I had a difficult time in phrasing my question. But I was wondering if there is an operation that, when repeated n times, results in the addition operator. Same way as repeating addition n times results in the multiplication operator, and repeating multiplication n times results in the exponentiation operator etc.

So $n$ times addition of a number $x$ results in $x\times n$.

And $n$ times multiplication of a number $x$ results in $x^n$

Then my question is $n$ times ...what... results into the number $x+n$.

Let's call this operator: $@$.

For example, the following would then hold:

$$a\times a=a^2$$

$$a+a=a\times 2$$

$$a@a=a+2$$

My question is, does it make any sense thinking of such an operator, is there anything known about it, can it be followed through even further like $a\sim a = a@2$?

  • 2
    A kind of unitary addition maybe ? Define $a @ a := a + 1$. – Zubzub Feb 21 '17 at 13:55
  • This question seems a bit ill-defined, because the natural operation that is recursed to get addition is a unitary operation (increment operator), so knowing only that $a@a=a+2$, you could define any number of operations that satisfy that condition, like $f(a, b) = a + 2$. – Paul Feb 21 '17 at 16:25

You want an operator $@$ such that for all $n\in\Bbb N$, $$\underbrace{a\mathop @a\mathop @\ldots\mathop @ a}_n = a+n.$$

The first problem with this is the case $n=1$ (not to mention $n=0$), where you demand $a=a+1$ (in analogy to $a = a\cdot 1$ and $a=a^1$). So we need drop the case $n=1$.

The next problem is that we did not specify whether $@$ should be associative. And if not: Do we mean $(a\mathop @ a)\mathop @ a = a+3$ or $a\mathop @ (a\mathop @ a) = a+3$? Let's attempt to actually achieve associativity. Then we need $$\begin{align}a\mathop @a&=a+2 \\ (a+n)\mathop @a &= a+n+1&\text{for }n\ge 2\\ a\mathop @(a+n) &= a+n+1&\text{for }n\ge 2\\ \end{align}$$ This in fact allows us to extract a complete definition for $@$: $$a\mathop @ b=\begin{cases} \max\{a,b\}+1&\text{if }a\ne b\\ a+2&\text{if }a=b\end{cases} $$ (where the cases with $b=a\pm 1$ are irrelevant).

  • not sure if I demand a = a + 1, I am not sure what the identity element is yet. I would call the identity element of the operator @ simply id, then a = a @ id. For the @ operator I would not know yet what a @ 1 would be – Bert van den Bosch Feb 21 '17 at 15:26
  • 1
    It is nice to see $\max$ here, since $+$ distributes over it. And actually, it seems that $c +(a\mathop @ b) = (c+a) \mathop @ (c+ b)$. – Pedro Sánchez Terraf Feb 21 '17 at 17:16
  • 2
    It's worth noting that $a\times a\times a\times\dots$ should technically be interpreted as $1\times a\times a\times a\times\dots$, so the equivalent for $n=1$ would actually be $1@a=a+1$, not $a=a+1$ (or perhaps a different number in place of $1$). – Glen O Feb 21 '17 at 17:20
  • 2
    Indeed, for your definition, the "identity" in question is actually $-\infty$, because $(-\infty)@a=a@(-\infty) = a+1$. – Glen O Feb 21 '17 at 17:58
  • 2
    @GlenO Cool observation. So $a^n=1\,\underbrace{\times a\ldots\times a}_n$, $a\times n = 0\,\underbrace{+a\ldots+a}_n$, $a+n=(-\infty)\,\underbrace{\mathop @a\ldots\mathop @a}_n$. – Hagen von Eitzen Feb 21 '17 at 22:13

Yes there is.

The successor $s$ function do that where:

$$\forall n\in\mathbb Z,\quad s(n)=n+1$$

and $s(0)=0$.

You can have more information here.

Thanks to a comment, you also might be interested in this.

Though you can not define this operator as a binary relation as you would like it to be defined.

  • 2
    The OP is looking for a dyadic operator, and $n$ incremented $n$ times doesn't yield $n+2$. – Yves Daoust Feb 21 '17 at 13:52
  • 6
    @YvesDaoust I think this is the operation the OP is looking for, even though he is asking for a binary operation. His last questions suggests he wants go on to be interested in en.wikipedia.org/wiki/Hyperoperation - perhaps that link should be edited into this answer. – Ethan Bolker Feb 21 '17 at 13:59
  • @EthanBolker: no, this is not the only post on this topic by the OP. He wants a dyadic operator, following the same pattern as $+$ and $\times$ and ^. – Yves Daoust Feb 21 '17 at 14:01
  • 2
    @EthanBolker I don't get why no one has written an answer using Hyperoperations. This question had been EXTENSIVELY studied on the Tetration Forums already. – Brevan Ellefsen Feb 21 '17 at 16:26
  • 1
    @Brevan: "I don't get why no one has written an answer using Hyperoperations." Be the change you wish to see in the world? – Rahul Feb 21 '17 at 17:30

This is a well studied topic in certain circles, and is referred to as Zeration. It is considered the $0$-th level of the extended Hyperoperation family. Often we simply see Zeration simply defined to be the Successor Function as defined in the Peano Axioms.

