Inverses of Hyperoperations

$H_{x}^{-1}(a,b)=H_{x}(a,H_{x+1}(b,-1))$

The formula above is a neat way I found to define constructive inverses of the first and second hyperoperations using other hyperoperations.

Hyperoperations

First, a bit of background.

The basic arithmetic operations (e.g. addition and multiplication) are defined in terms of each other. For example, multiplication is just repeatedly performing addition. This shouldn’t be news for you; you most certainly learned this in grade school. The technical math term for this is the hyperoperation sequence.

Successor

The zeroth hyperoperation, which is denoted $H_0$, is the successor function, which just increments its input by $1$. For example, $H_0(5)=6$.

Addition

The first hyperoperation, denoted $H_1$, is addition. It’s defined as passing $a$ through the successor function $b$ times. Here’s an example:

$\begin{aligned} 5+4 &= H_1(5, 4) \\ 1+1+1+1+5 &= H_0(H_0(H_0(H_0(5)))) \\ 1+1+1+6 &= H_0(H_0(H_0(6))) \\ 1+1+7 &= H_0(H_0(7)) \\ 1+8 &= H_0(8) \\ &= 9 \end{aligned}$

Multiplication

The second hyperoperation, denoted $H_2$, is multiplication. It’s defined as adding $a$ to itself $b$ times. Here’s an example:

$\begin{aligned} 5 \times 4 &= H_2(5, 4) \\ 5+5+5+5 &= H_1(5,H_1(5,H_1(5,5))) \\ 5+5+10 &= H_1(5,H_1(5,10)) \\ 5+15 &= H_1(5,15) \\ &= 20 \end{aligned}$

Exponentiation

The third hyperoperation, denoted $H_3$, is exponentiation. It’s defined as multiplying $a$ by itself $b$ times. Here’s an example:

$\begin{aligned} 5^4 &= H_3(5, 4) \\ 5 \times 5 \times 5 \times 5 &= H_2(5,H_2(5,H_2(5,5))) \\ 5 \times 5 \times 25 &= H_2(5,H_2(5,25)) \\ 5 \times 125 &= H_2(5,125) \\ &= 625 \end{aligned}$

Tetration

The fourth hyperoperation, denoted $H_4$, is tetration. It’s defined as raising $a$ to its own power $b$ times. Here’s an example:

$\begin{aligned} {}^4 5 &= H_4(5, 4) \\ 5^{5^{5^{5}}} &= H_3(5,H_3(5,H_3(5,5))) \\ 5^{5^{3125}} &= H_3(5,H_3(5,3125)) \\ 5^{(10^{2184.2813})} &= H_3(5,10^{2184.2813}) \\ &\approx 10^{10^{2184.1257}} \end{aligned}$

Unfortunately, I can’t show you the exact value of the result of the example above, because it’s simply too large.

That number is so large that there are more digits in that number than there are atoms in the observable universe.

In fact, if each atom in the observable universe contained its own observable universe, and each sub-atom in those observable universes contained its own observable universe, and this nesting process repeated for 25 times in total, the total number of atoms would still be less than the number of digits in that number. By over 100 orders of magnitude.

Tetration doesn’t mess around.

And beyond

There are more hyperoperations (infinitely more, in fact), but there’s really no practical reason to ever use the higher ones. Even with tetration, you run into “there aren’t enough atoms in the observable universe to represent the number of digits this number has” problems without even trying, and the higher hyperoperations only make it worse.

Inverses

Anyways, you might notice that the hyperoperation sequence doesn’t include subtraction or division. These operations are inverses of the hyperoperations. In other words, each inverse hyperoperation precisely undoes the actions of its corresponding hyperoperation.

However, unlike the hyperoperations themselves which follow a very predictable sequence, the inverses are much more chaotic.

Zeroth

The inverse of $H_0$ is the predecessor function. The successor function adds $1$ to a number, and the predecessor function subtracts $1$ from a number. Very simple.

