Hyper-Operations

This is an idea I had a while ago. Someone in 1982 BC probably had this idea before me but whatever, I couldn't find it. (I found something similar)

The Idea

The basic idea is to assign a number to a set of arithmetic operations and see where we can go with this.

I don't know the name of this set (maybe Hyper-operations?) but its items are Addition, Multiplication, Exponentiation, Tetration...

The syntax I will use to represent the operations is a (n) b where a is any number, b is any number, and n is a integer.

One way of defining x (n) y by saying that x (n) y is ((x (n - 1) x) (n - 1) x) (n - 1) x... b times. Or, to be more formal, x (n) y = (x (n) (y - 1)) (n - 1) x.

Here is a breakdown: x (n) y is just Iterate(LastIteration) y times. Iterate(value) is just value (n - 1) x. LastIteration is just x (n) (y - 1) because it is the current operation minus 1 operation. So Iterate(LastIteration) (x (n) y) is just Iterate(x (n) (y - 1)) which is just (x (n) (y - 1)) (n - 1) x.

A major pro of doing it like this is that you only need the last operation to define the current.

Defining operations "above" addition is cool and all but those are well known, the real question is what is below addition.

Below Addition

I will need additional names and symbols for the operations below addition.

It would make sense to use a naming system similar to that used in Hyperoperations, so Number Greek/Latin Prefix + -ation.

There is no Number Greek/Latin Prefix for 0 so I used null which means none, close enough to zero.

There is also no Number Greek/Latin Prefix for negative numbers so for those I use a (meaning not) + Positive Number Greek/Latin Prefix.

For the 0 operation, it would make sense to use the empty set symbol. But not so obvious for negative operations. I ended up choosing on using the Boolean symbol for negation (¬ or line on top) + the "normal" symbol.

So here is what the table would look like:

Operation 0

What is the Operation 0? We know that when repeated gives addition but not yet its properties.

Here is some info about how I will represent it:

Using our previous equation we can define addition in terms of nullation.

x (n) y = (x (n) (y - 1)) (n - 1) x
x (1) y = (x (1) (y - 1)) (0) x
x + y = (x + y - 1) Ø x

Ok, now we have addition defined using nullation but still not nullation.

Here is how I calculated nullation:

a = (x + y - 1)
b = x

a Ø b = x + y

now we just need the values of x and y in terms of a and b.

b = x
x = b

a = (x + y - 1) and x = b
so
a = (b + y - 1)
a + 1 = b + y
(a - b) + 1 = y
y = (a - b) + 1

now we have the values of x and y which are:

x = b
y = (a - b) + 1

now let's calculate the complete equation

a Ø b = x + y
a Ø b = (b) + ((a - b) + 1)
a Ø b = (b - b) + a + 1
a Ø b = a + 1

Now we have the full equation, a Ø b = a + 1.

Nullation is Successor, which makes a lot of sense when you think about it, if you compute 3 + 2 you get the successor of the successor of 3 which is 5!

Operation -1

Now that we have discovered the properties of nullation (successor), what comes below Successor?

As before, here is how I will represent it:

If you think about it, it makes not much sense what comes before Successor because the successor ignores the second argument which is the iteration count so what Iterated n times is Successor?

Using the previous equation we can define nullation in terms of amonation.

x (n) y = (x (n) (y - 1)) (n - 1) x
x (0) y = (x (0) (y - 1)) (-1) x
x Ø y = (x Ø (y - 1)) ¬+ x

Now that we know the properties of nullation we can simplify this even further.

x Ø y = (x Ø (y - 1)) ¬+ x
x Ø y = (x + 1) ¬+ x

Now we can use the same method we used before to calculate amonation in terms of nullation.

a = x + 1
b = x

x Ø y = a ¬+ b

now let's calculate a and b in terms of x and y

b = x
x = b

a = x + 1
x + 1 = a
x = a - 1

Oh no, we have two values for x and none for y.

But that is not a problem since nullation (successor) only takes a single argument.

So we can mark y as undefined.

For undefined I will use the Bottom Element (⊥) since it is the symbol used in Wheel Theory for undefined.

