Why is there a $p\in \mathbb{N}$ such that $mr - p < \frac{1}{10}$? I am reading the following part of the paper of Denef : 
Let $R$ be a commutative ring with unity and let $D(x_1,\dots , x_n)$ be a
relation in $R$. We say that $D (x_1,\dots , x_n)$ is diophantine over $R$ if there exists a polynomial $P(x_1,\dots , x_n,y_1,\dots  ,y_m)$ over $R$ such that for all $x_1,\dots , x_n$ in $R$ : $$D(x_1, \dots , x_n) \leftrightarrow \exists y_1, \dots , y_m \in R : P(x_1, \dots , x_n, y_1, \dots , y_m)=0$$ 
Let $R'$ be a subring of $R$ and suppose $P$ can be chosen such that its
coefficients lay in $R'$, then we say that $D (x_1,\dots , x_n)$ is diophantine over $R$ with coefficients in $R'$.
Proposition 1. 
Let $R$ be an integral domain of characteristic zero. Suppose
there exists a subset $S$ of $R$ which contains $\mathbb{Z}$ and which is diophantine over $R[T]$; then $\mathbb{Z}$ is diophantine over $R[T]$. 
In particular, this is true when $R$ contains $\mathbb{Q}$. 
A relation is diophantine over $\mathbb{Z}[T]$ if and only if it is recursively enumerable. 
Corollary (M. Boffa). 
Every subset $D$ of $\mathbb{N}$ is diophantine over $R[T]$. 
Proof. 
Let $r$ be the real number $r = \sum_{n=0}^{\infty}\frac{a_n}{10^{n+1}}$, where $a_n = 0$ for $n \in D$ and $a_n = 1$ for $n \notin D$. 
Then we have
$$n \in D \leftrightarrow n \in N \land \exists p, m \in N: \left (m = 10^n \land 0 \leq mr - p < \frac{1}{10}\right )$$ 
But $\mathbb{Z}$ is diophantine over $R[T]$ by Proposition $1$, and every recursively enumerable relation in $\mathbb{Z}$ is diophantine over $\mathbb{Z}$. Thus, using elementary algebra, we see that $D$ is diophantine over $R[T]$. 
$$$$ 
I haven't understood the equivalence: $n \in D \leftrightarrow n \in N \land \exists p, m \in N: \left (m = 10^n \land 0 \leq mr - p < \frac{1}{10}\right )$ 
When $n \in D$ we have that $a_n=0$. 
$r=\sum_{i=0}^{\infty}\frac{a_i}{10^{i+1}} \geq 0$ since the numeratoe is always $0$ or $1$. 
We take $m=10^n$ so $mr=\sum_{i=0}^{\infty}\frac{a_i}{10^{i+1-n}}$. 
Since $a_n=0$ we don't get the term $\frac{1}{10}$ at the sum. 
But how do we know that there is a $p\in \mathbb{N}$ such that $mr - p < \frac{1}{10}$ ? 
$$$$ 
EDIT: 
Could you explain to me the following part of the proof? 
"But $\mathbb{Z}$ is diophantine over $R[T]$ by Proposition $1$, and every recursively enumerable relation in $\mathbb{Z}$ is diophantine over $\mathbb{Z}$. Thus, using elementary algebra, we see that $D$ is diophantine over $R[T]$." 
By Proposition $1$ we have that $\mathbb{Z}$ is diophantine over $R[T]$ when there is a subset $S$ of $R$ which contains $\mathbb{Z}$ and which is diophantine over $R[T]$. 
Which is in this case the subset $S$ ? 
 A: We can split the sum $mr$ in $M+R$ where $M$ is the sum from $i=0$ to $n-1$ (that means, the part of the sum which terms are integers), and $R$ is the tail of the sum (from $i=n$ to infinity.) Define $p=M$. Then $mr-p=R$. You can estimate the sum $R$ taking account that is bounded by $\sum_{i\geq 2}1/10^n$.
A: We can think of $R$ sitting inside the polynomial ring $R[T]$ as the set of polynomials of degree 0, and because $R$ is an integral domain of characteristic $0$, we can think of $\mathbb{Z}$ as a subring of $R \subseteq R[T]$. What the first part of the proposition is saying is that, if there is some subset $S$ of $R[T]$ with $\mathbb{Z} \subseteq S \subseteq R$ that is diophantine over $R[T]$, then $\mathbb{Z}$ is diophantine over $R[T]$. In the second part, "$R$ contains $\mathbb{Q}$" means that every non-zero element of $\mathbb{Z} \subseteq R$ is invertible, so we can consider $\mathbb{Q}$ to be a subring of $R$ with $\mathbb{Z} \subset \mathbb{Q} \subseteq R$. The second part then says that, in this case, we can always define a subset $S$ of $R$ containing $\mathbb{Z}$ that is diophantine over $R[T]$. To do this, note that for $f, g \in R[T]$, $fg = 1$ implies that $\deg(f)=\deg(g) = 0$, i.e., that $f, g \in R$, so we can define $S$ by
$$
S = \{ f \in R[T] \mathrel{|} f = 0 \lor (\exists g \in R[T]\cdot fg = 1)\}.
$$
Then $S$ is clearly diophantine over $R[T]$, and, given that $R$ contains $\mathbb{Q}$, we have $\mathbb{Z} \subset \mathbb{Q} \subseteq S \subseteq R$.
