The Fundamental Theorems of Hyper-Operations

We examine the extensions of the basic arithmetical operations of addition and multiplication on the natural numbers into higher-rank hyperoperations also on the natural numbers. We go on to define the concepts of prime and composite numbers under these hyper-operations and derive some results about factorisation, resulting in fundamental theorems analogous to the Fundamental Theorem of Arithmetic. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 19 February 2019 © 2019 by the author(s). Distributed under a Creative Commons CC BY license. doi:10.20944/preprints201902.0176.v1


Introduction
Again-Again!

Teletubbies
Multiplication is repeated addition.Exponentiation is repeated multiplication.We can continue applying operations again and again, to come up with higher-rank operations, known as hyper-operations.
Definition 1.1.Hyper-operation: For a, b, n ∈ N we define the hyper-operation a [n] b recursively as follows: Taken from [1].
The reader may find it useful to refer to the Wikipedia article [1] to gain familiarisation with hyper-operations and their results.
Definition 1.2.n-operation: a hyper-operation a [k] b when k = n.We say that n is the rank of the hyper-operation, and that a and b are its operands.
Definition 1.3.n-product: the result of an n-operation.
Definition 1.4.n-factorisation: for m > 1, an n-factorisation is an expression of the form with k > 0. The individual numbers a i (i = 1 . . .k) are said to be the n-factors in the expression.The innermost (i.e.right-most) n-factor a k is said to be the final n-factor in the n-factorisation.
Unless stated otherwise, in this paper all values are taken to be natural numbers, i.e.∈ N = {0, 1, 2, . . .}.In this light, statements such as these are equivalent: Note that in this paper the hyper-operation in definition 1.1 is defined on natural numbers only; in some of the proofs we will have to pause and check that neither of the arguments have crept outside of the natural numbers (i.e. are not fractions nor negative integers).
Note that the n-operation has priority over multiplication, i.e.
very much similar to exponentiation having priority over multiplication: The 0-operation (a [0] b ≡ b + 1) is referred to as succession.For other small n, we will often use standard mathematical notation for n-operations: and this nicely-compact notation for the 4-operation, referred to as tetration: a [4] b ≡ b a tetration We shall often use these named operations (addition, multiplication, exponentiation and tetration) in examples.We shall switch between notations whenever it suits our purpose.The notation a [n] b is more cumbersome but is better suited to results which apply to n-operations in general.
Regarding (1), there is one small difference between the 3-operation and traditional exponentiation, in that (depending on the branch of mathematics) 0 0 is undefined but 0 [3] 0 = 1 by definition 1.1 This small difference in behaviour will not concern us.

Some Basic Equalities
Let us now prove our first results.The proofs are by induction, which is the approach taken for many of the proofs in this paper; this is to be expected with results that concern natural numbers and a function defined using recursion (definition 1.1).
Proof.Proof is by induction on n, starting at n = 2 (since we are given n > 1).
For n = 2, the n-operation is ordinary multiplication, so we have: and so the result holds true for n = 2.
Suppose the result holds true for n = k, i.e. that a [k] 1 = a.Then for the (k + 1)-operation we have: = a by the induction hypothesis and so the result holds for n = k + 1 and our proof by induction is complete.
Proof.Proof is by induction on n, starting at n = 3 (since we are given n > 2).
For n = 3, the n-operation is ordinary exponentiation, so we have: and so the result holds true for n = 3.
Suppose the result holds true for n = k, i.e. that 1 [k] b = 1.
For the (k + 1)-operation we have these cases according to the value of b: by the induction hypothesis and so for all values of b the result holds for n = k +1 and our proof by induction is complete.

Some Basic Inequalities
Big fleas have little fleas upon their backs to bite 'em, And little fleas have lesser fleas, and so, ad infinitum.And the great fleas, themselves, in turn, have greater fleas to go on; While these again have greater still, and greater still, and so on.

