Find identity element, invertible and inverses in $T=\mathbb Z \times \mathbb Q$ Let the following operation be defined on $T=\mathbb Z \times \mathbb Q$:
$$\begin{aligned}(a,b)\centerdot (c,d) = (-ac, b+d+2) \end{aligned}$$
in the commutative semigroup $(T, \centerdot)$, find the identity element, invertible elements and inverses.
I need to find the identity element, so I need an $(\alpha, \beta)$ so that
$$\begin{aligned}(a,b)\centerdot (\alpha,\beta) = (a,b) \end{aligned}$$
as $(T, \centerdot)$ is a commutative semigroup, I assume that $(\alpha,\beta)\centerdot (a,b) = (a,b)$ holds.
$$\begin{aligned}-a\alpha = a \Leftrightarrow \alpha = -1 \end{aligned}$$
$$\begin{aligned}b+\beta+2 = b \Leftrightarrow \beta = -2 \end{aligned}$$
so the identity element relative to this semigroup is $(-1,-2)$. 
In order to find the invertible elements and inverses:
$$\begin{aligned}(a,b)\centerdot (a',b') = (-1,-2) \end{aligned}$$
so
$$\begin{aligned}-aa' = -1 \Leftrightarrow a'=-\frac{1}{a} \in \mathbb Z \Rightarrow a = 1 \end{aligned}$$
$$\begin{aligned}b+b'+2=-2 \Leftrightarrow b'=-(b+4) \end{aligned}$$
in conclusion only the elements $(1,b)$ are actually invertible and their inverse is $(1, -(b+4))$.
Does this whole thing hold? Am I wrong in any part of it?
 A: In all of these problems, you are taking a structure with operations that you "already know", and are trying to test to see whether a new operation, defined in terms of the ones you know, satisfies certain properties.
It's usually best to denote the "new" operation with a symbol that is unlikely to generate confusion. You've been doing this elsewhere, but here you are using $\cdot$, which can easily be confused with regular integer multiplication. So I'm going to replace it with $\odot$, if you don't mind.
Let's approach it systematically, and essentially "following your nose." I'm going to check a few things that you seem to take for granted, just to get familiarity with the operation.
Now, we have the set $T=\mathbb{Z}\times\mathbb{Q}$, and the operation
$$(a,b)\odot (c,d) = (-ac, b+d+2).$$
First, the operation is commutative, since integer multiplication and addition of rationals is commutative. It is also associative:
$$\begin{align*}
\Bigl( (a,b)\odot (c,d)\Bigr)\odot (x,y) &= (-ac, b+d+2)\odot (x,y)\\
&=(-(-ac)x,(b+d+2)+y+2)\\
&= (acx, b+d+y+4)\\
(a,b)\odot \Bigl((c,d)\odot (x,y)\Bigr)&= (a,b)\odot (-cx, d+y+2)\\
&= (-a(-cx), b+(d+y+2)+2)\\
&= (acx, b+d+y+4).
\end{align*}$$
So the operation is commutative and associative, and we do indeed have a semigroup.
Now we are trying to see whether this is a monoid (semigroup with identity). So we are trying to see whether there exist $\alpha\in\mathbb{Z}$ and $\beta\in\mathbb{Q}$ such that, for all $a\in\mathbb{Z}$ and $b\in\mathbb{Q}$, we have
$$(\alpha,\beta)\odot(a,b) = (a,b)$$
(exactly as you did). This amounts to solving the equations you solved, which you did correctly, and so we discovered that $(T,\odot)$ is indeed a monoid, and that the identity element is $e_T = (-1,-2)$. Great!
To figure out what elements have inverses, you do set $(a,b)\odot (x,y) =e_T = (-1,-2)$, and figure out if you can determine necessary and sufficient conditions on $a$ and $b$ for $x$ and $y$ to exist. As you note, we have:
$$(-1,-2) = (a,b)\odot(x,y) = (-ax, b+y+2).$$
This requires $-ax = 1$, which means that $a$ has to be an integer that has a multiplicative inverse in the integers (namely, $-x$); there are only two possibilities: $a=1$ or $a=-1$.
On the other hand, for any rational number $b$ we can always find a rational number $y$ such that $b+y+2=-2$, namely, as you note, $y = -(b+4)$. So it would seem that provided that $a=\pm 1$, then $(a,b)$ has an $\odot$-inverse in $T$. And, indeed, if $a=\pm 1$, then we have $$(\pm 1,b)\odot (\pm1,-b-4) = (-(\pm1)(\pm 1), b-b-4+2) = (-1-2).$$
So a necessary and sufficient condition for $(a,b)\in T$ to be $\odot$-invertible is that $a=\pm 1$. 