$H_0^{-1}(b)=b-1$

There are a few reasons why we can’t use my formula to define the predecessor function, but the most obvious is that the predecessor function only takes one input, while my formula describes a function that takes two.

Onto the next one.

First

The inverse of $H_1$ is subtraction. You obviously know how subtraction works, though it’s possible that you might have to think for a bit in order to come up with a rigorous constructive definition (i.e. not just “a function that returns the value you’d need to add to $a$ in order to get $b$.”).

The best way I know of to do this is to use addition and multiplication:

$a-b = a+(b \times -1)$

Or, in hyperoperation notation:

$H_1^{-1}(a,b) = H_1(a,H_2(b, -1))$

This looks a lot like my formula! That’s because it is. If you set $x=1$, my formula directly produces the one above:

$\begin{aligned} H_{x}^{-1}(a,b) &= H_{x}(a,H_{x+1}(b,-1)) \\ &\downarrow x=1 \\ H_{1}^{-1}(a,b) &= H_{1}(a,H_{1+1}(b,-1)) \\ &\downarrow \\ H_{1}^{-1}(a,b) &= H_{1}(a,H_{2}(b,-1)) \end{aligned}$

Neat! Let’s see if the pattern holds.

Second

The inverse of $H_2$ is division. Again, you know how division works, even though you might have to think a bit to define it constructively.

The best definition I know of uses multiplication and exponentation:

$\frac{a}{b} = a \times (b^{-1})$

This seems eerily similar to our definition for subtraction. “Use the hyperoperation you’re inverting but pass the second value through the next hyperoperation along with $-1$”. And in fact, if we set $x=2$, my formula produces the expected result.

$\begin{aligned} H_{x}^{-1}(a,b) &= H_{x}(a,H_{x+1}(b,-1)) \\ &\downarrow x=2 \\ H_{2}^{-1}(a,b) &= H_{2}(a,H_{2+1}(b,-1)) \\ &\downarrow \\ H_{2}^{-1}(a,b) &= H_{2}(a,H_{3}(b,-1)) \end{aligned}$

If the formula has held twice now, it’ll probably hold for the rest too. Let’s see!

Third

The inverse of $H_3$ is… well… it’s complicated. Exponentiation actually has two inverses: roots and logarithms. We’ll get to why it has two later.

First, let’s try to apply my formula:

$\begin{aligned} H_{x}^{-1}(a,b) &= H_{x}(a,H_{x+1}(b,-1)) \\ &\downarrow x=3 \\ H_{3}^{-1}(a,b) &= H_{3}(a,H_{3+1}(b,-1)) \\ &\downarrow \\ H_{3}^{-1}(a,b) &= H_{3}(a,H_{4}(b,-1)) \end{aligned}$

Aaaand stop right there. We just tried to use $H_4(b, -1)$, which means tetrating with a height of $-1$. That’s not allowed.

There’s a case to be made that ${}^{-1}b = 0$ for any $b$, but you run into contradictions if you assume this, similar to what happens if you divide by $0$. Most mathematicians agree that, for $H_4$ and up, the domain of the second argument is $b \geq 0$ in most cases.

My formula clearly doesn’t work for the inverse(s) of exponentiation, so what does work?

Roots

As mentioned, the left inverse of $H_3$ is the root. As usual, you probably know how roots work, but a constructive definition might require a bit of thought.

You might remember that $a^{\frac{1}{2}}=\sqrt{a}$. You can generalize this to $a^{\frac{1}{b}}=\sqrt[b]{a}$. From there, you can rearrange the fractional exponent to get $a^{(b^{-1})}=\sqrt[b]{a}$. Which is, in hyperoperation notation:

$H_3^{-1L}(a,b) = H_3(a, H_3(b,-1))$

So rather than passing the second value through the next hyperoperation ($H_{x+1}$) along with $-1$, you just pass it through the same hyperoperation ($H_x$). Weird!

