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In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 5 that-

$\varLambda \leq \frac{1}{by^n}$ (see equation 7 on page 5)

But we get it from an equation. As I understand , it should be $\varLambda = \frac{1}{by^n}$.

How $\varLambda$ could be less than $\frac{1}{by^n}$?

If it is $\varLambda \leq \frac{1}{by^n}$ then why not $\varLambda \geq \frac{1}{by^n}$?

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$$a=b$$ implies both $$a\le b$$ and $$a\ge b.$$

It's the author's choice to weaken the comparison for the requirements of the exposition.


Quiz:

Are these propositions true ? ($\land$ is and, $\lor$ is or)

  • $a=b\implies a\le b\land a\ge b$ ?

  • $a=b\implies a\le b\lor a\ge b$ ?

  • $a=b\implies a< b\lor a=b\lor a> b$ ?

  • $a=b\implies a< b\land a=b\land a> b$ ?
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  • $\begingroup$ Seriously! I am surprised!! $\endgroup$ – Mike SQ Sep 5 '18 at 6:36
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    $\begingroup$ @MikeSQ: don't forget that $a\le b$ means "$a=b$ or $a<b$", whichever is true. $\endgroup$ – Yves Daoust Sep 5 '18 at 7:35
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From $$\tag6(b+1)x^n-by^n=1$$ we find after dividing by $by^n$ $$\left(1+\frac1b\right)\left(\frac xy\right)^n-1=\frac1{by^n}.$$ This certainly implies (as long as $b,y>0$) that also $$\left|\left(1+\frac1b\right)\left(\frac xy\right)^n-1\right|\le \frac1{by^n}.$$

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  • $\begingroup$ If you don't mind I will ask the same question again as it is yet to be clear to me! Are you saying $$\left|\left(1+\frac1b\right)\left(\frac xy\right)^n-1\right|\geq \frac1{by^n}.$$ is not possible? Do you disagree with Daoust's answer above? $\endgroup$ – Mike SQ Sep 5 '18 at 6:56

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