(9^9)^9 = 9^(9^2).
But that’s not the standard order of operations for exponents (partly because that order of operations is so easy to re-write in a different way).
Oh. I see. My apologies, msmith537, and thanks Chronos.
Would the book have been Mathematics and the Imagination by Edward Kasner and James R. Newman (published in 1940)? That’s the one that introduce the terms “googol” and “googolplex”. I’m not sure it counts as a children’s book but I read it when I was about 10 years old.
@BigT: No. To clarify: First of all, the rules about orders of operation aren’t “inherent” at all. They are arbitrary man-made rules intended to resolve the ambiguities of not having such rules. Alternatively, we could just as well have made a rule that all expressions must be fully parenthesized, so that, e.g.,
3 + 4 × 5
(with no parentheses) would be considered an invalid expression.
Second, given that the familiar rules specify left-to-right for like orders of operations, doesn’t mean it had to be defined that way.
And finally: In fact, for a sequence of exponentiations like m^n^p ( or equivalently, m[sup]n[sup]p[/sup][/sup] ), mathematicians decided to do it the other way: These associate to the right, so the computation would be m^(n^p) and NOT (m^n)^p — it’s an arbitrary rule, but that’s the agreed-upon rule.
Why? It turns out to be more useful that way. If m^n^p had been defined left-to-right like addition and multiplication, it would mean (m^n)^p. But this is just algebraically the same as m^(np) so there would be no need to ever write (m^n)^p. Thus, to make the expression m^n^p not only meaningful but useful and different from simply m^(np) they decided to make it mean m^(n^p).
No, you’re right. I entered the wrong operator.
In computer programming languages, exponentiation is almost always right associative unlike most common binary operations which are left associative. The rational is as given above.
Once in a while the idjit defines things elsewise and people wonder about the person.
Nitpick: 9[sup]9[sup]9[/sup][/sup] is always treated as 9sup[/sup] for the simple reason that (9[sup]9[/sup])[sup]9[/sup] can be written more readily as 9[sup]9·9[/sup]
With TREE(3) so vastly incredible huge, TREE(TREE(TREE(9))) seems rather self-indulgent, no? 
Graham’s “number”, BTW, is also a function. G(1) = 3 ↑↑↑↑ 3, G(2) Is similar but with G(1) arrows instead of just 4. The specific Graham number G(64) is defined as
G(64) = 3 ↑↑↑↑↑↑↑…↑↑↑↑↑↑ 3
where the number of ↑’s is G(63). Of course, there’s a G(65), or even G(G(64)) which are even larger!
I disagree - Graham’s number is not the function, it’s the specific result of plugging 64 into the G(n) function, and is a single number. If you’re going to say that Graham’s number is really not a number but is actually a function, you’ve got to say the same thing about 9^9^9 since exponentation and iterated exponentation are functions.
The most amusing thing to me about Graham’s number is that Graham came up with it as an upper bound for a question that mathematicians think the actual answer to is probably less than two dozen. So what he’s saying is ‘this answer has to be less than a number that’s so huge you couldn’t conceivably write it out in the observable universe’ when the answer itself is probably something completely bland like 17. It was actually a sensible thing to prove, since showing that there is an upper bound on a problem, even a huge one, answers certain other questions, but it’s just so hilariously large compared to what the answer probably is.
And if you dial 9[sup]9[sup]9[/sup][/sup] you get a lot of policemen.
In the UK, not the USofA.