Augustus De Morgan, Siphonaptera
In this section we look at the relative sizes of various n-products -i.e.how the value a

A hyper-product is larger than its operands
We confirm the intuitive result that combining a and b in an n-operation yields an n-product which is bigger than both a and b.For example: We shall now examine the following inequalities: We shall examine these inequalities in separate lemmas, as we shall find that different constraints apply to the ranges of b and n for the inequalities to hold.
Proof.We prove using double induction, the outer induction being on n.
For n = 1, we have a Then for the (k + 1)-operation we have: We now perform an inner induction on b.We start with b = 2 (since we are given b > 1): > a by the induction hypothesis on n and so the inequality a We continue the inner induction by assuming it holds true when b = j, i.e. a [k + 1] j > a.Then, for b = j + 1 we have: and so the inequality a [k + 1] (j + 1) > a holds for b = j + 1, completing our inner induction on b.
Returning to our outer induction, we have now shown that for every value b > 1, and thus our outer induction on n is complete.
We now turn to the inequality a [n] b > b, where we can prove a stronger result than for lemma 3.2, in that the ranges of values for b and n are unrestricted.Proof.We prove using double induction, the outer induction being on n.
For n = 0, we have a [n] b = b + 1, and so trivially b + 1 > b and the result holds.
(In fact, for n = 0 we do not require the restriction a > 1 but we have no need in this paper for such a slightly stronger result.Similarly, for n = 1 we would only require a > 0.) Suppose the inequality holds true for n = k, i.e. a [k] b > b.We examine the (k + 1)-operation: We now perform an inner induction on b.We start with b = 0: We continue the inner induction by assuming it holds true when b = j, i.e. that a [k + 1] j > j.Then, for b = j + 1 we have: by the induction hypothesis on n > j by the induction hypothesis on b a [k + 1] (j + 1) > j + 1 by lemma 3.1 and so the inequality holds for b = j + 1, completing our inner induction on b.
Returning to our outer induction, we have now shown that for every value of b, and thus our outer induction on n is complete.
We confirm the intuitive notion that the value of a [n] b increases if we increase the value of any of a, b or n, except for some corner cases.

A hyper-product grows with its operands
We show that increasing either operand also increases the result of an n-operation.
For example: and so on.
We shall examine the inequalities We first prove the result for increasing b -not only is it a simpler result to prove than the result for increasing a, but the result for increasing b is used in the proof for increasing a.

Proof.
Proof is by induction on n.For n = 0: Then for the (k + 1)-operation we have: and so the inequality holds for n = k +1, completing our proof by induction.
Proof.We first show the result for a = 1.We check the results for the various values of n > 0.
For n = 1 we have: For n = 2 we have: For n ≥ 3 we have: Now we deal with larger values of a.For a > 1 we can lean on previous results and we prove using double induction, the outer induction being on n.
For n = 1, we have trivially that Then for the (k + 1)-operation we have: We now perform an inner induction on b.We start with b = 1 (since we are given b > 0): and so the inequality (a + 1) We continue the inner induction by assuming it holds true when b = j, i.e. (a + 1) [k + 1] j > a [k + 1] j.Then, for b = j + 1 we have: We examine the RHS of (2) to find combining the above with (2) and so the inequality (a + 1) Returning to our outer induction, we have now shown that for every value b > 0, and thus our outer induction on n is complete.
We have thus proved the result for the cases a = 1 and a > 1. Hence the overall result holds.
Proof.By theorem 3.7, this sequence is strictly increasing 3.3 A higher-rank hyper-product is greater than a lowerrank hyper-product on the same operands We show that an (n + 1)-operation yields a larger result than an n-operation on the same operands.
For example, However, there is a particular corner case that causes us trouble.Observe We formalise this pattern in the following lemma.Freedom is the freedom to say that two plus two make four.If that is granted, all else follows.
Proof.We prove by induction on n.
For n = 1, the n-operation is addition, and so we have trivially that 2+2 = 4.
Suppose the result holds for n = k, i.e. 2 [k] 2 = 4. Then we have: by the induction hypothesis and so the result holds for n = k + 1 and the proof by induction is complete.Now we show that an (n + 1)-operation yields a larger result than an noperation on the same operands when the corner case above is excluded.
Theorem 3.10.Suppose a > 1 and b > 1 with not both a = 2 and b = 2. Then a

Proof.
Proof is by induction on b.We start with b = 2 since we are given b > 1.
We are given that when b = 2 then a = 2.We are also given a > 1 so we deduce that a > 2. We check the various cases for n.
For n = 0 we have: For n = 1 we have: For n > 1 we have: 2 by corollary 3.5 with a > 2 and so the result holds for b = 2 for all values of n.
Suppose the result holds for b = j ≥ 2, i.e. a [n + 1] j > a [n] j.Then for j + 1 we have: applying corollary 3.5 to (3) using ( 4) and so the result holds for b = j + 1 and the proof by induction is complete.Proof.By theorem 3.10, this sequence is strictly increasing Proof.This is an immediate consequence of corollary 3.11, since if not both We must have n > 0 since 2 [0] 2 = 3 but 2 [m] 2 = 4 by lemma 3.9 since m > 0.

