19
$\begingroup$

In natural numbers the unary successor operator $S$ is the most natural function which maps each number to the next one. Furthermore we may consider the binary relation $+$ as an iteration of $S$. Also $\times$ is an iteration of $+$ and $\exp$ is an iteration of $\times$ i.e.

$\forall m,n\geq 1$

$m+n:=\underbrace{S(S(S(\cdots S}_{n\ \text{times}}(m))))$

$m\times n:=\underbrace{m+m+m+\cdots+m}_{n\ \text{times}}$

$m^n:=\underbrace{m\times m\times m\times \cdots\times m}_{n\ \text{times}}$

We build the rich arithmetic of natural numbers via above three natural operators. Also they show many complicated mutual relations with each other. But, why do we stop here in arithmetic of natural numbers and don't go forward with continuing iterating operators again and again? i.e.

$m*n:=\underbrace{m^{m^{m^{.^{.^{.^{m}}}}}}}_{n\ \text{times}}$

$m\circledast n:=\underbrace{m*m*m*\cdots*m}_{n\ \text{times}}$

$m\circledcirc n:=\underbrace{m\circledast m\circledast m\circledast \cdots\circledast m}_{n\ \text{times}}$

$\vdots$

The point is that maybe there are rich interactions between these new natural operators and the ordinary arithmetic operators of natural numbers. These interactions may unfold some deep aspects of long standing open questions of number theory which hopefully can lead us to a solution itself.

Question: Why do we stop at exponentiation stage in arithmetic of natural numbers? Is there any mathematical or philosophical problem about defining such generalized operators and working with them as well as successor, sum, multiplication and exponentiation? Are these "unnatural" in any sense? If yes, what does this "unnatural" essence mean? Did these extended set of operators on natural numbers appeared in any text before? If yes, please introduce your references.

$\endgroup$
7
  • 15
    $\begingroup$ I guess that the main reason is that further iterations (tetration, $\uparrow$, whatnot) don't occur often enough to warrant that much attention. We don't do this stuff in a vacuum. $\endgroup$
    – Jyrki Lahtonen
    Oct 17, 2014 at 9:24
  • 4
    $\begingroup$ See tetration, Graham's number, and Knuth's up-arrow notation. $\endgroup$
    – Lucian
    Oct 17, 2014 at 9:35
  • 3
    $\begingroup$ See "my number is bigger" games among mathematicians. But part of the problem is that exponentiation already grows really fast: significantly faster growing functions quickly stop talking about physically realizable numbers. Which makes applications limited. $\endgroup$
    – Yakk
    Oct 17, 2014 at 14:51
  • $\begingroup$ Exponentiation isnt associative, and after this any other level is. This is a problem or a limitation. And too because bigger numbers are not too useful ofeten but some problems of theory number. $\endgroup$
    – Masacroso
    Oct 17, 2014 at 17:13
  • 3
    $\begingroup$ Why do we stop ? Because we're too afraid of what lies beyond$\ldots$ ;-$)$ $\endgroup$
    – Lucian
    Oct 17, 2014 at 23:34

7 Answers 7

Reset to default
21
$\begingroup$

Why do we stop at the exponentiation stage in the arithmetic of natural numbers?

We don't actually stop there. Knuth's up-arrow notation generalizes the operations up to any level you want and Conway's chained arrow notation even further. Note though that all these generalizations are applicable only for $n\in\mathbb{N}$.

Inability to generalize and hence stop at exponentiation, is a problem only when you try to extend the nomenclature to include definitions for REAL $n$. That's an entirely different problem and opens a rather big can of worms. See tetration on Wiki for details.

