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Of the following, why is a usually considered true, and for what reason other than "tradition" and "more convenient"?

a: ${x}^{y^z} = x^{(y^z)} \neq {(x^y)}^z$

b: ${x}^{y^z} = {(x^y)}^z \neq x^{(y^z)}$

Edit: I know a is correct, but what is the reason for this order of operations?

Stefan Hamcke
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toting
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    @Vadim123's answer is precise. Usually $x^{y^z}$ is taken as $x^{(y^z)}$ because you could just write $x^{yz}$ if you meant option b. – mikhailcazi Oct 21 '13 at 15:59
  • Aka, "tradition" and "more convenient"? – toting Oct 21 '13 at 16:08
  • I guess so. But this is possibly the only reason. Why do you want others? :) – mikhailcazi Oct 21 '13 at 16:48
  • Your question and vadim's answer have proven fortuitously useful to me, in the field of computing. I just realized that a N-ary exponentiation function that I designed as part of a programming language has in fact the less useful left-associative behavior of: (expt a b c) being (expt (expt a b) c), rather than the right associative (expt a (expt b c)). Thanks a lot! I may well just fix this. – Kaz Oct 21 '13 at 20:10
  • ^ Nope! I made it work in the useful way, in fact. I only documented it incorrectly as working left-associatively, haha. So the useful thing here is that we found a ref manual bug. – Kaz Oct 22 '13 at 00:20
  • See http://math.stackexchange.com/questions/359577/how-to-evaluate-powers-of-powers-i-e-234-in-absence-of-parentheses – Martin Sleziak Apr 13 '14 at 04:59
  • It is not just tradition, it is a convention ! But far lesser known than other conventions like what the square root of $4$ should be , for example. – Peter Nov 26 '22 at 12:13

3 Answers3

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Choice (b) is pointless since you could instead write $x^{yz}$.

To elaborate, sometimes we mean (a) and sometimes we mean (b). We already have a way to denote (b), but no other way to denote (a).

vadim123
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As for you title question,$\;$ x^y^z $\;$is ambiguous.

For the question in your post, as formatted:

$$(x^y)^z = x^{(yz)} \neq x^{\large y^z}$$

So option $a$ is correct.

amWhy
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Of course $x^{y^z} = x^{(y^z)} \ne (x^y)^z$. To outline a proof by contradiction - if you suppose that $x^{y^z} = (x^y)^z$, you can take the counter example of $2^{2^3} = 2^8 = 256 \ne (2^2)^3 = 4^3 = 64$.

DamianV
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  • But you did not follow the ${x}^{y^{z}} = {(x^y)}^z$ rule in step 2. If you apply it to both sides you do get $64 = 64$ – toting Oct 21 '13 at 15:52
  • You cannot use the rule in a proof (where you want to show, that such a rule is valid). Here, you only evaluate nested functions. You can also view it like this: Let $exp(a,x)=a^x$. then left side is $exp(x,exp(y,z))$ and right hand side is $exp(exp(x,y),z)$ – DamianV Oct 22 '13 at 14:26