Hyper-Prime and Hyper-Composite Numbers
Under multiplication, we have the familiar notion of prime and composite numbers.We say that m > 1 is composite if m = ab for some a > 1 and b > 1; otherwise m is said to be prime.The number 1 is said to be a unit, and is neither prime nor composite.
We extend these definitions to higher-rank hyper-operations.We ignore succession and addition as they do not yield any interesting results.
Under these definitions, the familiar notions of prime and composite become those of 2-prime and 2-composite.
We now investigate some properties of n-prime and n-composite numbers.

More n-primes as n increases
We will show that once a number m > 1 is n-prime, it stays hyper-prime for all operations of higher rank than n.
In particular, we will show that for n > 1: • all 2-prime numbers are n-prime -they are forever prime • one particular number, 4, is never n-prime -it is forever composite • all other numbers m > 1 are eventually prime -each m is n-composite for n < k and n-prime for n ≥ k, for some k depending on m.
Proof.We prove by contradiction.Suppose m is n-prime but (n + 1)-composite.Then for some a > 1 and b > 1, we have: We are given a > 1 and we now check that a [n + 1] (b − 1) > 1 for the possible values of b.
When b = 2: When b > 2, we have b − 1 > 1 and so: Hence ( 5) is an n-product of two numbers, each of which is > 1.By definition 1.4, (5) provides an n-factorisation of m, which contradicts the assumption that m is n-prime.
Theorem 4.5.4 is forever composite.
Proof.Already shown in lemma 3.9.
In our consideration of n-prime and n-composite numbers, we have excluded the number 1, dealt with the 2-primes 2 and 3, and mentioned the special case of the number 4. We now turn to larger numbers.
We now show that every number larger than 4 eventually becomes n-prime, for some high-enough ranked n-operation.
For example: • 15 = 3 × 5 is 2-composite, but is 3-prime because 15 is not a perfect power of any natural number.It follows that 15 is also 4-prime, 5-prime and so on.
and so the infinite sequence {c 2 , c 3 , . . ., c k , . . .} is strictly increasing and therefore the values in the sequence are eventually greater than any value m.
Hence there must be some n for which 4 < m < c n , and since every number d in the range 4 < d < c n is n-prime, then m is n-prime, as required.

Infinitely many n-prime and n-composite numbers
We confirm the intuitive result that for any n > 1 there are infinitely many n-prime and n-composite numbers.
By theorem 4.3 each 2-prime is also n-prime for any n > 2, hence there are infinitely many n-primes for any n > 1.Each element in the sequence is n-composite and the sequence is strictly increasing since 2 [n] (j + 1) > 2 [n] j by theorem 3.4.Hence all the elements of the sequence are distinct, giving that there are infinitely many n-composite numbers.
Clearly the sequence in the above proof does not contain all the n-composite numbers -for example, it does not contain 3 [n] 2. It is sufficient merely to show that there are infinitely many values in the sequence.