$\endgroup$
4
  • $\begingroup$ It seems you have done some research on the subject of extension of Knuth's up-arrow notation to $\mathbb{R}$. Is your literature available online ? $\endgroup$
    – Monk
    Oct 17, 2014 at 10:04
  • 3
    $\begingroup$ My research on that subject was very limited and did not produce anything interesting. For much more interesting results, try to find Andrew Robbins' extension or Gottfried Helms' Carleman Matrix method, which give more consistent results, based on the Abelization of the characteristic functional equation for tetration. I think Wiki has some links to both parties. $\endgroup$
    – user127032
    Oct 17, 2014 at 10:20
  • 2
    $\begingroup$ Does it have anything to say, that the operators stop being commutative? I.e. $2 + 3 = 3 + 2$, $2 * 3 = 3 * 2$, but $2^3 \neq 3^2$. $\endgroup$
    – kasperd
    Oct 17, 2014 at 15:07
  • $\begingroup$ I don't think anything changes in that direction. Anything above and including exponentiation ceases to be commutative (in general), hence many of the problems in the corresponding subject. That's why the problem of finding solutions to $x^y=y^x$ is so famous, as it is a notable exception to the general rule. $\endgroup$
    – user127032
    Oct 17, 2014 at 15:25
15
$\begingroup$

Others have done a good job of explaining how exponentiation can be generalized. I'm going to address the question of why mathematicians are not interested in these generalizations.

I think the main reason why Knuth's up arrow has not taken off in the same way that, say, exponentiation has is that some of the very nice properties enjoyed by addition and multiplication start to break down upon further iteration. Some of these nice properties are

  1. Associativity: $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$.
  2. Commutativity: $a+b = b+a$ and $ab=ba$.
  3. Existence of an identity: $a + 0 = a$ and $a\cdot 1 = a$.
  4. Existence of an inverse: $a - a = 0$ and $a\cdot \frac 1a=1$.
  5. Distributivity: $a(b+c) = ab + ac$ and $(a+b)c = ac + bc$.

Mathematicians have found it very fruitful to abstract out these properties and study other "number systems" that obey these rules, in addition to a few others. This train of thought leads to commutative algebra and the study of commutative rings, among other fertile branches of mathematics. We therefore have a good "moral" reason to be interested in binary operations that satisfy some subset of the five properties above: there's a lot to say about them.

Let's see how exponentiation stacks up: is exponentiation a "nice" binary operation?

  1. Associativity: $(a^b)^c\neq a^{(b^c)}$ in general.
  2. Commutativity: $a^b\neq b^a$ in general.
  3. Existence of an identity: $a^1 = a$ but there is no $x$ such that $x^a=a$ for all $a$. So exponentiation only has a partial identity.
  4. Existence of an inverse: it doesn't really make sense to talk about an inverse because there's only a partial identity. It is possible to solve the equation $x^b=c$ by taking the $b$th root of $c$, at least when $c$ is nonnegative. It is also possible to solve the equation $b^x = c$ by using the logarithm function, at least when $b$ and $c$ are positive and not equal to $1$.
  5. Distributivity: $a^{bc} \neq a^ba^c$, although $(bc)^a = b^ac^a$. So exponentiation is only partially distributive.

You can already see that exponentiation is a lot more complicated than addition and multiplication. There are really two exponentiation functions: $x\mapsto x^a$ and $x\mapsto a^x$, which behave very differently. The first is inverted by taking roots and the second by taking logarithms; both also start to behave badly if the arguments are negative. As you can imagine, the higher iterates of exponentiation are even more complicated.

I don't want to suggest that mathematicians never study objects that don't have "nice" behavior -- there are many counterexamples to such a claim. But we only tend to study badly behaved objects when there is some clear use for them, and Knuth's up arrow doesn't seem to be much more than a good way for talking about really large numbers.

$\endgroup$
1
  • $\begingroup$ Categorically, there are analogues of addtion, multiplication, and exponentiation. The good properties of addition and multiplication follow from the definition itself(which are easily seen to be commutative, associative.) But exponentiation is not. It is not commutative at first from its definition. I think that makes the difference, and it might not be fruitful categorically to discuss iterations of exponentiation and further successors. $\endgroup$
    – prime235711
    Nov 30, 2022 at 18:35
8
$\begingroup$

Sure, large numbers do appear moderately often, if you look up the research literature in Combinatorics. The examples I can think of are the Knuth's Up-Arrow Notation, Conway Chained-Arrow Notation, Hyperoperators, Tetration, Cutler's Bar Notation, Steinhaus-Moser Notation.

