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Does math have an incorrect notation / syntax? I don't mean writing misaligned notation (google), but when you take something like a number to powers to powers to powers, $${{2^2}^2}^3$$ (I was told this is incorrect notation by a teacher). Is it really incorrect, or does it just need to be simplified with parentheses? Do people write maths like this?

a radical expression with the root being a radical expression? $$\sqrt[\sqrt{2^3}]{2}$$

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    $\begingroup$ @GitGud $$\left(a^b\right)^c=a^{bc}$$ so by convention $$a^{b^c}=a^\left(b^c\right)$$ $\endgroup$
    – user5402
    Jul 8, 2013 at 18:37
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    $\begingroup$ @GitGud It'a a matter of convention, like $a\times b\div c$ which anyone does it (by convention) from left to right. The convention for $a^b^c$ is from right to left (up to bottom). $\endgroup$
    – user5402
    Jul 8, 2013 at 18:43
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    $\begingroup$ There is a Cuban writer, Onelio Jorge Cardoso, that in a tale made a humming bird say: "Things are not how they are called, but how we name them along the way." He said that to mama-bird who was complaining that her chicks were calling their nest a ship after throwing it into the river. Everyone uses your tower of exponentials and understands them as being $2^{(2^{(2^{3})})}$. So, it is not wrong. There is no wrong language, as long as it is understood. That is the ultimate purpose of language. $\endgroup$
    – OR.
    Jul 8, 2013 at 18:43
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    $\begingroup$ @StefanH. introducing in the context of the particuar questions qualms about what formal properties the notation might or not have in category theory seems rather disingenuous! $\endgroup$
    – Mariano Suárez-Álvarez
    Jul 8, 2013 at 19:56
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    $\begingroup$ Another example is $a+b\cdot c$ which by convention is $a+(b\cdot c)$ but that's just a convention; there's no inherent reason why it could not be $(a+b)\cdot c$. And BTW, that we use "+" for addition and $\cdot$ (or $\times$) for multiplication is pure convention, too. There's nothing "additive" in "$+$" and nothing "multiplicative" in "$\cdot$" or "$\times$". $\endgroup$
    – celtschk
    Jul 8, 2013 at 20:04

4 Answers 4

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Your teacher is mistaken. There is a well-established and universal convention about the meaning of an expression like $$2^{2^{2^3}}$$it is always understood to mean $$2^{\left(2^\left(2^3\right)\right)} =2^{2^8} = 2^{256}$$ People can and do write expressions like these. For example this paper, "Analog of the Skewes Number for Twin Primes", by Marek Wolf, contains the expressions $$10^{10^{10^{10^3}}}\qquad\text{and}\qquad 10^{10^{529.7}}$$on the first page, with no further explanation. Similarly "Some Rapidly Growing Functions" by Craig Smoryński has $$10^{10^{10^{34}}} < e^{e^{e^{e^{4.369}}}}$$ and similar expressions. (I picked these two papers arbitrarily; they were the first two hits in Google Scholar for "Skewes' Number".)

There is a good reason for the convention about what $a^{b^c}$ means: $a^{b^c}$ could be understood as either $a^\left({b^c}\right)$ or as $\left(a^b\right)^c$. But if it were understood as $\left(a^b\right)^c$, one would never need to write $a^{b^c}$, since it would be equal to $a^{bc}$. So it is always understood as $a^\left({b^c}\right)$.

Nobody ever writes $$\sqrt[\sqrt{2^3}]2$$ even though its meaning is clear. Partly this is because it would have been difficult to typeset with old-fashioned metal type, so there is a tradition of expressing this differently. And partly it is because it looks bad.

Since by definition, $$\sqrt[a]b = b^{1/a},$$ one would almost always write something like $$(2^{1/2})^{1/2^{3/2}}$$ instead, at which point it would become clear that the expression could be simplified to $$2^{(1/2)(1/2^{3/2})} = 2^{1/2^{5/2}} = 2^{2^{-5/2}}.$$ Good notation enables and encourages this sort of simplification; bad notation obscures and impedes it.

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  • $\begingroup$ Isn't it just $2^{1/2^{3/2}}$? Because $\sqrt[3]2=2^{1/3}\ne(2^{1/2})^{1/3}$. $\endgroup$
    – Akiva Weinberger
    Oct 24, 2014 at 15:51
  • $\begingroup$ Sorry, I don't follow your reasoning. There is no $\sqrt[3]2$ in my post. $\endgroup$
    – MJD
    Oct 24, 2014 at 15:56
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    $\begingroup$ Not sure where @columbus8myhw got the $\sqrt[3]{2}$, but I agree with her/him about $2^{1/2^{3/2}}$. By application of $\sqrt[a]{b}=b^{1/a}$, we have $\sqrt[\sqrt{2^3}]{2} = 2^{1/\sqrt{2^3}} = 2^{1/\left(2^{3/2}\right)}=2^{1/2^{3/2}}$. $\endgroup$
    – Ben Blum-Smith
    Jul 1, 2015 at 18:06
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Towers of exponents have been standard for ages. Cajori, in his book History of mathematical notations, §313, tells the story. He reproduces an image from a book by Waring published in 1785:

enter image description here

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    $\begingroup$ It is sad that we don't find &c used like that anymore! $x^{x^{x^{\&c}}}$. $\endgroup$
    – Mariano Suárez-Álvarez
    Jul 8, 2013 at 19:58
  • $\begingroup$ A translation might be helpful. $\endgroup$
    – Alraxite
    Jul 8, 2013 at 20:42
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    $\begingroup$ @Alraxite, it is just an explanation about how to compute derivatives of such beasts: the purpose of the image here is just pictorical. $\endgroup$
    – Mariano Suárez-Álvarez
    Jul 8, 2013 at 20:45
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    $\begingroup$ @Alraxite it is Latin: enjoy its power and elegance even without translation :) $\endgroup$
    – Avitus
    Jul 9, 2013 at 10:20
  • $\begingroup$ +1 for the brave Latin scan! ("Onus probandi incumbit ei qui dicit") $\endgroup$
    – Avitus
    Jul 9, 2013 at 10:25
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$a^{\large b^{\Large c}}$ means $a^{\large(b^{\Large c}\large)}$. This is not in dispute.

Note that $\left(a^{\large b}\right)^{\large c}=a^{\large bc}$, so there would be no reason for it to mean this.

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I think I'm starting to see the problem; exponentiation is right associative. Perhaps a more sensible notation (for $a^{b^c}$) would be $${}^{{}^c} {}^b a$$

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  • $\begingroup$ Like so: ${}^{{}^c} {}^b a$ $\endgroup$
    – Daniel McLaury
    Oct 24, 2014 at 21:55
  • $\begingroup$ Thank you Daniel, I just made the correction. $\endgroup$
    – John Joy
    Oct 26, 2014 at 11:24

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