1
$\begingroup$

Let $\frac m n$ be an irreducible fraction such that $1 \gt \sqrt 2 + \sqrt 3 - \frac m n \gt 0$. Show that there is a constant $c \gt 0 \space$ such that $$\sqrt 2 + \sqrt 3 - \frac m n \gt \frac {1} {c \times n^4}$$ for every $n \gt 1.$

  • 0
    from where do you got this? – 2017-02-16
  • 0
    A Romanian magazine – 2017-02-16
  • 2
    I am not an expert on this topic, but by guess would be that this a consequence of [Liouville's theorem on diophantine approximation](https://en.wikipedia.org/wiki/Liouville_number#Liouville_numbers_and_transcendence). – 2017-02-16
  • 1
    Of course, $\sqrt{2}+\sqrt{3}$ is an algebraic number of degree $4$ over $\mathbb{Q}$. – 2017-02-16
  • 0
    @JackD'Aurizio Is there a simpler proof for this particular case? – 2017-02-17
  • 0
    @M.Stefan: I do not think so. The continued fraction of the algebraic number $\sqrt{2}+\sqrt{3}$ is not that nice and its convergents have little to do with the convergents of $\sqrt{2}$ and $\sqrt{3}$. But maybe I am wrong. – 2017-02-18

0 Answers 0