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


  • 6
    $\begingroup$ 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.) $\endgroup$ Sep 27, 2011 at 4:37
  • 1
    $\begingroup$ Up-arrow is related to tetration, en.wikipedia.org/wiki/Tetration and tetration.org. There has been work to extend it to non-natural arguments. $\endgroup$ Sep 28, 2011 at 18:30

2 Answers 2


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 $$ \begin{align*} 2 \uparrow \uparrow 3 &= 2 \uparrow 2 \uparrow 2\\ &= (2 \uparrow 2) \uparrow 2\\ &= (2^2) \uparrow 2\\ &= 2^{2^2}. \end{align*} $$

  • $\begingroup$ 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 $2\uparrow\uparrow{-2}$? $\endgroup$
    – JShoe
    Sep 27, 2011 at 10:50
  • 4
    $\begingroup$ 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. $\endgroup$ Sep 27, 2011 at 16:49

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.


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