Logarithms

The right inverse of $H_3$ is the logarithm. Logarithms are way weirder than any other hyperoperation or inverse hyperoperation below tetration.

For starters, logarithms often produce transcendental numbers as results. While you can construct equations that use logarithms and only include normal numbers (e.g. $\log_4(64) = 3$), their default behavior is to spew transcendental numbers.

The typical definition for logarithms is this:

$\log_b(a) = c \Leftrightarrow b^c = a$

This works, but it’s an implicit definition. If you want to actually solve for the value of a given logarithm, you have to guess-and-check, which sucks. We want a constructive definition.

Here’s my preferred constructive definition of logarithms, derived using the integral definition of the natural logarithm and the change of base formula:

$H_3^{-1R}(a,b) = \log_b(a) = \frac{\int_{1}^{a} \frac{1}{t} \, dt}{\int_{1}^{b} \frac{1}{t} \, dt}$

Eww. Integrals? In the definition of an arithmetic operation? Unfortunately, because logarithms are transcendental, you by definition can’t constructively define them using finite algebra. So we have to use at least some calculus. Still gross though.

Fourth

Like exponentiation, tetration has two inverses. They’re very creatively named super-root and super-logarithm.

Because tetration gets really weird really fast (and also it’s only defined for natural numbers), it’s difficult to create general constructive definitions.

And besides, it should be clear by now that my formula isn’t going to apply to the higher hyperoperations.

Why does the pattern break?

Well, for a lot of reasons, but the most prominent reason is that addition and multiplication are commutative, but exponentiation is not:

$\begin{aligned} a + b &= b + a \\ a \times b &= b \times a \\ a^b &\neq b^a \end{aligned}$

There’s also the aforementioned “tetration is not defined for $b=-1$” problem, but non-commutativity is the root (ba-dum-tss) of the issue.

For example, non-commutativity is the reason why the higher hyperoperations have two inverses. Inverses are about calculating the value of one input using only the other input and the output. If the same inverse function could calculate either input of a non-commutative function, the inputs of the function must be structurally the same. If the inputs must be structurally the same, they must be interchangeable. If the inputs must be interchangeable, the function must be commutative, which is a contradiction. Therefore, non-commutative functions must have multiple inverses.

Why is exponentiation non-commutative?

Because it is.

I wish I had a more satisfying explanation, but I don’t. Why exactly the pattern of commutativity holds for $H_1$ and $H_2$ but breaks at $H_3$ is difficult to explain without either going in circles or just kicking the “because it is” can slightly further.

For example, one explanation goes like so:

Below is a table of what effect each input of each hyperoperation has when varied.

Hyperoperation	First Input ($a$)	Second Input ($b$)
Addition	Additive effect	Additive effect
Multiplication	Linear effect	Linear effect
Exponentiation	Polynomial effect	Exponential effect

Because the inputs of exponentiation have different effects, they aren’t interchangeable. Therefore, exponentiation isn’t commutative.

And while this is true, it just begs the question “but why do the inputs of exponentiation have different effects?”, for which the answer is either “because exponentiation isn’t commutative” or “because they do”. In other words: either circular logic or an unjustified assertion.

See why this is tricky?

If you have a satisfying explanation for why exponentiation is non-commutative that isn’t circular or assertive, I would really really love to hear it. Please email me if you do.

Conclusion

I’m 99.99% sure that I’m not the first person to discover “my” formula, though I did at least discover it independently and I couldn’t find my exact formulation anywhere online.

The fact that you can construct each of the arithmetic operations in terms of the others is delightful to me (examples below), and I thought that the pattern in the definitions of subtraction and division was interesting, so I formalized it.

$\begin{aligned} a + b &= a-(0-b) \\ a - b &= a+(b \times -1) \\ a \times b &= a/(1/b) \\ a / b &= a \times (b^{-1}) \end{aligned}$

Anyways, thanks for reading!

~Ethan