Dealing with

large numbers can be

confusing, once the numbers get so big you need to use exponents to express their exponents. I recently dealt with a number about the size of 10

^{2.1 x 10343}, call it

*N*, a number whose

*exponent* has 343 digits. I wanted to see what it was when raised to its own power.

Firing up *hypercalc* (*"just try and MAKE me overflow!"*) I was surprised to see that my number (*N*) when raised to its own power, gave the result 10^{N}.

(That's 10^{102.1 x10343}.)

But in fact hypercalc is correct, and, *to normal levels of accuracy*, *N*^{N} = 10^{N} when *N* is very large, as in the present case.

To raise to the power of *N*, we multiply a number by itself *N* times. If the number is a power of 10, with exponent *exp*, then we add *exp* digits for each such multiplication. So 10^{N} will have *N* digits, since 10 = 10^{1} and 1 times *N* is *N*.

*N* itself has an exponent of log_{10}(*N*), and so *N*^{N} has log_{10}(*N*) * *N* digits.

So *N*^{N} has log_{10}(*N*) times as many digits as 10^{N}.

So the *exponent* of *N*^{N} (putting *N*^{N} into the form 10^{exp}) has log_{10}(log_{10}(*N*)) times as many digits as *N*.

Which means it has around 343 times as many digits as *N* (because *N* = 10^{2.1 x 10343}.)

So the **second exponent** has less than 3 times as many digits as the second exponent of 10^{N} (because log_{10}(343) < 3.)

So the *third exponent* is less than log_{10}(3) out.

Which means that increasing the third exponent of 10^{N} by one would give a value **significantly larger** than *N*^{N}, since log_{10}(3) < 1.

So, given *N* large enough to force us to use these 'third exponents', 10^{N} is pretty much "equal" to *N*^{N}, which seems **surprising**, shocking even, especially considering how large *N* is in the first place.

Now I'm not saying that *N*^{N} and 10^{N} are converging - the difference is plainly growing.
What's happening is that the difference is growing more slowly than the numbers themselves are, with the result that at some point, our notation gives out, and to record the difference, we'd have to include more digits than it will allow.

If you're having problems believing this (and I did myself) here's a different proof, showing the same thing in more detail, and more understandably, from *hypercalc* author mrob27 (whose large number pages at `http://home.earthlink.net/~mrob/pub/math/largenum.htm` will be of *undoubted* interest to anyone still reading. Go there.)

Your explanation of how the exponents work is fine (based on an intuitive
notion of "number of digits") but I would prefer something more like this:
Given R2 = 10 ^ (2 x 10^343), compare 10^R2 to R2^R2.
We need to use the fact that (a^b)^c = a^(b*c), and also (a^b)*(a^c) =
a^(b+c)
10^R2 = 10 ^ [10 ^ (2 x 10^343)]
R2^R2 = [10 ^ (2 x 10^343)] ^ [10 ^ (2 x 10^343)]
= 10 ^ [(2 x 10^343) x 10 ^ (2 x 10^343)]
= 10 ^ [2 x 10^(343 + 2 x 10^343)]
= 10 ^ [10 ^ (log(2) + 343 + 2 x 10^343)]
Now notice that log(2) + 343 + 2*10^343 is only a little bit more than
2*10^343. In fact, when the addition is performed, the log(2) and the
343 will get completely lost in the roundoff error unless at least 342
digits of precision are used in the calculation. So the most accurate
value we can calculate for R2^R2 is 10^[10^(2x10^343)], which is
10^R2. (q.e.d.)

(end quote)

This is a particular instance of the general rule that as the exponent gets larger and larger, *the value of the mantissa becomes increasingly irrelevant for the size of the result*.