@DanielFischer It is my dream to produce, in the distant future, a set of books covering all major branches of mathematics, doing what Bourbaki or Lang tried to do in my own way.
My last creation: Prove that $$\int_0^1\sqrt{x}\tan^{\large 1/2^2}(2\arctan(x))\tan^{\large 1/2^3}(2^2\arctan(x))\tan^{\large 1/2^4}(2^3\arctan(x))\cdots \ dx = \log(2)$$
The scientists at SETI finally decoded a message which appears to be proof of intelligent life in the universe. Unfortunately, it doesn't make sense: Itu, the Eye-Pie, and won snot.
Nice. I just figured that for each $x$, there are $n$ $y$ to choose from and there are $n$ $x$, not to mention for each $x$, there is one value $x+y$ which will be the same as the former selection from that $x$.
your $A + A$ is just $A \times A$ but with $(x, y)$ identified with $(y, x)$. So just use the inclusion-exclusion business to count $n^2$ ordered pairs with $n$ identifications, thus $n^2 - n$ elts.
This should be patently obvious if you are not as sleepy as a rotten moldy old dishrag.
Does it make sense to say that the number of elts inside $A + A$ is equal to the number of elts inside $A \times A$, modulo some congruence, like $(x, y) \sim (y, x)$?
@Khallil i mean you include all the stuff you think there is in $A \times A$ and exclude all the stuffs which gets chopped off by the equivalence $\sim$
'modulo some congruence' doesn't make much sense, but I understand that if you identify $(x,y)$ with $(y,x)$, there'll be $n$ such cases, so you can remove that from the cardinality of $A \times A$ due to the fact that $x+y = y+x$.
