# Commutative binary operation integers such as x∘y is unique

I'm basically looking for an commutative binary operation which, given a couple of two strictly positive integers $(a,b)$, would provide me with another unique strictly positive integer $c$ (which I want to use to uniquely identify the initial couple). The order in the couple should not be important, hence the commutative property.

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Let's try in a more Mathematical way... (Probably not perfect, sorry in advance ^^)

Let $\circ$ be the operation I'm looking for, it should verify that:

$\forall(a,b,c,d) \in \Bbb {N^{+*}}^4$
$a \circ b = b \circ a$
$\ \ \ \ \ \ \; \not=a \circ c$
$\ \ \ \ \ \ \; \not=c \circ d$

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An example on the set of prime numbers would be the multiplication. Indeed, given two prime numbers $a$ and $b$, then $a \times b$ gives a unique integer.
I know the input and output sets are different in this example (which is not what I'm looking for) but it was just as an example to make it a bit more clearer it not already enough ^^

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Another example (which is sadly not commutative =/):
Given that I'm in $\Bbb {N^{+*}}$, I can do concatenate $a$ and $b$, and using the $0$ to separate them.

e.g. : $11 \circ 2 = 1102$

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If someone could found such an operation, I would be so grateful =D
Well, if such an operation exists...

First order all pairs of positive integers $(a,b)$ such that $a \ge b$.

In reading order. . . .

$(1,1)$

$(2,1) \ \ \ (2,2)$

$(3,1) \ \ \ (3,2) \ \ \ (3,3)$

$\vdots$

Your product should assign each pair to its place in this order.

i.e. $1 \circ 1= 1$, $2 \circ 1 = 2$, $2 \circ 2 = 3$ and so on.

Then extend the product to when $a < b$ by defining $a \circ b$ to be equal to $b \circ a$.

Now let's try and find an explicit formula. . .

Where is $(a,b)$ located on the triangle? Well it's on row $a$ position $b$. How many numbers come before it? There are all the numbers in column $1$, column $2$, and so on all the way up to column $a-1$, along with everything left of $(a,b)$. Putting it all together we get. . .

$a \circ b = 1 + 2 + 3 + \ldots + (a-1) + b = \displaystyle \frac{(a-1)a}{2} + b$

using the well-known formula $1 + 2 + 3 + \ldots + n = \displaystyle \frac{n(n+1)}{2}$