Finding a presentation of $A$-algebra $B$ 
Find a presentation of the $A$-algebra $B$, where $B=\mathbb{Z}[1/2]\subseteq \mathbb{Q}$ and $A= \mathbb{Z}$.

I want to prove it but I can't understand what want to me! Please describe to me. 
 A: Finding a presentation in this context means writing down a surjective $\mathbf{Z}$-algebra (i.e. ring) homomorphism $\mathbf{Z}[X_1,\ldots,X_n]\rightarrow\mathbf{Z}[1/2]$ and giving generators for the kernel of this map, i.e., expressing $\mathbf{Z}[1/2]$ as a quotient of a polynomial ring over $\mathbf{Z}$ (a priori this might require infinitely many generators but it doesn't). Every element of $\mathbf{Z}[1/2]$ is, by construction, of the form $m\cdot 2^n$ for $n\in\mathbf{Z}$. Consider the homomorphism $\mathbf{Z}[T]\rightarrow\mathbf{Z}[1/2]$ got by sending $T$ to $1/2$. This is visibly surjective, and the kernel clearly contains $2T-1$, so you get an induced map $\mathbf{Z}[T]/(2T-1)\rightarrow\mathbf{Z}[1/2]$. Now consider the map $\mathbf{Z}\rightarrow\mathbf{Z}[T]/(2T-1)$. The image of $2$ is a unit with inverse the residue class of $T$, so the universal property of localization furnishes a unique ring map $\mathbf{Z}[1/2]\rightarrow\mathbf{Z}[T]/(2T-1)$ with $1/2\mapsto T\pmod{2T-1}$. Now verify that these maps are mutually inverse. The composite beginning in $\mathbf{Z}[1/2]$ is a $\mathbf{Z}$-algebra endomorphism of $\mathbf{Z}[1/2]$, and there is only one of these, namely the identity (as follows from the universal property of localizations). If you consider the composite beginning on $\mathbf{Z}[T]/(2T-1)$, it sends the residue class of $T$ to $1/2$, and then $1/2$ to the residue class of $T$, so, because it is a $\mathbf{Z}$-algebra map, it must be the identity. So you win.
In general, if $A$ is any commutative ring with identity and $f\in A$, then the natural $A$-algebra map $A[T]/(fT-1)\rightarrow A[1/f]$ with $T\pmod{fT-1}\mapsto 1/f$ is an isomorphism. The proof is the argument above verbatim.
