This post is attempting to answer one of the questions at the end of Let’s build an operator which introduces the idea of hyperoperations. It’s not strictly necessary to read that post first, but I’d recommend it.

\[\newcommand{\tetration}[2]{ {}^{#2}#1 }\]

Repeated addition turns into multiplication. Repeated multiplication turns into exponentiation. What does repeated exponentiation turn into? Well, that’s tetration.

$3+3+3+3 = 3\cdot 4$
$3 \cdot 3 \cdot 3 \cdot 3 = 3^4$
$3^{(3^{(3^3)})} = \tetration{3}{4}$. This is called tetration and note the parenthesis: $(3^3)^3 \neq 3^{(3^3)}$

So here’s the main question of this post: What is $\tetration{3}{\frac{1}{2}}$? As in, I’d like you to make an exponential tower of 3’s. How many 3s in the tower? $\frac{1}{2}$ of one.

Nonsense! is your first reaction. But then you remember that you can do that for addition and multiplication. Sure, but adding three “one half times” is totally fine - that’s just $\frac{3}{2}$. And multiplying by three “one half times” is… oh right, that’s just $\sqrt{3}$.

For this post I’m going to ask you to do something that’s honestly pretty hard. Try your best to forget that you know about square roots. Try to put yourself into the head of someone who knows how to multiply, but has not yet learned about square roots. Ok ready?

Someone asks you to multiply 3 by itself 4 times, no problem. I can write that out as $3 \cdot 3 \cdot 3 \cdot 3$ and work it out. Nothing complicated about that. Now someone asks you to multiply 3 by itself 1 time. Also easy. Now someone asks you to multiply 3 by itself $\frac{1}{2}$ times. Huh?

Like honestly, this is pretty weird (and stepping out of character for a second: just because we’ve given a name to this idea “square root” doesn’t make it any less weird). What does the question even mean?

Well, it’s not obvious, but one way to make progress is by noticing the following: If I multiply $3 \cdot 3$ and then multiply that by $3 \cdot 3$, that’s the same as multiplying 3 by itself 4 times (since $2 + 2 = 4$): $3 \cdot 3 \cdot 3 \cdot 3$. So maybe whatever the answer to “3 multiplied by itself $\frac{1}{2}$ times” should have the property that if you multiply that by itself, it should be the same as “3 multiplied by itself 1 time” (since $\frac{1}{2} + \frac{1}{2} = 1$). And 3 multipled by itself 1 time is just 3.

That’s something I can work with. Now I’m just looking for some other number that, when multiplied by itself produces 3. Using guess and check, I find that it’s something around 1.7.

Armed with that line of reasoning, what if someone asks “What is 3 multiplied by itself $\frac{1}{3}$ times?”. I’d find some number $x$ such that $x \cdot x \cdot x = 3$.

Before moving forward to tetration, let’s actually go backwards to addition. Does this sort of thinking work for addition? If you want to add 3 to itself $\frac{1}{2}$ times, you need to find some other number $x$ such that $x + x = 3$. It works!

Now back to our original question: What is 3 raised to itself $\frac{1}{2}$ times? Well, I think I just need to find some number $x$ such that $x^x = 3$. I’m not sure how to do that efficiently, but using a computer and guess and check I find that $x \approx 1.82$.

Not so fast…

Success! We did it. You made it seem like this was going to be hard.

Well, first off, it was hard! Realizing how to extend what we already know with addition and multiplication took some real creativity. But yes - unfortunately we’re about to see some problems.

Back to addition: We found that $3 \cdot \frac{1}{2} = 1.5$ because $1.5 + 1.5 = 3$. That was the most natural way think about it, but we could have used this line of reasoning: I need to find $x$ such that $x + x + x + x = 3 + 3$. Right? 4 halfs is 2, so adding 4 of these weird “half 3s” should be the same as adding 2 3s. Luckily for us, if we solve for $x$, we still get $1.5$.

Now let’s check multiplication: We found that $3^{\frac{1}{2}} \approx 1.7$ because $1.7 \cdot 1.7 \approx 3$. That was the most natural way think about it, but we could have used this line of reasoning: I need to find $x$ such that $x \cdot x \cdot x \cdot x = 3 \cdot 3$. Right? 4 halfs is 2, so multiplying 4 of these weird “half 3s” should be the same as multiplying 2 3s. Luckily for us, if we solve for $x$, we still get about $1.7$.

Now let’s check tetration: Instead of finding $x$ such that $x^x = 3$ (which was about $1.82$), let’s find $x$ such that $x^{(x^{(x^x)})} = 3^3$. Unfortunately, we get something different: $x \approx 1.806$.

What gives?

Why did this “just work” for repeated addition and multiplication but not exponentiation? I think the reason is because addition and multiplication are associative but exponentiation is not. Here is what I mean:

\[\begin{aligned} a + (a + (a + a)) &= (a + a) + (a + a) \\ a \cdot (a \cdot (a \cdot a)) &= (a \cdot a) \cdot (a \cdot a) \\ a ^ {(a ^ {(a ^ a)})} &\neq (a ^ a) ^ {(a ^ a)} \end{aligned}\]

What this means is that the solution to $x^x = 3$ is not necessarily the same as the solution to $x^{x^{x^x}} = 3^3$.

How to resolve?

I’m not sure and a quick google suggests maybe there is no consistent way to extend tetration to work with non-integer “exponents” (for lack of a better word). Maybe you can think of another way to define $\tetration{3}{\frac{1}{2}}$ that generalizes in a way that my version didn’t.

Even though this attempt failed, I learned a lot and found the entire process quite satisfying. At the risk of sounding overly preachy: I find this sort of exploratory work to be such a beautiful side of math that we don’t really get to experience growing up. In particular, taking an idea that is familiar and extending it into a domain that feels unnatural is a great way to generate interesting open-ended problems. My first introduction to this way of thinking was in the book flatland and it’s still my favorite book to this day.

Taking this further

Earlier I said “4 halfs is 2”. Of course that’s true. But… what if it wasn’t? We’ve discovered that the reason it’s true is that addition is associative.

\[\begin{aligned} \frac{1}{2} + (\frac{1}{2} + (\frac{1}{2} + \frac{1}{2})) = 2 \\ (\frac{1}{2} + \frac{1}{2}) + (\frac{1}{2} + \frac{1}{2}) = 2 \\ 1 + 1 = 2 \\ \end{aligned}\]

But what if addition wasn’t associative? Then:

\[\begin{aligned} \frac{1}{2} + (\frac{1}{2} + (\frac{1}{2} + \frac{1}{2})) \neq (\frac{1}{2} + \frac{1}{2}) + (\frac{1}{2} + \frac{1}{2}) \end{aligned}\]

I think the implication of this is that we would not be able to simplify fractions:

\[\begin{aligned} \frac{4}{2} \neq \frac{2}{1} \end{aligned}\]

It sounds impractical and strange to not be able to simplify fractions. Each rational number would be like it’s own little island and it would be harder to relate one to another, but… maybe it could work? I think that’s the sort of world we live in when it comes to tetration:

\[\begin{aligned} \tetration{3}{\frac{2}{4}} \neq \tetration{3}{\frac{1}{2}} \end{aligned}\]