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 $\varLambda \geq \frac{1}{by^n}$?


$$a=b$$ 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.


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

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

  • $a=b\implies a\le b\lor a\ge b$ ?

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

  • $a=b\implies a< b\land a=b\land a> b$ ?
  • $\begingroup$ Seriously! I am surprised!! $\endgroup$ – Mike SQ Sep 5 '18 at 6:36
  • 1
    $\begingroup$ @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

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 $$\left|\left(1+\frac1b\right)\left(\frac xy\right)^n-1\right|\le \frac1{by^n}.$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.