Is the number represented by the decimal

\[0.001001002003005008013021034055089144233377610...\]

rational, or irrational? Please justify your answer.

Is the number represented by the decimal

\[0.001001002003005008013021034055089144233377610...\]

rational, or irrational? Please justify your answer.

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TopNewestLet \(x=\sum _{ n=1 }^{ \infty }{ F(n){ 10 }^{ -3n } } \). A little foodling with this, using an approach you (yes, you, Matt) used in another problem similar to this, gets us the equation to solve: \(1000(1000x-x-1)=x\), which means x is rational. I leave it to the reader as an exercise to find that value x. Thanks for the tip on how to do this, Matt! – Michael Mendrin · 3 years, 1 month ago

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The other version that I like: Is the decimal

\[0.1123581321345589144...\]

rational or irrational?

It simplifies the consideration of "What happens if the fibonacci number has 4 digits or more?" – Calvin Lin Staff · 3 years, 1 month ago

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@Calvin Lin Hmmm i remember this one from somewhere else not sure where,

to proof The periodic sequence, It can be shown that fibonacci sequence is never eventually periodic Which goes to show that it is irrational.

I think we discussed this on brilliant a while back. – Beakal Tiliksew · 3 years, 1 month ago

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– Calvin Lin Staff · 3 years, 1 month ago

Yup, I believe that there was a past discussion about it. This is a gem, and worth revisiting.Log in to reply

\(F=0.{ f }_{ 1 }{ f }_{ 2 }{ f }_{ 3 }.........{ f }_{ n }\), \(F(n)\equiv f(n)mod10\), Then \( { f }_{ n+2 }={ f }_{ n+1 }+{ f }_{ n }mod10\)

We can develop on this to show the existance of a fibonacci sequence congruent to \(-1 mod{ 10 }^{ k }\) but we will have to show that for any modulus p -- \({ F }_{ n }mod\quad p for\quad n(-\infty \quad to\quad \infty )\), is periodic.

I will have to think about it a bit more, My number theory is a rough. – Beakal Tiliksew · 3 years, 1 month ago

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– Eddie The Head · 3 years, 1 month ago

Can you post the link to the previous discussion?Log in to reply

@Calvin Lin , i won't post it here,so as not to spoil it.and for others to try it, i could mail you the link if you want.

i found the answer before i could figure it out, since it is a solution ofYou can try by proving that

For every \(k\) there exists a fibonacci number whose decimal representation ends in \(k\) 9's , – Beakal Tiliksew · 3 years, 1 month ago

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Would I want to try proving this? Ah, maybe too much for me now. – Michael Mendrin · 3 years, 1 month ago

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– Rohit Singh · 3 years, 1 month ago

u r ans r complez why is math complexLog in to reply

As a matter of fact, Calvin, your suggested decimal number is similar to Champernowne's constant, which is 0.123456789101112131415..., which was shown to be transcendental. Not easy to do that. – Michael Mendrin · 3 years, 1 month ago

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It equals \(\frac{1000}{998999}\), a rational number (see my response to Eddie the Head). – Cody Johnson · 3 years, 1 month ago

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The first question I'd have is, what does the number look like after you get to four digit Fibonacci numbers? Would the thousand's digit of one segment add to the one's digit of the previous? – Dan Krol · 3 years, 1 month ago

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@Calvin Lin 's comment.) – Matt Enlow · 3 years, 1 month ago

Yes, great question. And I would say that your suggestion as to how to resolve this ambiguity is the one that makes the most sense. (Also seeLog in to reply

irrational – Kiran Kumar · 3 years, 1 month ago

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rational because no is not repeated till now – Rohit Singh · 3 years, 1 month ago

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since it can be converted into p/q.form...it should be rational number..what do u say..??? – Max B · 3 years, 1 month ago

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– Eddie The Head · 3 years, 1 month ago

How can you convert it to p/q form??It has no repeating parts as far as I see :p ..But I would love to see a rigorous solutionLog in to reply

generating function for the Fibonacci sequence is

The\[F(x)=f_0+f_1x+f_2x^2+f_3x^3+\dots\]

Since \(f_n=f_{n-1}+f_{n-2}\), we have to cancel some terms

\[F(x)=f_0+f_1x+f_2x^2+f_3x^3+\dots\] \[xF(x)=f_0x+f_1x^2+f_2x^3+f_3x^4+\dots\] \[x^2F(x)=f_0x^2+f_1x^3+f_2x^4+f_3x^5+\dots\]

Gathering the coefficients of \((1-x-x^2)F(x)\), we get

\[(1-x-x^2)F(x)=f_0+(f_1-f_0)x+(f_2-f_1-f_0)x^2+(f_3-f_2-f_3)x^3+\dots=f_0+(f_1-f_0)x\]

Hence,

\[F(x)=\frac{f_0+(f_1-f_0)x}{1-x-x^2}=\frac{x}{1-x-x^2}=f_0+f_1x+f_2x^2+f_3x^3+\dots\]

If we want to encode this into a decimal, we just let \(x=\frac{1}{1000}\). The answer is

\[f_0+f_1\cdot\frac{1}{1000}+f_2\cdot\frac{1}{1000^2}+f_3\cdot\frac{1}{1000^3}+\dots=\frac{\frac{1}{1000}}{1-\frac{1}{1000}-\frac{1}{1000^2}}=\frac{1000}{998999}\]

But beware of convergence issues! \(\frac{1}{1000}\) is sufficiently small, so I assumed it converged, which it did. Challenge: find \(0.001002003004\dots\) and \(0.001004009016025\dots\). – Cody Johnson · 3 years, 1 month ago

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\[F(x) = x + 2x^{2} + 3x^{3} + 4x^{4}+...............\] \[xF(x) = x^{2} + 2x^{3} + 3x^{4}+........\]

\[(1-x)F(x) = x + x^{2} + x^{3} + x^{4}+.........\] \[F(x) = \frac{x}{(1-x)^{2}}\] Hence we have \(f(\frac{1}{1000}) = \frac{1000}{998001}\)

We can clearly see that the terms converge.

\(\textbf{Challenge 2.}\)

In this problem we must use the same approach twice,the resulting generating function \[F(x) = \frac{1+x^{2}}{(1-x)^{3}}\]

Hence \(F(\frac{1}{1000}) = \frac{1000001000}{998001}\) – Eddie The Head · 3 years, 1 month ago

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– Eddie The Head · 3 years, 1 month ago

But what will your logicbe for convergence? I think it is because the denominator is incresing exponentially and the numerator is not.Log in to reply

Just do partial fraction decomposition on the LHS and you can find the interval of convergence. – Cody Johnson · 3 years, 1 month ago

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– Eddie The Head · 3 years, 1 month ago

Nice...I'm familiar with the usage of generating functions in association with the terms of a series but in this case it didn't come to my mind at a first glance....,.Log in to reply

– Beakal Tiliksew · 3 years, 1 month ago

What value of p and q would you say they are??Log in to reply