amilapsn
  • amilapsn
An advanced challenge for math enthusiasts:Given any 9 reals show that there will be at least 2 distinct reals a, b such that \(\left|\frac{a-b}{1+ab}\right|\leq\sqrt{2}-1\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amilapsn
  • amilapsn
@rational @amistre64 @AngusV
amistre64
  • amistre64
why 9 of them? does it fail with 8 of them?
mathmate
  • mathmate
pigeon-hole?

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amistre64
  • amistre64
sqrt(2) - 1 is less then 1/2
amilapsn
  • amilapsn
@mathmate si
amistre64
  • amistre64
good luck
amilapsn
  • amilapsn
@amistre64 yes
amilapsn
  • amilapsn
\[\Large\sf{\text{Here is the answer:}\\ \forall x \in \mathbb{R}\ \exists\ \theta\in(-\frac{\pi}{2},\frac{\pi}{2})\ such\ that\\ x=tan\theta\\ \text{For 9 reals there'll be 9 corresponding }\theta s\\ \text{Let's divide the interval }(-\frac{\pi}{2},\frac{\pi}{2})\\ \text{8 equal parts. So out of 9 }\theta s\\ \text{there'll be at least 2 such that}\\ \theta_1-\theta_2<\frac{\pi}{8}\\ \Rightarrow tan(\theta_1-\theta_2)
mathmate
  • mathmate
@amilapsn Brilliant. I went as far as splitting the circle in eight parts, but went nowhere. Didn't think of mapping to tan theta. Thank you!
amilapsn
  • amilapsn
That idea suddenly came to my mind with \(\Large\sf{\tan{\frac{\pi}{8}}=\sqrt{2}-1}\)

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