Irrational Bases 2.

Level 2

Let ϕ = 1 + 5 2 \phi = \dfrac{1 + \sqrt{5}}{2} and α = 2 + 1 \alpha = \sqrt{2} + 1 .

Which one of the five choices below are true?

1: ϕ − 1 = ( 0.010101010101... ) ϕ \quad \phi - 1 = (0.010101010101 ... )_{\phi} and α − 2 2 = ( 0.010101010101... ) α \dfrac{\alpha - 2}{2} = (0.010101010101 ... )_{\alpha}

2: ϕ − 1 = ( 0.101010101010... ) ϕ \quad \phi - 1 = (0.101010101010 ... )_{\phi} and α − 2 2 = ( 0.010101010101... ) α \dfrac{\alpha - 2}{2} = (0.010101010101 ... )_{\alpha}

3: ϕ − 1 = ( 0.100100100100... ) ϕ \quad \phi - 1 = (0.100100100100 ...)_{\phi} and α − 2 2 = ( 0.001001001001... ) α \dfrac{\alpha - 2}{2} = (0.001001001001 ... )_{\alpha}

4: ϕ − 1 = ( 0.101010101010... ) ϕ \quad \phi - 1 = (0.101010101010 ... )_{\phi} and α − 2 2 = ( 0.101010101010... ) α \dfrac{\alpha - 2}{2} = (0.101010101010 ... )_{\alpha}

5: ϕ − 1 = ( 0.010101010101... ) ϕ \quad \phi - 1 = (0.010101010101 ... )_{\phi} and α − 2 2 = ( 0.101010101010... ) α \dfrac{\alpha - 2}{2} = (0.101010101010 ... )_{\alpha}

4 5 1 2 3

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Feb 9, 2019

ϕ \phi is one solution to x 2 − x − 1 = 0 x^2 - x - 1 = 0

⟹ ϕ 2 = ϕ + 1 ⟹ ϕ 2 − 1 = ϕ ⟹ 1 ϕ 2 − 1 = 1 ϕ = 2 5 + 1 = 5 − 1 2 = ϕ − 1 \implies \phi^2 = \phi + 1 \implies \phi^2 - 1 = \phi \implies \dfrac{1}{\phi^2 - 1} = \dfrac{1}{\phi} = \dfrac{2}{\sqrt{5} + 1} = \dfrac{\sqrt{5} - 1}{2} = \phi - 1

⟹ ϕ − 1 = 1 ϕ 2 − 1 = 1 ϕ 2 1 − 1 ϕ 2 = \implies \phi - 1 = \dfrac{1}{\phi^2 - 1} = \dfrac{\dfrac{1}{\phi^2}}{1 - \dfrac{1}{\phi^2}} = ∑ n = 1 ∞ ( 1 ϕ 2 ) n = ( 0.010101010101... ) ϕ \sum_{n = 1}^{\infty} (\dfrac{1}{\phi^2})^{n} = \boxed{(0.010101010101 ...)_{\phi}}

α \alpha is one solution to x 2 − 2 x − 1 = 0 x^2 - 2x - 1 = 0

⟹ α 2 = 2 α + 1 ⟹ α 2 − 1 = 2 α ⟹ 1 α 2 − 1 = 1 2 α = 1 2 ( 2 + 1 ) = 2 − 1 2 = α − 2 2 \implies \alpha^2 = 2\alpha + 1 \implies \alpha^2 - 1 = 2\alpha \implies \dfrac{1}{\alpha^2 - 1} = \dfrac{1}{2\alpha} = \dfrac{1}{2(\sqrt{2} + 1)} = \dfrac{\sqrt{2} - 1}{2} = \dfrac{\alpha - 2}{2}

⟹ α − 2 2 = 1 α 2 − 1 = 1 α 2 1 − 1 α 2 = ∑ n = 1 ∞ ( 1 α 2 ) n = ( 0.010101010101... ) α \implies \dfrac{\alpha - 2}{2} = \dfrac{1}{\alpha^2 - 1} = \dfrac{\dfrac{1}{\alpha^2}}{1 - \dfrac{1}{\alpha^2}} = \sum_{n = 1}^{\infty} (\dfrac{1}{\alpha^2})^{n} = \boxed{(0.010101010101 ...)_{\alpha}}

∴ \therefore Choice (1) is true.