Hyper-Products and Equality
We are used to the elementary properties of addition and multiplication that allow the same result to be obtained in more than one way, e.g.: It is also possible for different exponentiation operations to yield the same results, e.g.: We observe that in the equality 4 3 = 8 2 , the bases 4 and 8 are not perfect powers of each other.Indeed, 8 = 4 3 2 and 4 = 8 2 3 .However, both 4 = 2 2 and 8 = 2 3 are perfect powers of 2. This relationship will feature in the proofs later in this section.
We now ask whether non-trivial equalities are possible for higher-rank operations, such as tetration.
Let's look at a couple of examples.Firstly, does 2 4 = b 3 for some b?
We see that 2 4 = 4 4 is a power of 4, but b 3 = 3 b−1 3 is a power of 3. By 2-prime factorisation, no non-trivial power of 4 can equal a power of 3, and so we can see that there are no solutions to 2 4 = b 3.
So let us try to use a base which shares its 2-prime factors with 4, namely 2. Does 2 4 = b 2 for some b?
Expanding the LHS gives We can see that the final 4-factor of 3 cannot be expressed as a tetration of 2 (since 3 is not even a power of 2, let alone a tetration of 2) and thus there is no b for which 2 4 = b 2.
Note that the somewhat laboured steps in (6) give a flavour of the steps in the proof that we will see below.
As a further confirmation, we can check the size of 2 4 against the sizes of these tetrations of 2: Informally, we can see there is no room between the neighbouring tetrations 3 2 and 4 2 to fit another tetration of 2 that equals the required value of 2 4. So, we cannot express 2 4 as a tetration of 2. We shall now extend this example into a theorem which shows that it is impossible to express a nontrivial tetration of two numbers as a non-trivial tetration of two other numbers.
Before we prove the main theorem, we assemble some tools that we shall need to do the job.
Proof.We prove by induction on u.
For u = 0, we have: and so the result holds for u = 0.
Suppose the result holds for u = k, i.e. r k + k < r k+1 , then for k + 1 we have: Thus, the result holds for u = k + 1, and the proof by induction is complete.Lemma 5.2.Suppose r u < s < r u+1 with r > 1.Then s is not a power of r.
Proof.By theorem 3.4 the powers of r strictly increase and we deduce that between the two neighbouring powers r u and r u+1 there cannot lie any other power of r.
Definition 5.3.Least root: Given m > 1, then r is said to be the least root of m if r is the least value such that m = r g .
Lemma 5.4.Suppose m > 1.Then there is a unique r which is the least root of m.Furthermore, 1 < r ≤ m.
Proof.We calculate r as follows.Let the 2-prime factorisation of m be for distinct 2-primes p i and indices b i > 0 (i = 1 . . .k), with p i < p j for i < j.
Such a factorisation of m exists and is unique by the Fundamental Theorem of Arithmetic.Observe that k > 0 because m > 1 and so m has at least one prime factor.
We set g as the greatest common divisor of the indices: We now specify the value of r: We now check that r has the properties we require.
We have that r ∈ N and r > 1 because r is a product of positive integers (since we know that each bi g is an integer, by the definition of g as a common divisor).
We have that r ≤ m because we know that g ≥ 1 by the properties of a greatest common divisor [4].
We have that r is a root of m, since (and in the trivial case where g = 1, then m = r, i.e. m is its own least root).Lastly, we show that r is the least of all roots of m.
Assume that there is a smaller root u < r with m = u v .Then v > g and We have that u is a natural number and so each index bi v must be a natural number, and thus we have v|b i for each i = 1 . . .k. Thus v is a common divisor of (b 1 , b 2 , . . ., b k ) greater than g, contradicting the definition of g as greatest common divisor.Thus there can be no root of m less than r, confirming that r is the least root of m.
We now show that any root of any non-trivial power of m must be a natural power of the least root of m.
Lemma 5.5.Suppose m > 1, d > 0, and u is a root of m d .Then u = r v for some v ∈ N, where r is the least root of m.
Proof.We have that r is the least root of m, so by continuing to use the notation in lemma 5.4 we have r g = m.
We have that u is a root of m d , so u w = m d for some w, and thus We set v = gd w and show that v is a natural number, which will prove that u = r v is a natural power of r.
As in the proof of lemma 5.4, we factorise m into its prime factors: with bid w ∈ N for i = 1 . . .k and thus w|b i d for i = 1 . . .k.
By the properties of gcd [4] we have: We thus have w|dg giving v = dg w ∈ N as required.Lemma 5.6.Suppose m > 1 with least root r.Then r is 3-prime.
Proof.Let r be the least root of m = r g .
Trivially we have r > 1, and so r is not the unit 1.Thus r must be either 3-prime or 3-composite (by definition 4.1 and definition 4.2).
Suppose r is 3-composite.Then by definition 4.1, r = u v for some u and v, with 1 < u < r and v > 1.
Thus we have m = r g = (u v ) g = u vg and so u is a root of m.But u < r, contradicting the definition of r as least root of m.
Thus r cannot be 3-composite and must instead be 3-prime.
So, with our tools laid out, we can now prove our theorem that all non-trivial tetrations result in different 4-products.We are given b > 1 and thus b − 1 > 0 and b−1 a > 0. We deduce that b a = a b−1 a is a non-trivial power of a.
Similarly, d c = c d−1 c is a non-trivial power of c.
We now set r to be the least root of c = r h (which exists and is unique by lemma 5.4).
From lemma 5.5, a must also be a power of r (with the possibility that a = r).We set a = r g for some g > 0, and we have: Now let us return to the equation in the statement of this theorem: We pause to check that step ( 8 From (9) we have that h g is an integer power of r, and (7) gives us that h g > 1.We thus have that h g is a natural power of r.
We set Proof.Proof is by induction on n.
The result for n = 4 is proved in theorem 5.7.
Assume the result holds for n = k, i.e. there are no non-trivial solutions to a Suppose we have an equation using the [k + 1] operation for a, b, c, d > 1: We put We pause to check that the expansion in ( 14) is valid: we are given b > 0 and d > 0, hence b − 1 > 0 and d − 1 > 0 and the expansion in ( 14) is indeed valid.
Similarly we have h > 1.We already have a > 1 and c > 1, so it is valid to apply the induction hypothesis to (15) to show that a = c.We then apply corollary 3.6 to (13) to show that b = d.
Thus the result holds for n = k + 1, and the proof by induction is complete.

