# Existence of a simultaneous rational approximation of real numbers in (0,1)

I have a simple question the rational approximation of real vectors.

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).

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The theorem doesn't say "there exists $N$"; it says, "for every $N$". – Gerry Myerson Mar 21 '14 at 22:37
Thanks! I just changed it in the theorem. You might notice that I did not change this in my proposed simplifications. This is because I would be happy knowing that there exists $N$ instead of knowing that the result holds for every $N$ (though it would be cool if the latter were true). – Elements Mar 21 '14 at 23:04

To restrict to real numbers between $(0,1)$, notice that if $\alpha>1$, then $\exists \beta,k$ 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, then there exists a closer approximation.
Thanks for this answer! I see what you did. However, I think that this only shows that we can use the theorem to produce a rational approximation for a real number $\alpha \in (0,1)$ - right? What I was asking is if knowing that $\alpha \in (0,1)$ also guarantees that there exists a rational approximation $\frac{p'}{q} \in (0,1)$ (i.e. $p' \leq q$). If not, is there some way that we can modify the theorem so as to obtain a rational approximation in $(0,1)$ as well? – Elements Mar 24 '14 at 21:06