0 Helpful 1 Interesting 1 Brilliant 0 Confused

In general, if we let ϕ n = n 2 + 4 + n 2 \phi_n = \frac {\sqrt{n^2+4}+n}2 be one solution to x 2 − n x − 1 = 0 x^2-nx-1=0 , then

ϕ n 2 − n ϕ n − 1 = 0 ⇔ ϕ n 2 − 1 = n ϕ n ⇔ 1 ϕ n 2 − 1 = 1 n ϕ n = 1 n n 2 + 4 + n 2 = 2 n n 2 + 4 + n 2 ⇔ 2 n n 2 + 4 + n 2 = 1 ϕ n 2 − 1 = 1 ϕ n 2 1 − 1 ϕ n 2 = ∑ k = 1 ∞ ( 1 ϕ n 2 ) k = ( 0.01010101 … ) ϕ n \begin{aligned} \phi_n^2-n\phi_n-1 &= 0 \\ \Leftrightarrow \phi_n^2-1 &= n\phi_n \\ \Leftrightarrow \frac 1{\phi_n^2-1} &= \frac 1{n\phi_n} \\ &= \frac 1{n\frac{\sqrt{n^2+4}+n}2} \\ &= \frac 2{n\sqrt{n^2+4}+n^2} \\ \Leftrightarrow \frac 2{n\sqrt{n^2+4}+n^2} &= \frac 1{\phi_n^2-1} \\ &= \frac {\frac 1{\phi_n^2}}{1-\frac 1{\phi_n^2}} \\ &= \sum_{k=1}^\infty \left( \frac 1{\phi_n^2} \right)^k \\ &= (0.01010101\ldots)_{\phi_n} \end{aligned}

Here are the first cases

n 2 n n 2 + 4 + n 2 1 2 5 + 1 2 1 8 + 4 3 2 3 13 + 9 4 1 4 5 + 16 ⋮ ⋮ ∞ 0 \begin{array}{cc} n & \frac 2{n\sqrt{n^2+4}+n^2} \\ 1 & \frac 2{\sqrt{5}+1} \\ 2 & \frac 1{\sqrt{8}+4} \\ 3 & \frac 2{3\sqrt{13}+9} \\ 4 & \frac 1{4\sqrt{5}+16} \\ \vdots & \vdots \\ \infty & 0 \end{array}

Henry U - 2 years, 4 months ago

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Nice. You generalized the problem.

I would state the problem as follows:

Let n ∈ N n \in \mathbb{N} and ϕ n = n 2 + 4 + n 2 \phi_{n} = \dfrac{\sqrt{n^2 + 4} + n}{2}

Which is a representation of ϕ n − n n \dfrac{\phi_{n} - n}{n} in base ϕ n \phi_{n} . (Give choices)

Solution:

