Mathematics

I've heard that 2 to the power of 2, to the power of 2, etc... 6 times is a number so huge that we could never figure it out. Would that qualify as being infinite? And how would we be able to intelligibly come to that conclusion, or is it a "rough estimate" that we could never figure it out? Thank you for your time. ~Kris S.

Comments

If I understand well the number you have in mind, 2^2^2^2^2^2, it is not all that large:

2^2 = 4

4^2 = 16

16^2=256

256^2=65,536

65,536^2=4,294,967,296

The person you heard this from may have had another number in mind, namely:

2^(2^(2^(2^(2^2)))). Let's construct this one:

2^2 = 4

2^4 = 16

2^16 = 65,536

2^65,536 = ???

.... and this fifth step (bringing in the fifth "2") already goes beyond most ordinary spreadsheets and calculators. Still, since 2^10 is about 10^3, we can estimate the result to be around 10^19661, i.e. a "1" with nearly 20,000 zeros. A good computer could probably do the calculation and could print out the resulting number on perhaps 12 pages or so.

The sixth step, taking 2 to the power of this number, would really go beyond what most of us can even imagine. It would bring us to a number -- let's call it "K" in your honor -- that, written in the decimal notation, would have so many digits that this number of digits would have nearly 20,000 digits. This could not be calculated or written down in decimal notation.

And yet, this would get us nowhere near infinite. Just think of the seventh step, and the eighth, and so on. Each of these steps goes vastly beyond the preceding one, and yet they all, no matter how far you may go, bring you to finite numbers. Mathematicians can deal with such numbers in their own special notations and can do calculations with them. And we ordinary mortals can keep up with them at least some of the way by thinking in terms of number of digits, number-of-digits of the number of digits, and so on.

So there are two great lessons in your question. No matter how large the leaps you take, you cannot get to infinity in a finite number of finite steps. Taking such steps, you can go indefinitely far (beyond any number anyone might specify). But you will always still have vastly (infinitely!) more numbers ahead of you than behind.

The second lesson is that there are some mind-bogglingly large finite numbers out there. The large numbers we're acquainted with from real life -- cost of the Iraq occupation in US$, age of the world in seconds, electrons in the universe -- are ridiculously small in comparison with even the number we get from your fifth step. Your number, K, is vastly vastly greater. Now imagine going further, not merely taking the seventh step, and the eighth, but the Kth step in your sequence. Even that number is still very much a finite one.

A very good and accessible book on the subject is Shaughan Lavine: Understanding the Infinite.

This question concerns, in effect, number-theoretic functions that grow very fast. We can say a lot about them.

The operation in play here is called "superexponentiation", and is also known as "tetration". We can define it as follows:

superexp(0) = 1
superexp(n+1) = 2^(superexp(n))

So superexp(4), e.g., is: 2^(2^(2^2))), and K is superexp(7).

Noting that superexponentiation is just repeated exponentiation, we can now define superduperexponentiation as follows:

supdupexp(0) = 1
supdupexp(n+1) = superexp(supdupexp(n))

supdupexp(4) is already an enormous number: It is superexp(2^16) = superexp(65,536) = 2^(2^...), where there are 65,536 2s in the tower. supdupexp(5) is unimaginably huge. It is 2^(2^...), where there are supdupexp(4) 2s in the tower.

But there is no need to stop there. Let's rename superexponentiation 2-exponentiation, or exp2 for short; and let's call superduperexponentiation 3-exponentiation, or exp3 for short; and let's just write 2^x as exp1(x). Then the foregoing definitions take the form:

exp2(0) = exp1(0) = 20 = 1
exp2(n+1) = exp1(exp2(n))
exp3(o) = exp2(0) = 1
exp3(n+1) = exp2(exp3(n))

And now we might notice a general pattern here. So we define:

exp4(0) = exp3(0) = 1
exp4(n+1) = exp3(exp4(n))

and more generally:

expk+1(0) = expk(0) = 1
expk+1(n+1) = expk(expk+1(n))

The functions in this sequence grow faster, and faster, and faster. The first three values are always the same:

expk(0) = 1
expk(1) = 2
expk(2) = 4

But after that, things start to grow quickly.

exp1(3) = 2 x 4 = 8
exp2(3) = exp1(exp2(2)) = exp1(4) = 2^4 = 16
exp3(3) = exp2(exp3(2)) = exp2(4) = exp1(exp2(3)) = 2^16
exp4(3) = exp3(exp4(2)) = exp3(4) = exp2(exp3(3)) = exp2(2^16) = 2^(2^...), with 65,536 twos
exp5(3) = exp4(exp5(2)) = exp4(4) = exp3(exp4(3)) = exp3(2^(2^...)), with 65,536 twos = exp2(exp3((2^(2^...) - 1))), where there are again 65,536 twos

This last number is mind-bogglingly huge. And as Thomas mentioned, we are only getting started. What's more, the growth of all of these functions (they are all "primitive recursive") is utterly dwarfed by that of the Ackermann function, which you can read about on MathWorld or Wikipedia.

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