×

# Rational or Irrational?

Is the number represented by the decimal

$0.001001002003005008013021034055089144233377610...$

Note by Matt Enlow
3 years, 3 months ago

Sort by:

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! · 3 years, 3 months ago

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?" Staff · 3 years, 3 months ago

@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. · 3 years, 3 months ago

Yup, I believe that there was a past discussion about it. This is a gem, and worth revisiting. Staff · 3 years, 3 months ago

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. · 3 years, 3 months ago

Can you post the link to the previous discussion? · 3 years, 3 months ago

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 , · 3 years, 3 months ago

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. · 3 years, 3 months ago

u r ans r complez why is math complex · 3 years, 3 months ago

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. · 3 years, 3 months ago

It equals $$\frac{1000}{998999}$$, a rational number (see my response to Eddie the Head). · 3 years, 3 months ago

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? · 3 years, 3 months 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 see @Calvin Lin 's comment.) · 3 years, 3 months ago

irrational · 3 years, 3 months ago

rational because no is not repeated till now · 3 years, 3 months ago

since it can be converted into p/q.form...it should be rational number..what do u say..??? · 3 years, 3 months 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 solution · 3 years, 3 months ago

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 $$0.001004009016025\dots$$. · 3 years, 3 months ago

$$\textbf{Challenge 1.}$$

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

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

But what will your logicbe for convergence? I think it is because the denominator is incresing exponentially and the numerator is not. · 3 years, 3 months ago

The Fibonacci does increase exponentially with ratio $$\frac{\sqrt5+1}2$$, but this is less than $$1000$$.

Just do partial fraction decomposition on the LHS and you can find the interval of convergence. · 3 years, 3 months 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....,. · 3 years, 3 months ago