ϕ n \phi_{n} is one solution to x 2 − n x − 1 = 0 x^2 - nx - 1 = 0

⟹ ϕ n 2 = n ϕ n + 1 ⟹ ϕ n − 1 − n ϕ n ⟹ 1 ϕ n 2 − 1 = 1 n ϕ n = ( n 2 + 4 − n 2 ) n \implies \phi_{n}^2 = n\phi_{n} + 1 \implies \phi_{n} - 1 - n\phi_{n} \implies \dfrac{1}{\phi_{n}^2 - 1} = \dfrac{1}{n\phi_{n}} = \dfrac{(\dfrac{\sqrt{n^2 + 4} - n}{2})}{n} = ϕ n − n n ⟹ ϕ n − n n = 1 ϕ n 2 − 1 = 1 ϕ n 2 1 − 1 ϕ n 2 = = \dfrac{\phi_{n} - n}{n} \implies \dfrac{\phi_{n} - n}{n} = \dfrac{1}{\phi_{n}^2 - 1} = \dfrac{\dfrac{1}{\phi_{n}^2}}{1 - \dfrac{1}{\phi_{n}^2}} = ∑ m = 1 ∞ ( 1 ϕ n 2 ) m = ( 0.010101010101... ) ϕ n \sum_{m = 1}^{\infty} (\dfrac{1}{\phi_{n}^2})^{m} = \boxed{(0.010101010101 ...)_{\phi_{n}}} .

Rocco Dalto - 2 years, 4 months ago

And even more general, if ϕ n , m \phi_{n,m} is a solution to x m − n x − 1 = 0 x^m - nx -1 = 0 (which I can't find explicitly), then the base ϕ n , m \phi_{n,m} representation of 1 n ϕ n , m \frac 1{n\phi_{n,m}} is ∑ n = 1 ∞ ( 1 ϕ n , m m ) n \displaystyle \sum_{n=1}^\infty \left( \frac 1{\phi_{n,m}^m} \right)^n , which is a 1 at every m m -th digit.

Henry U - 2 years, 4 months ago

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Very nice!

Let α = ϕ n . m \alpha = \phi_{n.m} be solution to x m − n x − 1 = 0 ⟹ α m = n α + 1 ⟹ α m − 1 = n α ⟹ 1 α m − 1 = 1 n α ⟹ x^m - nx - 1 = 0 \implies \alpha^{m} = n\alpha + 1 \implies \alpha^m - 1 = n\alpha \implies \dfrac{1}{\alpha^m - 1} = \dfrac{1}{n\alpha} \implies

1 n α = 1 α m 1 − 1 α m = ∑ j = 1 ∞ ( 1 ϕ n , m m ) j \dfrac{1}{n\alpha} = \dfrac{\dfrac{1}{\alpha^m}}{1 - \dfrac{1}{\alpha^m}} = \displaystyle\sum_{j = 1}^{\infty} (\dfrac{1}{\phi_{n,m}^m})^j .

Note: For m = 3 m = 3 and n = 2 n = 2 we obtain: x 3 − 2 x − 1 = 0 ⟹ ( x + 1 ) ( x 2 − x − 1 ) = 0 ⟹ x = − 1 , 1 ± 5 2 x^3 - 2x - 1 = 0 \implies (x + 1)(x^2 - x - 1) = 0 \implies x = -1, \dfrac{1 \pm \sqrt{5}}{2} , so that ϕ \phi is a root of above cubic.

Problem: Which is a representation of ϕ − 1 2 \dfrac{\phi - 1}{2} in base ϕ \phi (State Choices)

ϕ − 1 2 = ( 0.001001001001... ) ϕ \dfrac{\phi - 1}{2} = (0.001001001001 ...)_{\phi} .

Of course you could just do the above using ϕ \phi as one root of x 2 − x − 1 = 0 x^2 - x - 1 = 0 .

I.E; ϕ 2 = ϕ + 1 ⟹ ϕ 3 = 2 ϕ + 1 ⟹ 1 ϕ 3 − 1 = 1 2 ϕ = ϕ − 1 2 = ∑ j = 1 ∞ ( 1 j 3 ) j = ( 0.001001001001... ) ϕ \phi^2 = \phi + 1 \implies \phi^3 = 2\phi + 1 \implies \dfrac{1}{\phi^3 - 1} = \dfrac{1}{2\phi} = \dfrac{\phi - 1}{2} = \sum_{j = 1}^{\infty} (\dfrac{1}{j^3})^{j} = (0.001001001001 ...)_{\phi}

Rocco Dalto - 2 years, 4 months ago

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