# Extremely Large Numbers

## Extremely Large Finite Numbers

The easiest way to make large finite numbers is by exponentiating small finite numbers: e.g. 1 googol = 10^100. The easiest way to make really large finite numbers is by exponentiating large finite numbers: e.g. 1 googolplex = 10^googol. In some peculiar sense, googol^googol isn't really that much bigger, as googol^googol= 10^(100 googol).

The naming of small powers of 1000 follows a regular pattern of prefixes. This pattern is different in different parts of the world, so 1 billion in some places is different than 1 billion in others. I'll follow the practice in the U.S. Suppose we want the name for 10^(3N). The small ones are agreed upon universally: 10^3 is 1 thousand and 10^6 is 1 million. In the U.S., 10^9 is 1 billion, whereas in other places, 10^9 is called 1 thousand million and 1 billion is 1 million million, or 10^12. In the U.S., the name for 10^(3N) uses a latinized prefix for N-1, so 10^9 is 1 billion, 10^12 is 1 trillion, 10^15 is 1 quadrillion, 10^18 is 1 quintillion, and so on. In the U.S., 1 centillion is 10^303. The other common practice is to assign a name with a latinized prefix for N to the number 10^(6N). So 10^12 = 1 billion, 10^18 = 1 trillion, 10^24 = 1 quadrillion, and so on reaching 1 centillion at 10^600. I've seen the terms "trigintillion", "quinto-quadragintillion" and "millillion" in print. I would read these as having prefixes corresponding to 30, 45 and 1000, respectively, but I think it's much more clear to write these as powers of 10.

After reading the net news recently, the following scheme for names of huge numbers was proposed. We use the term "k-oogol" to indicate 10^(10^k). So a googol is also a two-oogol while a fiveoogol would be 10^(10^5)=10^100000. Also, we'll use the construction "k-plex" to indicate raising 10 to the k. So, a googolplex is 10^googol and a googolplexplex is 10^googolplex, etc. Thus, 100 is also a twoplex, which makes a googol a twoplexplex and a googolplex a twoplexplexplex. Thus, a k-oogol is also a k-plexplex. Finally, let's get the Greek and/or Latin prefixes into the game so instead of writing plexplex, let's write duplex, and plexplexplex should be triplex, etc. Now you can wow your friends by saying things like sevenoogoldodecaplex.

### The Moser

Well, one way of making extremely large finite numbers is to repeat this process to silly extremes. Here's one approach which manufactures the largest number I know of which has it's own (short) name, the Moser (presumably named after Leo Moser):

Define n| (that's n followed by a line segment) to be n^n. So, 2| = 2^2 = 4, 3| = 3^3 = 27, etc. Notice what happens if we add more line segments: 2||| = 2^2|| = 4|| = 4^4| = 256| = 256^256. Now, define n< (that's n followed by a wedge) to be n followed by n line segments. So 3< = 3||| = 27|| = 27^27| = (27^27)^(27^27). Fairly big already. Now, we continue: n followed by a triangle is the same as n followed by n wedges, n followed by a square is the same as n followed by n triangles, and, in general, n followed by a k+1 sided polygon is the same as n followed by n k-sided polygons. Let's just see what 2(triangle) is:

2(triangle) = 2<< = 2||< = 4|< = 256< = 256|||...256 lines...||||| = a very very large number which we'll call zelda (the infamous 25th letter of the Greek alphabet).

2(square) = 2(triangle)(triangle) = zelda(triangle) = zelda<<<...zelda wedges...<<< = something enormous (call it a-ooga, the 26th letter of the Greek alphabet).

The Moser, named for Leo Moser who is purportedly responsible for this construction, is defined as 2(zelda-gon), unarguably an unbelievably large number. Question: It's easy to see that the last digit in the base ten expansion of the Moser is 6. What's its second to the last digit? If that's too easy, keep trying to find previous digits until you get bored.

### Ackermann's Function and generalized exponentiation

Here, with apologies for any deviations from the original, is an adaptation of Ackermann's function A(n,k) where n and k are non-negative integers:

• A(1,n) = n+2.
• A(k,1) = 2, for k>1
• A(k,n) = A(k-1,A(k,n-1))
Looks innocent, doesn't it? I'll leave it as an exercise to the reader in induction to prove that

• A(2,n) = 2n
• A(3,n) = 2^n
• A(4,n) = 2^(2^(2^...2))), where there are n 2's in the stack

Of course, the fun really begins with A(5,n). By definition, A(5,1) = 2. A(5,2) = A(4,A(5,1)) = A(4,2) = 2^2 = 4. A(5,3) = A(4,A(5,2)) = A(4,4) = 2^(2^(2^2)) = 65536. And, now, A(5,4) = A(4,A(5,3)) = A(4,65536) which is a stack of 65536 2's, which is a large number, indeed. In a sense, this Ackermann function is quite similar to the process which derives the Moser.

## Extremely Large Infinite Numbers

Okay, it seems silly to put all those modifiers in front of ``infinite'' since infinity already seems extreme and large in its own right. On the other hand, using Cantor's diagonalization proof, it is possible to show that there are ``more'' real numbers than natural numbers, even though both sets are infinite. By ``more'', we mean no matter how one chooses to pair off real numbers with natural numbers, there will always be some real numbers which are not paired off.

In the future, I hope to describe various classes of transfinite cardinals in this section. More later....