This question already has an answer here:

I was thinking "What is before addition", and came up with this. This analogy describes it: Addition is to multiplication as [operation] is to addition. On wikipedia it says it is just "1+b". I came up with an alterantive; the symbol that I'll use for it will be $@$.

$b@b=b+2$

$b@b@b=b+3$

$b@b@b@b=b+4$...

For example:

$3@3=5$

If you try to compute it you run into problems. For example:

$3@2=(0@0@0)@(0@0)$ but it also equals $(-1@-1@-1@-1)@(-1@-1@-1)$

$(0@0@0)@(0@0)=(-1@-1@-1@-1)@(-1@-1@-1)$

$0@0@0@0@0=-1@-1@-1@-1@-1@-1@-1$

$0+5=-1+7$

$5\neq6$

So, you can't take away or add parentheses.

1)Is this new?

2)How do you compute it?

marked as duplicate by Gottfried Helms, Claude Leibovici, user91500, астон вілла олоф мэллбэрг, user223391 Feb 22 '17 at 15:01

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  • 2
    You defined $b@b$ and longer sequences, but you didn't define $b@c$, and even less $b@c@d$... – Yves Daoust Feb 20 '17 at 23:42
  • 1
    How do you define b@c? If it only takes one operand you shouldn't write it as a binary operation. what you have is an operand on a term b and an operation based solely on how many times you write b. In other words b@b@.....@b = f(b, n) = b + n -1. " you can't take away or add parentheses" is more properly stated as "is not associative. – fleablood Feb 20 '17 at 23:44
  • @Chappers: this $@$ operator does not denote the successor function. – Yves Daoust Feb 20 '17 at 23:50
  • You might do better with $b@c=b+1$ and $b@c@d=(b@c)@d$ which would then be $b+2$ and $b@c@d@e=((b@c)@d)@e=b+3$ etc. – Henry Feb 20 '17 at 23:51
up vote 3 down vote accepted

You're extremely confused about what it means to define an operation.

2) How do you compute it?

You tell me! You invented it. The question is like the inventor of chess asking how bishops move - it's your game, dude. Or, to take a more mathematical example, suppose I were to say "I'm going to define a new class of numbers called Pretty numbers. For example, by definition $3$ and $12$ are pretty numbers. Now, can you help me figure out if $20$ is a pretty number?".

This is what a definition of an operation looks like:

Define the operation $\star$ on integers by $a\star b=a^2 +\text{the last digit of b}$.

Now I have a rule that lets me calculate $a\star b$ for any pair of integers $a$, $b$. What you've done is defined $a@a$ for any integer $a$. There is no "correct" answer to what $3@2$ equals, you have to define it. Until you've defined it, the answer doesn't exist.

Furthermore, you're not really "allowed" to just define $b@b@b$. If $@$ is a binary operation, then it has to have two arguments, not three. When you write down $b@b@b$, do you mean $(b@b)@b$, or $b@(b@b)$? What if there is no operation $@$ such that $x@x=x+2$ for all $x$, but also $x@(x@x)=x+3$ for all $x$?

Here's a way of reformulating your question to give it meaning:

Is there any operation $@$ on $\mathbb Z$ such that for all $x$, we have: $$x@x=x+2$$ $$x@(x@x)=x+3$$ $$x@(x@(x@x))=x+4$$ and so on?

Notice that we're not defining some object and then asking about its properties, we're listing some properties and asking if such an object exists.

Well, let's see. If such an operation existed, then the conditions would imply that for all $x$, and all $n\geq 2$. $x@(x+n)=x+n+1$. This determines $x@y$ whenever $y\geq x+2$: $x@y=x+(y - x)+1=y+1$. In fact, by induction, you can show that as long as an operation verifies $x@y=y+1$ for pairs $(x, y)$ that verify $y\geq x +2$, the required condition will be satisfied, and you can define it in any way you like for other pairs $(x, y)$.

Noting the comments, I think the answer you might be trying to define is actually a unary operation, it is the successor function. $s(x) = x + 1$.

The reason I think this is because multiplication (of naturals) is defined to be iterated addition. That is to say, to do $3 \times 4$, we do $4 + 4 + 4$ or alternatively $3 + 3 + 3 + 3$. Let's just stick to one of these for the sake of simplicity, and we'll choose the second one. So now what you want to do is come up with an operation, which when iterated, yields addition. So if we considered the expression $3 + 4$, we want something which allows us to add $4$ to $3$. Can we rig up an operation which is iterative in the same way? Well, yes and no.

If we interpret the express $4 + 3$ as "take $4$ and add $3$ to it, we can then think of this as $(4) + 1 + 1 + 1$, which is an interative process, as we want, but it's not a binary operation any more - it depends on $4$ and the number of times we apply it, but in each iteration, it depends only on it's current input. In this sense, addition is the binary operation which is a kind of short-hand for this successor function. Similarly, multiplication can be defined recursively in terms of addition, and exponentiation in terms of multiplication, etc.

So in some sense, this successor function is the foundation of arithmetic, and indeed it is, as all that I've said can, and probably should, be thought of as the groundwork for the underlying ideas of the Peano Axioms.

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