Curious decimal behaviour

Algebra Level 3

Define x = max { m Z m x } \lfloor x \rfloor = \max\{m\in\mathbb{Z} \, \big| \, m \le x\} for all x R . x\in \mathbb{R}.

Now, let ρ \rho be some real number—but not an integer—such that ρ ρ = ρ 2 ρ 2 . \rho - \lfloor \rho \rfloor =\rho^2 - \big\lfloor \rho^2 \big\rfloor. Can such a number exist?

Yes Yes, but only in base 8 and base 10 Yes, but only in base 2 No, because it would be complex No

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2 solutions

Vitor Juiz
Mar 11, 2018

You're absolutely right! I forgot about specifying that you have to assume also ρ Z \rho\notin\mathbb{Z} .

Vittorio Bertoletti - 3 years, 3 months ago

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I've updated the text of the problem.

Vittorio Bertoletti - 3 years, 3 months ago

Loosely speaking the problem is asking us if we can find at least one ρ \rho with this property: all the digits following the dot of ρ \rho are the same digits following the dot of ρ 2 \rho^2 . Now let's write this ρ ρ = ρ 2 ρ 2 ρ 2 ρ = ρ 2 ρ . \rho - \lfloor \rho \rfloor=\rho^2 - \lfloor \rho^2 \rfloor \quad \Leftrightarrow \quad \rho^2 - \rho= \lfloor \rho^2 \rfloor - \lfloor \rho \rfloor. φ = 1 + 5 2 \varphi=\frac{1+\sqrt{5}}{2} The golden ratio φ \varphi will do the trick for ρ \rho because it's the positive solution for ρ 2 ρ 1 = 0 \rho^2-\rho-1=0 so 1 < φ < 2 φ = 1 1<\varphi<2 \Rightarrow \lfloor \varphi \rfloor=1 φ 2 = φ + 1 = 2 \lfloor \varphi^2 \rfloor=\lfloor \varphi+1 \rfloor=2 and hence φ 2 φ = 1 = φ 2 φ = 2 1 = 1. \varphi^2-\varphi=1=\lfloor \varphi^2 \rfloor - \lfloor \varphi \rfloor=2-1=1.

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