Commutativity and Associativity
I'm playing all the right notes, but not necessarily in the right order.

Eric Morecambe
The commutative and associative properties of addition and multiplication are well known, i.e. for any a, b and c: We take a look at other hyper-operations.

Commutativity
Here we look at whether commutativity holds for other hyper-operations.We ignore the trivial cases where a = b since clearly any operation which takes in two identical operands will yield the same result if those identical operands are swapped.We also ignore cases where either operand is 0 or 1.
First we show, as a mere curiosity, that the succession operation (n = 0) is not commutative: The proofs for commutativity under addition and multiplication (for all values a and b) are well known and are not repeated here.
For the moment we skip the 3-operation and look at higher-rank operations.Returning now to the 3-operation, let's see if there are any non-trivial commutative operations.We can see at once that not every 3-operation is commutative by way of a simple counter-example, 2 3 = 3 2 .However, after some searching, we do find a non-trivial commutative 3operation: Are there any others?The answer is no, which we shall now prove.Proof.We set r to be the least root of a = r g for some r > 1, g > 0 by lemma 5.4.
Then by lemma 5.5, r is also the least root of b = r h for some h > 0. We also have that h > g since we are given b > a. a b = b a given (r g ) (r h ) = r h (r g ) substituting for a and b r gr h = r hr g gr h = hr g by corollary 3.6, since r > 1 h g = r h−g rearranging (which shows that h g is a natural number > 1, since h > g gives that the RHS is a non-zero power of r) However, applying corollary 3.5 with n = 2 gives us that r ( h g ) > r h g contradicting ( 17), unless r = 2 and h g = 2.
We duly set r = 2 and h g = 2 and substitute into (16) to give as required.

Associativity
Now we turn to whether associativity holds for other hyper-operations.Again we ignore the trivial cases where a = b = c and also ignore cases where any of the operands is 0 or 1.
giving a contradiction, so there are no associative 0-operations (even trivial ones).
As for commutativity, the proofs for associativity under addition and multiplication (for all values a and b) are well known and are not repeated here.7 Unique n-Factorisation: The Fundamental Theorems There are often many ways of using multiplication to arrive at the same result.For example, It is natural to ask whether there is a unique canonical representation of a number, arrived at through multiplication of other numbers.
The Fundamental Theorem of Arithmetic [3] is a well-known result that every m > 1 can be factorised as a product of powers of 2-primes, i.e.
Due to the commutativity of multiplication, these powers of primes can be put in any order to arrive at the product m.In order to provide a unique factorisation of m, we choose a canonical order for the primes, the most natural being to place them in ascending order, i.e. p i < p j for i < j.
Note that there may be gaps in the 2-primes, for example 20 = 2 2 × 5 1 , which makes no mention of the 2-prime 3.In this paper (for the purposes of our theorems) we do not permit an index of 0, and so we cannot write 20 = 2 2 × 3 0 × 5 1 .We now explore factorisation for higher-rank operations, arriving at results that we shall name as the Fundamental Theorems of Hyper-Operations.We shall prove two distinct but related theorems, one for the 3-operation and another for ≥ 4-operations.

