I have a simple question the rational approximation of real vectors.
Dirichlet's simultaneous approximation theorem states:
Given any $d$ real numbers $\alpha_1,\ldots,\alpha_d$ and for every natural number $N \in \mathbb{N}$, there exists an integer denominator $q \leq N$, and $d$ integer numerators $p_1,\ldots,p_d \in \mathbb{Z}$, such that:
$$ \Bigg|\alpha_i - \frac{p_i}{q}\Bigg| < \frac{1}{qN^{1/d}} \text{ for } i=1,\ldots,d $$
I am hoping to simplify this theorem for the case that $\alpha_i \in (0,1)$.
My questions are:
1) If $\alpha_i \in (0,1)$, then can we say that there is a value of $N$ that is large enough so that the rational approximation $\frac{p_i}{q}$ is also confined to the interval $(0,1)$. If so, what is this value?
2) Assuming that there is a value of $N$ such that the rational approximation $\frac{p_i}{q}$ is within $(0,1)$, then it should follow that $p_i < q$ for all $i$. Since the initial theorem states that $q \leq N$, we know that $p_i < q \leq N$ for all $i$. In light of this, can I simply omit the $N$ and restate the theorem as follows:
Given $d$ real numbers $\alpha_1,\ldots,\alpha_d$ such that $|\alpha_i|<1$ for $i=1,\ldots d$, then there exist an integer denominator $q \in \mathbb{N}$ and $d$ integer numerators $p_1,\ldots,p_d < q$ such that:
$$ \Bigg|\alpha_i - \frac{p_i}{q}\Bigg| < \frac{1}{q^{1+\frac{1}{d}}} \text{ for } i=1,\ldots d $$
The key difference here is that I have used the fact $q \leq N$ in order to replace $\frac{1}{qN^{1/d}}$ with $\frac{1}{q^{1+1/d}}$.
Alternatively if anyone has a proof of the simultaneous approximation theorem (for reals), that would be helpful as well (I can't seem to find one anywhere).
To restrict the approximation theorem to real numbers between $(0,1)$, notice that if $\alpha>1$, then $\exists \beta\in\mathbb{R},k\in\mathbb{Z}$ such that $\beta\in (0,1)$, and $\beta+k=\alpha$. In this case, we have: $$\begin{align} \left|\alpha-\frac{p}{q}\right| &=\left|\alpha-\frac{p}{q} +k-k\right| \\ &=\left|(\alpha+k)-\left(k+\frac{p}{q}\right)\right| \\ &=\left|\beta-\frac{p+kq}{q}\right|\\ \\ &=\left|\beta-\frac{p'}{q}\right| \end{align}$$
This is allowed since the denominator doesn't change and the numerator is allowed to depend on $\alpha_i$
To see that we can always find a rational approximation $\frac{p_i}{q}$ within $(0,1)$ simply pick $q$ large enough such that $0<\alpha-q^{-1}<\alpha+q^{-1}<1$, as if this holds and $\frac{p_i}{q}$ is not in the desired range than there exists a closer approximation.