Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How would I solve something like
$2\uparrow\uparrow n$? when n ≤1?


share|cite|improve this question
To be precise, there is nothing to solve in $2\uparrow\uparrow n$, it's just an expression with some meaning. (The same way there is nothing to "solve" in the expression $2^3$; you can only evaluate it.) – ShreevatsaR Sep 27 '11 at 4:37
Up-arrow is related to tetration, and There has been work to extend it to non-natural arguments. – Ross Millikan Sep 28 '11 at 18:30
up vote 3 down vote accepted

Here is a possible answer:

We know that for every positive integer n, m↑↑n = logm[m↑↑(n+1)]. If we assume the same equation to hold for nonpositive n, we get:

m↑↑0 = logm(m↑↑1) = logmm = 1

m↑↑-1 = logm(m↑↑0) = logm1 = 0

m↑↑-2 = logm(m↑↑-1) = logm0 = -infinity (in other words, it is undefined)

This seems to be the only logical extension of the double-arrow operator to nonpositive integer "exponents".

As for the extension to a "nonpositive number of arrows", it is even simpler:

m↑0n = m*n (ordinary multiplication)

m↑-1n = m+n (ordinary addition)

But the next one is somewhat surprising:

m↑-2n = (m+n+4)/2 (you might want to try and verify on your own that this is correct, by proving that m+n is indeed equal to this function iterated n times on m)

So the expression in your original question, 2↑-22, evaluates to (2+2+4)/2 = 4. Of-course, this is hardly surprising, since 2[any-number-of-arrows]2 always evaluates to 4.

share|cite|improve this answer

I'm not aware of an interpretation with a negative number of arrows, so I'll just address the case where $n \geq 1$.

With up arrow notation, you can "strip away" an arrow and then write $n$ copies of 2, each separated by one fewer arrows than you had previously. In your case, we'd have $$ 2 \uparrow \uparrow n = \underbrace{2 \uparrow 2 \uparrow \cdots \uparrow 2}_{n \text{ copies of } 2}. $$

A single uparrow is just regular exponentiation, so we'll get an exponent tower of 2's having height $n$.

For example, when $n = 3$, this all works out to $$ \begin{align*} 2 \uparrow \uparrow 3 &= 2 \uparrow 2 \uparrow 2\\ &= (2 \uparrow 2) \uparrow 2\\ &= (2^2) \uparrow 2\\ &= 2^{2^2}. \end{align*} $$

share|cite|improve this answer
But how would I "strip away" an arrow and then write $n$ copies of 2, when $n$ is a negative number? How do you solve $2\uparrow\uparrow{-2}$? – JShoe Sep 27 '11 at 10:50
I don't think the up arrow notation is defined for negative values of $n$, but it's possible I've just never seen it (though I can't imagine a reasonable interpretation for it). If you can find an example of its use online, I'd be interested to see it. – Austin Mohr Sep 27 '11 at 16:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.