Tetration is often stated to be the next step after exponentiation (see for example Wikipedia): $$\large a^{a^{a^{...^a}}}$$

Where the exponents are taken $b$ times from the top. I refer to the Wikipedia article and many questions on this site for clarification and references.

However, it seems to be incorrect, if we 'honestly' continue the sequence of operation. Let's do that.

For now we consider only $a,b \in \mathbb{N}$.

$$a \cdot b=a + \ldots + a \tag{b times}$$

$$a^b=a \cdot \ldots \cdot a \tag{b times}$$

And here is how I think we should continue:

$$\large a^{a^b}=\left(\left(\left(a\right)^a\right)^a \dots \right)^a \tag{b times}$$

This is the only 'honest' form of the next hyperoperation, in my opinion. We take the operation from the previous step, replace $b$ by $a$, then repeat the operation $b$ times. In no way tetration appears on this step (or not yet)!

Then for the next step we should go the same way, replace $b$ by $a$ (we get $a^{a^a}$), and repeat $b$ times:

$$\large a^{a^{ab}}=\left(\left(\left(a\right)^{a^a}\right)^{a^a} \dots \right)^{a^a} \tag{b times}$$

And here our 'power tower' stops growing for a while! Becasue of the usual rules of exponentiation, we have on the next step:

$$\large a^{a^{a^2b}}=\left(\left(\left(a\right)^{a^{a^2}}\right)^{a^{a^2}} \dots \right)^{a^{a^2}} \tag{b times}$$

Thus, to obtain even $$\large a^{a^{a^a}}$$ we have to make many more steps.

I think the reason people call tetration the 'next' operation after exponentiation is the pure visual appeal of it, when expressed in the usual notation. It 'kind of' looks like repeated exponentiation, but it's really not.

Why is tetration considered the next step after exponentiation? Is there some flaw in my logic?


closed as primarily opinion-based by Adam Hughes, Vladhagen, астон вілла олоф мэллбэрг, Daniel W. Farlow, Namaste Feb 1 '17 at 18:59

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  • $\begingroup$ It's like your trying to confuse polynomials and exponential functions. Let $f(x)=a^x$ and $g(x)=x^a$. One is a polynomial and the other an exponential function. However,$$f(f(x))=a^{a^x}$$ $$g(g(x))=(x^a)^a=x^{a^2}$$ $\endgroup$ – Simply Beautiful Art Feb 11 '17 at 17:53
  • $\begingroup$ @SimplyBeautifulArt, I did not try to consider operations in terms of functions at all. In my world operations come first, and then come functions. I know that's not the official position (and there are some functions which do not fit this restricition), but that's how it is for me. So I don't see how your point relates here $\endgroup$ – Yuriy S Feb 11 '17 at 17:55
  • $\begingroup$ I was thinking that there is nothing unique about $(a^b)^b$ since it's equal to $a^{b\cdot b}$, but on the contrary, $a^{a^b}$ is not comparable to $a^b$. $\endgroup$ – Simply Beautiful Art Feb 11 '17 at 18:23
  • $\begingroup$ @SimplyBeautifulArt, that's subjective and depends on the notation. $a^b$ is just a number, it's not any better or worse (or indeed, unique) than $b^2$. Multiplication is just a shorthand way of writing multiple additions. Addition is a shorthand of writing multiple increment operations. In this way, tetration is a useful shorthand as well, but I was trying to see if it always directly follows exponentiation. $\endgroup$ – Yuriy S Feb 11 '17 at 19:13
  • $\begingroup$ Well, I think the ultimate goal of Hyperoperations is to represent repeated application of the previous in a helpful way, no more than that. $\endgroup$ – Simply Beautiful Art Feb 11 '17 at 22:10

It comes down to left or right bias and loss of symmetry.

Rewrite your binary operation as $f(x, y)$ instead. The 3 times repetition to get the 'next' operation on $a$ with 3 could be $f(f(a, a), a)$ or it could be $f(a, f(a, a))$. This doesn't make a difference at first, but it does once you try to move up from powers. One way gives tetration and the other gives your result.

  • $\begingroup$ If we think this way, we get an infinite tree of operations with tetration and my case as only two possible paths out of many $\endgroup$ – Yuriy S Jan 30 '17 at 14:24
  • $\begingroup$ Indeed, you could have arbitrary pattern of switching left/right biases, producing a different 'next' operation for any infinite binary sequence. $\endgroup$ – G. H. Faust Jan 30 '17 at 14:38

If we define the function $$f(n)=a^n$$ then tetration arises very natural. We have $$a\uparrow\uparrow b=f^b(1)$$

So, we start with $1$ and apply $f$ $b$ times. This way, we get $a$ after the first step , $a^a$ after the second step, $a^{(a^a)}$ after the third step and so on.

Finally, we arrive at the power tower containing $b$ $a's$.

Moreover, the usual convention to handle nested exponentiation is that $a^{b^c}$ is considered to be $a^{(b^c)}$ rather than $(a^b)^c$. This is also the way to get extremely large numbers (but this is , I admit, not a very convincing argument).

  • $\begingroup$ This different orders of evaluation made me to use the two terms "powertower" (for the left-associative operation in the OP) and "exponentialtower" (for the right-associative and in context to tetration) because "exponentiation" of a value z with a given base (and repeatedly) is different to raising a value z to some power (and repeatedly) $\endgroup$ – Gottfried Helms Jan 31 '17 at 11:53

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