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

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.

• Then how about considering the function $f(x) = x/(x-\beta)$? – xbh Sep 12 '18 at 5:18
• Define 'close to' $1$. Exactly how far away $\frac{\alpha}{\alpha-\beta}$ is from $1$ (or at least an upper bound) will depend closely on the particular values of $\beta$ and $\eta$. – Hayden Sep 12 '18 at 5:23
• @xbh detail plz! – Michael Sep 12 '18 at 5:23
• @Hayden Plz, check the post again, I have edited. – Michael Sep 12 '18 at 7:41
• I don't get the sentence "Observe that $\alpha/(\alpha-\beta)$ is close to $1$ when $\alpha/\beta$, ..." what does "when $\alpha/\beta$" mean? $\alpha/\beta$ isn't a condition it is just a term. Do I get something wrong? – Nathanael Skrepek Sep 14 '18 at 9:23

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.

• Note, $\eta \leq 1/2$ – Michael Sep 14 '18 at 9:19
• @MikeSQ: Okay, that doesn't affect the estimate above. We still need $|\beta|^\eta$ to be large. – robjohn Sep 14 '18 at 9:32

$$\frac\alpha{\alpha-\beta} = 1+\frac{\beta}{\alpha -\beta}$$ 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".

• Your first equation is not right: $\frac{\alpha}{\alpha-\beta} =1+\frac{\beta}{\alpha-\beta}$ – robjohn Sep 14 '18 at 8:53
• @robjohn oops. Thanks – Calvin Khor Sep 14 '18 at 8:54
• do u really need s? – Michael Sep 14 '18 at 9:07
• @MikeSQ: not if we use the triangle inequality – robjohn Sep 14 '18 at 9:10
• @MikeSQ no, but if you knew $s=-1$ then there's a slight improvement. Given that its not clear how tight a bound you want, I left it in – Calvin Khor Sep 14 '18 at 9:11

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 $\beta>0.$

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 $1$

• $\dfrac{\alpha}{\alpha-\beta} > 1$ as $\alpha>(\alpha-\beta)$ – Michael Sep 17 '18 at 20:51
• Yes. That's right and $\alpha\over\alpha-\beta$ is close enough to $1$ from above as I said – Mostafa Ayaz Sep 22 '18 at 6:42