Is there a use for this binary operation? For any $a, b \in \mathbb{Z}$ define
$a * b = a + b + 2ab$
I've seen this pop up a few times in exercises, and I'm wondering if there's a name for this (or any similar variants), and if it's ever useful.
 A: Define $f\colon \mathbb Z\to\mathbb Z$ by $f(n) = 2n+1$. Then $f(a)$ and $f(b)$ are odd for all integers $a,b$, hence their product is odd as well. Therefore the expression
$$
f^{-1}\big( f(a)f(b) \big)
$$
makes sense (even though $f^{-1}$ doesn't make sense in general). Since $f^{-1}(m) = (m-1)/2$, we conclude that
$$
f^{-1}\big( f(a)f(b) \big) = f^{-1}\big( (2a+1)(2b+1) \big) = 2ab+a+b.
$$
In other words, your binary operation is the result of mapping the integers to a different monoid (namely the odd integers), doing the multiplication there, and then mapping back. It follows quickly that this binary operation has all the nice properties needed to make $\mathbb Z$ a monoid (associative, for example).
One can use $f(n) = wn+1$ for any integer $w$ - that just turns the binary operation into $a\ast b = a+b+wab$. One can also use these binary operations on rational numbers or real numbers (in which cases $f^{-1}$ really does make sense).
So I think this is just a nice way to construct exercises, but otherwise isn't particularly useful or natural: it's just a relabeled monoid in disguise.
A: I don't know if this has any name, but up to affine transformation, it's the usual product (on odd integers). Actually, for any $\lambda \in \mathbb{Z}$, you could define
$$a \, \underset{\lambda}{*} \, b = a + b + \lambda ab$$
and check that
$$1 + \lambda (a \, \underset{\lambda}{*} \, b) = (1 + \lambda a)(1 + \lambda b)$$
Which means that $f_{\lambda} : x \mapsto 1 + \lambda x$ defines a morphism from $(\mathbb{Z} , \underset{\lambda}{*})$ to $(\lambda \mathbb{Z}+1, \times)$ which is bijective provided $\lambda \ne 0$.
As for applications, a similar operation on $\mathbb{R}$ with $\lambda = \frac{1}{100}$ pops up when dealing with percentages : an increase of $a$ percents followed by one of $b$ percent amounts to an increase of $a \, \underset{0,01}{*} \, b = a + b + \frac{ab}{100}$ percents (which comes from the fact that an increase of $a$ percents, corresponds to a multiplication by $f_{0,01}(a) = 1 + \frac{a}{100}$).