However, other definitions of Zeration have been studied. A great start is to look on the Tetration Forums; in particular, I reccomend this post. Some snippets from that page include:

At the very simple common sense level, zeration is an attempt for finding an operation filling the gap in the following operations’ sequence:

$a ^ a = a [3] a = a [4] 2 = {}^2a \implies$ exponentiation $\leftrightarrow$ tetration
$a\times a = a [2] a = a [3] 2 = a ^ 2 \implies$ multiplication $\leftrightarrow$ exponentiation
$a + a = a [1] a = a [2] 2 = a \times 2\implies$ addition $\leftrightarrow$ multiplication

which should be logically completed by a new operation that we may call zeration (indicating it by the infixed operation sign “$\circ$”) and that should at least have the following “strange” property:

$a \circ a = a [0] a = a [1] 2 = a + 2\implies$ zeration $\leftrightarrow$ addition.

Note that the notation $a[n]b$ is common Hyperoperation notation, ${}^2a$ is Tetration. I have edited the quotes to add MathJax formatting. The symbol "$\circ$" is your @ symbol. Continuing in that link, we find the following:

The theoretical way for justifying such new operation is provided by the Ackermann Function. In fact, the definition of Ackermann Function (AF) can be summarised [sic] as follows:

$A(0, n) = n + 1$
$A(s, 0) = A(s-1, 1)$
$A(s, n) = A(s-1, A(s, n-1))$

Continuing further, we get to the crux of the post

With the provisional exception of row $s=0$, we could re-define Ackermann’s Function as follows:

$A(s, n) = 2 [s] (n + 3) – 3$, or:
$2 [s] n = A(s, n-3) + 3$

For $s=0$ we have:
$A(0, n) = 2 [0] (n + 3) – 3 = n + 1$ (zeration)
which gives: $2 \circ (n+3) = n + 4$ therefore: $2 ° n = n + 1$, (for: $n \ge 3$)
to which we may add: $2 \circ 2 = 2 + 2$ and: $n \circ n = n + 2$

We can start using these expressions in order to find out the first properties of the “zeration” operation, which can be described as follows:

$a \circ b = a + 1$ , if $a > b $
$a \circ b = b + 1$ , if $a < b $
$a \circ b = a + 2 = b + 2$ , if $a = b$

If you continue reading the page I linked you will find an extensive discussion of other interpretations and definitions of Zeration, including the different properties such a function could be expected to hold (in particular, having an inverse!). Note that much of the Tetration Forums are a contentious subject at best, but seems to be the best treasure trove of information on this topic you will find. Other proposed definitions include $$ a\circ b = \begin{cases} b+1, & b>a+1 \\ a+2, & b \le a+1 \end{cases}$$
As well as the classic $$ a\circ b = \begin{cases} \max(a,b)+1, & a \neq b \\ a+2=b+2, & a=b \end{cases}$$
The second of which is covered in @Hagen's post

  • Thanks for the extensive answer, and the link, that sounds interesting. – Bert van den Bosch Feb 22 '17 at 12:16
  • Great answer. I've provided more links in my answer. – MphLee Mar 6 '17 at 21:18

If we could use multiplication operator '$\times$' and division operator '$/$' to get this @ operator, then

$\underbrace{x@x@...@x}_n=n \times ((x/n) +1)=x+n$ for $n \neq 0,1$.

Discrepancies will occur at $n=0$ and $n=1$ as for $n=1$ we will get $x=x+1$ and at $n=0$, $ x/n$ will be undefined.

There are already amazing answers!

But I'll try to give some additional info since this question was asked a lot of time in the past (and it will be asked again in the future right?) and I gave some answers too.


As Hagen von Eitzen and Brevan Ellefsen said $$a\circ b:= \max (a,b)+1+\delta_{ab}$$ is one of the solutions of the equation $a*(a+n)=a+n+1$ , where $a$ and $n$ range over some restricted domains.

This particular solution was called Zeration by Rubtsov and Romerio: you can easily find the details of the story in the Zeration thread that was linked by Brevan Ellefsen.

Another solution is $\max(a,b)+1$ which is commutative and is not discontinuous but in the linked thread an user, Tetration Forum's founder, also noticed that it is possible to find a non-commutative solution too.

You can find an updated discussion of the topic in this thread:

Zeration Update @Tetration Forum

Also here at MSE similar questions question were answered: for example this one is almost identical

Does anything precede incrementation in the operator “hierarchy?”

and also remark 1 of this answer I wrote 2 years ago:

More info on Zeration - Go to remark 1

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.