2

I know how to show that, for example $1/4$ is an element of the Cantor set using tertiary expansion but how do you use this method for an irrational number?

Keralf2020
  • 47

2 Answers2

2

Exactly the same principle applies as in the rational case. You go down one digit in the ternary expansion at a time; if any digit is $1$ the number is not in the Cantor set. For $\frac\pi9$, the first digit is already $1$…

Almost all irrational numbers contain $1$ in their ternary expansions, and are thus not part of the set.

Parcly Taxel
  • 103,344
  • Amazing! I mean, unless you build it on purpose, all the irrationals I tried have a $1$ in their ternary expansion. I built $$\sqrt[8]{3} \vartheta _2\left(0,\frac{1}{\sqrt{3}}\right)-2$$ which is $$\sum _{n=1}^{\infty } 2 \left(\frac{1}{3}\right)^{\frac{1}{2} n (n+1)}$$ where the digit $2$ is only in triangular integer positions. – Raffaele Oct 24 '20 at 11:49
1

$\frac{\pi}{9}\in \left(\frac{1}{3},\frac{2}{3}\right)$, so it cannot be in the Cantor set.

Laars Helenius
  • 7,965
  • 1
  • 23
  • 35
  • 1
    It is more difficult to find a number IN the Cantor set than OUT, actually... Nevertheless it is more than countable. Amazing! – Raffaele Oct 24 '20 at 12:03
  • Yes. The Cantor set has a lot of counterintuitive properties and if you aren’t blown away by it, you don’t really understand it. – Laars Helenius Oct 24 '20 at 12:46