Prove : $ \frac{\alpha}{\alpha -\beta} \quad \text{close to 1.}$

Question

If $\alpha \geq |\beta^{1+\eta}|,$ how to prove that-

$$ \frac{\alpha}{\alpha -\beta} \quad \text{close to 1} ?$$

Attempt: If, $\alpha= |\beta^{1+\eta}|$,

then, $$ \frac{\alpha}{\alpha -\beta}= \frac{\beta^{1+\eta}}{\beta^{1+\eta} -\beta}= \frac{\beta^{\eta}}{\beta^{\eta} -1} \approx \frac{\beta^{\eta}}{\beta^{\eta} -0} \approx 1$$

but if, $\alpha> |\beta^{1+\eta}|$ then how we extend the argument?

Note that we can not write directly $ \frac{\alpha}{\alpha -\beta}> \frac{\beta^{1+\eta}}{\beta^{1+\eta} -\beta}$ as $\alpha -\beta > \beta^{1+\eta} -\beta$

Edit: Note, $\eta \leq 1/2$.

Source of the problem :

Full Paper: linear forms in the logarithms of real algebraic numbers close to 1, page 11.

Answer 1

Assuming $|\beta|^\eta\gt1$, then the triangle inequality guarantees $$ \begin{align} \left|\,\frac{\alpha}{\alpha-\beta}-1\,\right| &=\left|\,\frac{\beta}{\alpha-\beta}\,\right|\\ &=\left|\,\frac1{\alpha/\beta-1}\,\right|\\ &\le\frac1{|\beta|^\eta-1} \end{align} $$ which is small if $|\beta|^\eta$ is large.

Answer 2

$$\frac\alpha{\alpha-\beta} = 1+\frac{\beta}{\alpha -\beta}$$ with the basic error bound $$\left|\frac\beta{\alpha-\beta}\right| \le \frac{|\beta|}{|\beta|^{1+\eta}-s|\beta|} = \frac1{|\beta|^\eta-s} \le \frac1{|\beta|^\eta - 1}$$ where $s=\pm 1$ is the sign of $\beta$.

So if at least one of $\eta>0$, and $\beta>1$ is "large enough", then $\alpha/(\alpha+\beta)$ is "close to one".

Answer 3

In the proof of $\bf{Theorem\; 13}$, $\alpha$ and $\beta$ are considered integers, large enough.

$$\frac{\alpha}{\alpha-\beta}=\frac{1}{1-\frac{\beta}{\alpha}}\leq \frac{1}{1-\beta^{-\eta}},$$ from where the result.

I wrote only the RHS, and this just for $\beta>0.$

Answer 4

Since you have proven the inequality for $\alpha=|\beta^{1+\lambda}|$ notice that the function $\dfrac{\alpha}{\alpha-\beta}$ is a decreasing function of $\alpha$ when $\alpha>\beta>0$ so we have $$1<\dfrac{\alpha}{\alpha-\beta}<\dfrac{|\beta^{1+\lambda}|}{|\beta^{1+\lambda}|-\beta}\approx 1$$which means that $\dfrac{\alpha}{\alpha-\beta}$ is close to $1$ because $\dfrac{|\beta^{1+\lambda}|}{|\beta^{1+\lambda}|-\beta}$ is close to $1$