# Irrationality measure of $\alpha+\beta$, $\alpha\beta$

Let $\alpha$ and $\beta$ be real numbers with finite irrationality measures. My question is:

Are the irrationality measures of $\alpha+\beta$ and $\alpha\beta$ also finite?

I tried using triangle inequality $$\left| \alpha - \frac {p_1}{q_1} + \beta - \frac{p_2}{q_2}\right| \geq \left| \left| \alpha - \frac{p_1}{q_1}\right|-\left|\beta-\frac{p_2}{q_2}\right|\right|$$for the sum, but this doesn't lead me anywhere. For the product, I am not even sure where to start.

I wonder if this is well-known. If so, I am looking for a reference. If this is true, we would have results as $e+\pi$ has a finite irrationality measure. But, I could not find any reference for such results.

My progress so far is:

If one of $\alpha$, $\beta$ is a rational number, then the result is true.

From Hata's result, we have $\pi/\sqrt{k}$ has finite irrationality measure for any integer $k\geq 1$. So, $\pi/\sqrt{k} +1$ has a finite irrationality measure. But, that would not necessarily imply that $\pi+\sqrt{k}$ has finite irrationality measure.

Added on 2017 May 22

In fact $\pi + \sqrt{k}$ has a finite irrationality measure. This can be done by Baker's theorem.

[Baker's Theorem]

If $\alpha_1, \ldots, \alpha_n$ are algebraic numbers, not $0$ or $1$. If $\log\alpha_1, \ldots, \log\alpha_n$ are linearly independent over $\mathbb{Q}$, then $1, \log\alpha_1, \ldots, \log\alpha_n$ are linearly independent over $\overline{\mathbb{Q}}$. Moreover, there is an effectively computable constant $C>0$ such that for any algebraic $\beta_0, \ldots, \beta_n$ not all zero, $$|\beta_0+\beta_1\log \alpha_1 + \cdots + \beta_n \log \alpha_n |\geq H^{-C}$$ where $H=\max(h(\beta_i))$.

For positive integers $p, q$, we have $$|q(\pi+\sqrt k)-p|\geq H^{-C}$$ where $H=\max(h(q\sqrt k-p) , q^2)\ll q^{A}$ for some effectively computable constant $A>0$. This shows that $\pi+\sqrt k$ has a finite irrationality measure.

This can be generalized that $\pi+\alpha$, $\pi/\alpha$ has a finite irrationality measure for any nonzero algebraic $\alpha$.

## 1 Answer

By Khinchin's theorem, the set of real numbers $\alpha$ such that $\mu(\alpha)>2$ has Lebesgue measure $0$. It follows that the sets $\{\alpha \in \mathbb R \mid \mu(L-\alpha)>2\}$ and $\{\alpha \in \mathbb R\setminus\{0\} \mid \mu(L/\alpha) > 2\}$ also have Lebesgue measure $0$, where $L$ is Liouville's constant.

So for almost all real $\alpha$, both $\alpha$ and $L-\alpha$ have irrationality measure $2$, yet their sum has irrationality measure $\infty$. Similarly, for almost all real $\alpha$, both $\alpha$ and $L/\alpha$ have irrationality measure $2$, yet their product has irrationality measure $\infty$.

• This answers my question negative. Thank you. – i707107 May 23 '17 at 2:36
• Although we have the existence of such $\alpha$ but finding it explicitly will be difficult. I have no idea on finding $\alpha$ explicitly. – i707107 May 23 '17 at 2:43