Is the number represented by the decimal

\[0.001001002003005008013021034055089144233377610...\]

rational, or irrational? Please justify your answer.

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## Comments

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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!

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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?"

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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.

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

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

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@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 ,

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One idea that we can try is proving that given any digit, and any number n, there always exist some Fibonacci number that has a sequence of that digit n times. If we were to prove that we can always find in some Fibonacci number a sequence like 11111.. as arbitrarily long as we'd like, that would pretty much prove that the decimal never repeats, and therefore it'd be irrational.

Would I want to try proving this? Ah, maybe too much for me now.

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u r ans r complez why is math complex

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If I were to make a snap judgement on this one, I'd say not only it's probably irrational, it's probably even transcendental. I'm almost afraid to try touching this one.

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.

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It equals \(\frac{1000}{998999}\), a rational number (see my response to Eddie the Head).

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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?

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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 see @Calvin Lin 's comment.)

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irrational

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rational because no is not repeated till now

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since it can be converted into p/q.form...it should be rational number..what do u say..???

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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 solution

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The generating function for the Fibonacci sequence is

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

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 \(0.001004009016025\dots\).

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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 \(F(\frac{1}{1000}) = \frac{1000001000}{998001}\)

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Just do partial fraction decomposition on the LHS and you can find the interval of convergence.

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What value of p and q would you say they are??

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