This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

Let $x=\sum _{ n=1 }^{ \infty }{ F(n){ 10 }^{ -3n } }$. 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!

@Calvin Lin
–
Ok lets see
If i remember we write the above sequence like

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

@Eddie The Head
–
i found the answer before i could figure it out, since it is a solution of @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.

You can try by proving that

For every $k$ there exists a fibonacci number whose decimal representation ends in $k$ 9's ,

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.

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.

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?

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.)

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
–
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....,.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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!

Log in to reply

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?"

Log in to reply

@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.

Log in to reply

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.

Log in to reply

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 ,

Log in to reply

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.

Log in to reply

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.

Log in to reply

u r ans r complez why is math complex

Log in to reply

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?

Log in to reply

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.)

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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 $0.001004009016025\dots$.

Log in to reply

$\textbf{Challenge 1.}$

$F(x) = x + 2x^{2} + 3x^{3} + 4x^{4}+...............$ $xF(x) = x^{2} + 2x^{3} + 3x^{4}+........$

$(1-x)F(x) = x + x^{2} + x^{3} + x^{4}+.........$ $F(x) = \frac{x}{(1-x)^{2}}$ 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 $F(\frac{1}{1000}) = \frac{1000001000}{998001}$

Log in to reply

Log in to reply

$\frac{\sqrt5+1}2$, but this is less than $1000$.

The Fibonacci does increase exponentially with ratioJust do partial fraction decomposition on the LHS and you can find the interval of convergence.

Log in to reply

Log in to reply

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

Log in to reply

rational because no is not repeated till now

Log in to reply

irrational

Log in to reply