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When $ \alpha_1,\alpha_2 \cdots \alpha_m$ algebraic numbers, with rational $b_1, b_2 \cdots b_m$ it is stated in LINEAR FORMS IN LOGARITHMS, on page $3$ that-

We apply this with $w := \alpha_1^{b_1} \cdots \alpha_m^{b_m}-1$. If $\left| w\right|>1/2$ we are done, so we suppose that $\left| w\right|\leq1/2$.

Question: I couldn't get how it is possible $\left| w\right|\leq1/2$ when $w := \alpha_1^{b_1} \cdots \alpha_m^{b_m}-1$ where$ \alpha_1,\alpha_2 \cdots \alpha_m$ algebraic numbers, with rational $b_1, b_2 \cdots b_m$. Because $\alpha_1^{b_1} \cdots \alpha_m^{b_m}-1$ could be a big number.

How is it possible $\left| w\right|\leq1/2$? Can you provide an example?

Source of the problem: enter image description here

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    $\begingroup$ The number $w$ is not necesarily "big". For example, taking $\alpha_1=\cdots=\alpha_m = (3/4)^{2/m}$ then they are algebraic and taking $\beta_1=\cdots=\beta_m=1/2$ which are rational integers, you obtain that $w=3/4-1=1/4$ and then $|w|\leq 1/2$. $\endgroup$ – Albert Sep 20 '18 at 15:00
  • $\begingroup$ You don't know that $|w| \le \frac 12$. It's just that you do two different arguments: One if $|w| \le \frac 12$ and one if $|w| > \frac 12$. If $|w| > \frac 12$ then the result is supposedly obvious and you are done. So you only have to worry about showing it for the cases when $|w| \le \frac 12$. $\endgroup$ – fleablood Sep 20 '18 at 16:48
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The algebraic numbers are large and small, like the rationals. The powers and products of them are also large and small. A simple example would be $$2^13^{-1/2}-1\approx 2\cdot \frac 1{1.732}-1\approx 1.155-1=0.155\lt \frac 12$$

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I think that you have a misunderstanding about the nature of the proof. In fact this may be a misunderstanding about proofs in general.

To see the point consider the following-

The author(or proof-writer) wants to prove a certain statement (in this case, your Corollary 1.6). In this process, he assumes a certain number (here, $w$).

Now, if the number satisfied a certain property (here, $w\gt 1/2$), then the statement is already proved. So, he considers the other case ($w\leq 1/2$).

Now,

  1. If that case is impossible, than no harm is done. Indeed, we may get a contradiction to the hypothesis, allowing us to use reductio ad absurdum to rule out the case.

  2. If, on the other hand, this case is possible, he proves the statement for the case, as he should for the proof to be correct.

Thus, we see that as long as we use reductio ad absurdum upon hitting a contradiction, and stick to the rules of deductive logic at all times, we are all good doing the procedure you mentioned.

In fact, proceeding on a case-by-case basis is a very important logical method in mathematical reasoning.

Finally, as pointed out by @Carlos Ajila in the comments, $w$ can in fact be lesser than $1/2$ and the fact pointed out by you (that it can also be greater than $1/2$) does not change this. An example would be ${2 \over {\sqrt 3}} - 1$, as given in Ross Millikan's answer.

Thus, the proof is correct :D.

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