**Hyperoperations and huge numbers**

If you repeat multiplication you get exponentation:

2↑4 = 2 × (2 × (2 × 2))

If you repeat exponentiation you get

**tetration**, which gives you a 'power tower':

2↑↑4 = 2 ↑ (2 ↑ (2 ↑ 2))

This is 2 to the power 2 to the power 2 to the power 2. If you repeat tetration, you get

**pentation**:

2↑↑↑4 = 2 ↑↑ (2 ↑↑ (2 ↑↑ 2))

Let's see how big this is!

2 ↑↑ 2 = 2 ↑ 2 = 2 × 2 = 4

so

2 ↑↑ (2 ↑↑ 2) = 2 ↑↑ 4 = 2 ↑ (2 ↑ (2 ↑ 2)) = 2 ↑ (2 ↑ 4) = 2 ↑ 16 = 65536

so

2 ↑↑ (2 ↑↑ (2 ↑↑ 2)) = 2 ↑↑ 65536

In short, 2↑↑↑4 is a power tower with 65536 twos in it!

This is a staggeringly large number. It can't be written in binary on all the atoms in the observable universe. Its number of binary digits can't either. The number of binary digits in its number of binary digits can't either. And so on... for over 65,000 rounds.

But when we hit the next operation,

**hexation**, all hell breaks loose.

2↑↑↑↑4 = 2 ↑↑↑ (2 ↑↑↑ (2 ↑↑↑ 2))

Let's see how big this is! By definition,

2 ↑↑↑ 2 = 2 ↑↑ 2 = 2 ↑ 2 = 2 × 2 = 4

So, by our previous calculation:

2 ↑↑↑ (2 ↑↑↑ 2) = 2 ↑↑↑ 4 = 2 ↑↑ 65536

and then

2 ↑↑↑ (2 ↑↑↑ (2 ↑↑↑ 2)) = 2 ↑↑↑ (2 ↑↑ 65536)

Can you comprehend this number? This is

2 ↑↑ (2 ↑↑ (2 ↑↑ (2 ↑↑ (2 ↑↑ (2 ↑↑ (2 ↑↑ (2 ↑↑ (2 ↑↑ .....

where there are 2 ↑↑ 65536 twos. The mind boggles, but here's a good way to think of it:

2

2 ↑↑ 2

2 ↑↑ (2 ↑↑ 2)

etc.

We start with some number, namely 2. Then we replace that number with a bigger one: a power tower consisting of that many 2s. Then we replace

*that*number with a power tower consisting of

*that*many 2s. And we do this process 2 ↑↑ 65536 times!

The result can't be written in binary on all the atoms in the observable universe. Its number of binary digits can't either. The number of binary digits in its number of binary digits can't either. And so on...

*for a number of rounds that can't be written in binary on all the atoms of the observable universe!*

These operations ↑, ↑↑, ↑↑↑ etc. are called

**hyperoperations**:

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

and we're using Knuth's

**up-arrow notation**to describe them:

http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

I hope you use these ideas to impress your friends and make new enemies. But soon I'll show you notations for vastly larger numbers. And later I'll show you that

*infinity infects the finite*. The best, most systematic notations for truly enormous but still finite and computable numbers use

*infinite ordinals*of the sort I've been discussing lately!

I'm using the word 'computable' in a sense explained in the attached article by Scott Aaronson. He ran some contests to see who could describe the biggest number. If you ever do this, make sure to require that the descriptions be computable: otherwise it may be impossible to tell who won!

#bigness

View 33 previous comments

- Wow, so they're all associative!? Do they all have unit elements too? Then we'd get a sequence of monoids, each distributing over the previous one. That's literally true only if they're all defined on some domain that contains all their units... or at least a finite list of them. But the question of their domain of definition can perhaps be finessed somehow. For example, we could work in the category of sets and partially defined maps.Nov 30, 2012
- The unit for the nth hash is exp(n):

eⁿ #ⁿ x = expⁿ(lnⁿ(eⁿ) * lnⁿ(x)) = expⁿ(1 * lnⁿ(x)) = x

The unit for the nth @, if it exists at all, is minus infinity:

-∞ @ⁿ x = lnⁿ(expⁿ(-∞) + expⁿ(x)) = lnⁿ(0 + expⁿ(x)) = xNov 30, 2012 - Cool. By the way, I love the self-referential link in blue in this comment!Nov 30, 2012
- :D Yay for automatic markup!Nov 30, 2012
- Ok, dumb question here.

Repeated addition gives us multiplication, and repeated multiplication gives us exponentation ↑ and then repeated exponentation gives us tetration ↑↑ and then even higher repeated functions ↑↑↑, ↑↑↑↑, ↑↑↑↑↑, etc.

But what about going in the opposite direction?

For instance, logs change exponents to multiplication.

And they also change multiplication to addition.

Is there any function "below" addition?Dec 4, 2012 - +David Foster: +Mike Stay's post with the # and @ operations has operations below addition (his @ operations), but that belongs to a different sequence (since the # operations are not the ↑ operations). There can be no operation below addition that works in this sequence for the reasons in my comment in this thread. In particular, if ⊙ is repeated ⊡ in this sense, then
*a*⊙ 1 =*a*, but*a*+ 1 ≠*a*.Dec 4, 2012

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