Hi all, I was wondering whether you could help me understand a part in a proof that is assumed to be an "obvious detail" by the author, but that I can't wrap my head around. Essentially, as far as I can understand, there are two values $A$ and $B$, and we want to prove that:


$\Bigg|A-B \; \Bigg| > \frac{1}{p}$ 


for some polynomial $p$ (the actual value of $p$ is unimportant).


We reach a point in the proof where we have shown that 


$\Bigg |\alpha \cdot A- \beta \cdot B \; \Bigg| \geq \frac{2}{p}$ 


and that 


$\Bigg|\; \alpha - \beta\; \Bigg| < \frac{1}{3p^2} $


The next line of the proof (where I stop understanding) concludes that


$\Bigg |\beta \cdot A- \beta \cdot B \; \Bigg| > \frac{1}{p}$


from which point we can make several conclusions.


Can anyone explain how to arrive at the last inequality from the previous ones? Is it possible, or am I missing some additional information somewhere? 


Thanks!



Problem with making an estimate when values of many variables are unknown?
If we truly have (as you say in the comments) that $0 \leq \alpha,\beta, A, B \leq 1$ then 
$$
\frac2p \leq |\alpha A - \beta B| \leq 1
$$
implying that $p \geq 2$.


This in turn means that $\frac{A}{3p^2} \leq \frac1{3p^2} \leq \frac1{6p}$. 


Letting
$$
\epsilon=\beta-\alpha
$$
and using
$$
\left| \beta \cdot A - \beta \cdot B \right| = \left| \alpha \cdot A + \epsilon \cdot A- \beta \cdot B \right|
$$
plus the triangle inequality shows that your condition is easily satisfied.