A good starting point for entry into the study of such operators would be this article on Wikipedia : Knuth's Up-Arrow Notation

Also, if you interested in large numbers, you might want to check this out : Googol

You could also read Lore of Large Numbers published by American Mathematical Society.

$\endgroup$
5
$\begingroup$

Instead of thinking of multiplication as repeated addition, we can define it as the unique commutative, associative operator that distributes over addition and has 1 as the identity. This suggests another approach to generalizing beyond multiplication, by asking what operator distributes over multiplication. Define the operator $$ a *_n b = \exp^n (\log^n(a) + \log^n(b)) $$ where $\exp^n$ denotes repeated application of the $\exp$ function, and similarly for $\log^n$. The first few cases are: $$ a *_0 b = a + b \\ a *_1 b = ab \\ a *_2 b = a^{\log(b)} = b^{\log(a)} $$ There are known as Bennett's commutative hyperoperations. The operator $*_2$ distributes over multiplication, in the sense that $$ a *_2 (bc) = (a *_2 b)(a *_2 c) $$ In general, the operator $*_n$ distributes over $*_{n-1}$: $$ a *_n (b *_{n-1} c) = (a *_n b) *_{n-1} (a *_n c) $$ In contrast to exponentation and tetration, these operators are all commutative, associative, and can be applied to complex numbers. We can even extend the sequence in the other direction, by asking "what does addition distribute over?". The answer is the operator $*_{-1}$, defined as: $$ a *_{-1} b = \log(\exp(a) + \exp(b)) \\ a + (b *_{-1} c) = (a + b) *_{-1} (a + c) $$

$\endgroup$
1
3
$\begingroup$

A logician's reason:

Every recursive function (and much more) is definable in the (first-order) structure $(\mathbb{N}, +, \times)$ but not in $(\mathbb{N}, S, +)$. This is due to Godel. To appreciate why this theorem is non-trivial try finding a formula $\phi(x, y, z)$ in the language of arithmetic that expresses $x^y = z$.

$\endgroup$
1
$\begingroup$

As others pointed out, we don't stop at exponentiation. We do stop at it in primary, secondary, and early post-secondary school though. Which I think is why your question arises.

Simply we stop in school because teacher's have a hard enough time teaching multiplication, and they teach it incorrectly. You yourself may not understand multiplication when you say it is merely iterated addition, which it is not. The simplest example of the difference in multiplication is that 1 cm x 1 cm != 1 cm (centimetres). Multiplication can simulate repeated addition, but it is not iterated addition. (Yes I understand that when working just with Natural Numbers, it always simulates repeated addition but try fixing that misinformation in university students or adults who've had that misinformation in their heads for decades and don't understand a plethora of concepts as a result)

$\endgroup$
3
  • 6
    $\begingroup$ Whether multiplication 'is' repeated addition or not depends on whether you believe there is a platonic ideal of multiplication or not. The definition of multiplication by $a$ as the 'inductive closure' of the add-$a$ function is a perfectly legitimate one and immensely useful in mathematical logic. $\endgroup$
    – Steven Stadnicki
    Oct 17, 2014 at 18:39
  • $\begingroup$ @StevenStadnicki Depending on the day I am either a fictionalist or Platonist. Today is a platonist day. $\endgroup$
    – Lan
    Oct 17, 2014 at 23:10
  • $\begingroup$ Living in a universe where mass increases as velocity approaches the speed of light, and matter dilutes time and bends space, it seems sensible to stop teaching Newtonian physics and the Pythagorean Theorem. Multiple views of the same concept just confuses people. Lets just study Abstract algebra in grade 1 and forget about Natural, Rational, Real and Complex numbers entirely. $\endgroup$
    – John Joy
    Nov 7, 2014 at 16:03
0
$\begingroup$

In another question I gave an answer which even discusses such operations in fractional order. This might be interesting as a side-view. See here

But I also copied a short citation of an article of A Bennet, where he gives one argument why we actually do not increase the order of operations beyond exponentiation - except for research on very special numbertheoretic properties. I paste it to this answer too.
picture

$\endgroup$

You must log in to answer this question.