The Fundamental Theorem of the 3-operation
It is easy to come up examples of multiple ways of using the 3-operation to arrive at the same result.1 is an all-3-prime factorisation of m.
We see that c 1 is a root of m (which is a natural power of a 1 ), and by lemma 5.5, c 1 is a natural power of the least root r of m.We chose a 1 to be the least root of m and so c 1 = a 1 u for some u > 0.
However, by definition 7.1, c 1 is 3-prime which forces u = 1 and thus c 1 = a 1 , giving c with this last expression containing 3-factors which are all 3-prime.
We observe that, in comparison to a 2-prime factorisation, the order of the factors in an all-3-prime factorisation is critically important.
Under multiplication, these permutations all produce the same result: By contrast, under exponentiation, the 6 permutations each produce a number which is different from the other permutations: This difference in the effect of ordering between a 2-prime factorisation and an all-3-prime factorisation is related to commutativity (see section 6.1).

The Fundamental Theorem of the (≥ 4)-operations
Having established a unique factorisation theorem for the 3-operation, we now turn to higher-rank operations.Using tetration as an example, we have: Observe that in the expression 2 3 4, not all the 4-factors are 4-prime.In particular, 4 is 4-composite, since it can be 4-factored as 4 = 2 2. However, due to theorem 5.7, we know we cannot equate a tetration of 4 with a tetration of any other number, and so we are stuck with the number 4 being the first 4-factor in any non-trivial 4-factorisation of 18, 014, 398, 509, 481, 984.
Clearly then, we cannot prove a unique factorisation theorem which insists that every 4-factor is a 4-prime.This contrasts strongly with theorem 7.2 for all-3-prime factorisation where every 3-factor is a 3-prime.However, we were able express the index 27 as 2 3, so that both 27 4 and 2 3 4 are expressions that use only the tetration operation, but evaluate to the same result.
Observe that in 27 4, the final 4-factor 27 is 4-composite, whereas in 2 3 4 the final 4-factor 2 is 4-prime.It is this final factor that is important in our uniqueness theorem.
Definition 7.3.Final-n-prime factorisation: an n-factorisation where the final n-factor is n-prime.
Theorem 7.4.Suppose m > 1 and n ≥ 4. Then there is a unique final-n-prime factorisation of m.
Proof.We take a similar approach to that used in the proof for theorem 7.2.
By corollary 5.8, we must have c 1 = a 1 and d 1 = b 1 .We repeat the check on the n-factorisation of b 1 and d 1 to deduce that c i = a i for i = 1 . . .k, also forcing t = k.
This shows that the final-n-prime factorisation thus proving our uniqueness requirement.
[n] b grows in relation to the values of a, b and n.First we establish a tiny result about natural numbers.Lemma 3.1.Suppose c > b > a. Then c > a + 1.Proof.Since we are restricted to N, we have: [n] b = a + b, and so trivially a + b > a we are given b > 1 Suppose the inequality holds true for n = k, i.e. a [k] b > a.
Theorem 5.7.Suppose b a = d c with a, b, c, d > 1.Then a = c and b = d.Proof.If a = c, then we have b = d by corollary 3.6.Otherwise, a = c, and we shall show that this eventually leads to a contradiction.Without loss of generality, assume a < c.This gives b > d, since if b ≤ d then by lemmas 3.8 and 3.5 we would have b a < d c contradicting b a = d c.Given c, what values are possible for a?
) is valid, i.e. that both b − 2 and d − 2 are natural numbers; we are given that d > 1 and have deduced that b > d, hence b − 2 > d − 2 ≥ 0 and thus step (8) is indeed valid.
Suppose a [n] b = c [n] d with a, b, c, d > 1 and n ≥ 4. Then a = c and b = d.
number of iterations, b k = 1 for some k > 0, halting the procedure, and giving us an all-3-prime factorisation m = a show that this all-3-prime factorisation of m is unique.