rightarrow 3 \rightarrow 3 = 3 \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow 3 \uparrow \uparrow 3 = 3 \uparrow \uparrow 7625597484987 = \underbrace{ 3^{3^{\cdot^{\cdot^{\cdot^3}}}}}_{7625597484987 \, \textrm{times}}$$

http://en.wikipedia.org/wiki/Conway_chained_arrow_notation