Now we can build the full equation

x = b
x = a - 1
y = ⊥

x Ø y = a ¬+ b

a ¬+ b = x Ø y
a ¬+ b = b Ø y = (a - 1) Ø y
a ¬+ b = b Ø ⊥ = (a - 1) Ø ⊥
a ¬+ b = b + 1 = (a - 1) + 1
a ¬+ b = b + 1 = a

This makes a lot of sense!
a ¬+ b = b + 1 is the summation of the second argument
a ¬+ b = a is the identity of the first argument

both of these solutions work since identity repeated N times is always going to be identity and the summation of the second argument repeated N times is always going to be the summation of the second argument since the second argument is constant no matter the number of times you repeat it.

Anti-operations

We have only defined a few operations. Operations like subtraction, division, radiation and logarithmization are not included in our set.

Those operations are what I call anti-operations as they are the reverse of operations.

The name I will give for an anti-operation is the name of the operation prefixed with anti-. Example: The anti-operation of addition (subtraction) is going to be called anti-addition.

The symbol I will give for anti-operation is the operation symbol and a -1 subscript. Example: The symbol for anti-addition (subtraction) is going to be +₋₁.

To calculate anti-operations I will use the equation (x (n)₋₁ y) (n) y = x

Let's try to define anti-addition.

Using the equation from before,

(x (n)₋₁ y) (n) y = x
(x (1)₋₁ y) (1) y = x
(x +₋₁ y) + y = x
x +₋₁ y = x - y

And look at that! Just by rearranging the terms we get to a solution.

And, as you probably expected the solution is subtraction.

We can also try to define anti-amonation (or anti-succession).

Using the same method as before,

(x (n)₋₁ y) (n) y = x
(x (0)₋₁ y) (0) y = x
(x Ø₋₁ y) Ø y = x
(x Ø₋₁ y) + 1 = x
x Ø₋₁ y = x - 1

And as before, just by rearranging the terms we get to a solution.

And, as you probably expected the solution is predecession.

Neutral Element

What is a neutral element? The way I will define them is that the neutral element for operation n is such that x (n) y = x where x is any number and y is the neutral element.

The neutral element for addition (Operation 1) is 0 (x + 0 = x)
The neutral element for multiplication (Operation 2) is 1 (x * 1 = x)
The neutral element for exponentiation (Operation 3) is 1 (x¹ = x)

Is there an equation to find neutral elements?

Yes, x (n) 0 (where x is any number) is the neutral element for operation n.

If you think about it kind of makes sense, x (n) 0 would be operation n repeated on x 0 times and repeating 0 times should not affect the output, or, in other words, the neutral element.

If x (n) 0 is really the neutral element for operation n - 1 than x (n) (y (n + 1) 0) should always equal to x because y (n + 1) 0 is the neutral element for operation n.

Now, there are some cases where there obviously aren't any neutral elements such as for Nullation as Nullation ignores the second element.

Nullation: x Ø y = x + 1

So what happens if we try to calculate the neutral element?

x (1) 0
x + 0
x

We get a variable! Normally we would get a constant such as 0 for addition or 1 for multiplication.

Let's test our conjecture from before.

x (n) (y (n + 1) 0) = x
x (0) (y (1) 0) = x
x Ø (y + 0) = x
x Ø y = x
x + 1 = x

We get a contradiction! So x (n) (y (n + 1) 0) will only be x when the operation n has a neutral element which is y (n + 1) 0.

Repetition

First, let's define the Composition Operator, (f ∘ g)(x) = f(g(x)).
And then, Repetition,
fₙ(x) = (fₙ₋₁ ∘ f)(x)
f₀(x) = x

Given f(x) = x (n) y, how could you define fₘ(x)?

Well, fₘ(x) is just x (n) y (n) y (n) y (n) y ... m times or x (n) (y (n) y (n) y (n) y ...) m times and y (n) y (n) y (n) y ... is just y (n + 1) m.

So now we have our full equation, fₘ(x) = x (n) (y (n + 1) m).

I don't see any practical use for this (maybe aside from m not being limited to only integers?)