Question

which number is large

Answer 1

\[m = [1;\overline{1}]=1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\ddots}}}\]
and
\[n = [1;1,2,\overline{1}]=1+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\ddots}}}\]

Answer 2

I'm back!

Answer 3

I am going to evaluate the truth value of this statement since it seems likely true, just a guess: \[n<m\]\[1+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\ddots}}} < 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\ddots}}}\] Subtracting both sides of the inequality by 1 and then multiplying both sides by the fractions is exactly the same as just flipping the direction of the inequality and shifting the number up! So we can do this flip flop climb: \[1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{1+\ddots}}} > 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\ddots}}}\] \[2+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\ddots}}} < 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\ddots}}}\] \[2<1\] Hey that's false! So it turns out that m<n! Fun!

Answer 4

how did that happen xD
i was thinking n>m lol

Answer 5

Actually I am just realizing this flip flop thing is quite interesting in that it implies that the parity of how far down it is placed determines which is greater than the other... Kinda uncomfortable. Maybe I have some flaw in my logic here let me just work out the recursion since I know that'll get the right answer I just wanted to try something new.

Answer 6

thats a very cool trick, im still trying hard to identify the flaw

Answer 7

there is no flaw! it looks perfect to me

Answer 8

Here's something interesting and probably flawed.
\[n =1+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\ddots}}}\] \[n =1+\dfrac{1}{1+\dfrac{1}{1+m}}\]

Answer 9

`(...)the parity of how far down it is placed determines which is greater than the other...`

i think this is true, keeping everything constant, the parity of position whose value we increase determines whether the value be greater or lesser...

Answer 10

Interesting, I wonder if we could extend this with modular arithmetic in mind and consider perhaps 3 different numbers and depending on how they're arranged we could look at 3n, 3n+1, and 3n+2 down the stack to determine their order a<b<c ?

Answer 11

\[a=[a_0;a_1,a_2,a_3\ldots ]=a_0+\dfrac{1}{a_1+\dfrac{1}{a_2+\dfrac{1}{a_3+\ddots}}}\]
\[b=[b_0;b_1,b_2,b_3\ldots ] =b_0+\dfrac{1}{b_1+\dfrac{1}{b_2+\dfrac{1}{b_3+\ddots}}} \]

Answer 12

\(a_i = b_i\) for all \(i\), except for \(i=\) some position ?

Answer 13

then we want to compare a and b ?

Answer 14

Ahhh this sounds good, I think this is a great generalization and should work exactly identically.

Answer 15

Okay conjecture :
\(b\gt a\) if we keep \(a_i=b_i\) for all \(i\) but increase one particular even index, \(i=2j\)

In initial example, we kept everything same except \(a_2\) and we got \(n\gt m\)

Answer 16

So if \[\large a_{2i}<b_{2i} \implies a<b \\ \large a_{2i+1}<b_{2i+1} \implies b<a\]

Answer 17

that looks good to me! xD

Answer 18

We could definitely do some sort of inductive proof to show this, to give a quick example of how we can construct it, I am satisfied that this is true already. \[\Large a_2 < b_2 \\ \Large \frac{1}{b_2}<\frac{1}{a_2} \\ \Large a_1+ \frac{1}{b_2}<a_1 + \frac{1}{a_2}\]

Answer 19

but we should changing only one value and keep everything else same

Answer 20

Now this is pretty confusing to think about, so I'm not exactly sure if this is the right way to try to generalize up a step, but I want something of this form: \[\large a_{3i}<b_{3i} <c_{3i}\implies ??? \\ \large a_{3i+1}<b_{3i+1} <c_{3i+1} \implies ??? \\ \large a_{3i+2}<b_{3i+2} <c_{3i+2}\implies ???\] So maybe we have to change more than one value, and have one base number 'a' that gets changed at two different spots to create 'b' and 'c'.

Answer 21

Nice you're working the fraction up

\[\Large a_2 < b_2 \\ \Large \frac{1}{b_2}<\frac{1}{a_2} \\ \Large a_1+ \frac{1}{b_2}<a_1 + \frac{1}{a_2}\\

\frac{1}{a_1+ \frac{1}{b_2}}\gt\frac{1}{a_1 + \frac{1}{a_2}}\\~\\
a_0+\frac{1}{a_1+ \frac{1}{b_2}}\gt a_0+\frac{1}{a_1 + \frac{1}{a_2}}\\~\\
n\gt m
\]

Answer 22

Of course we could have also started with the infinite thing and build on top of it, but since they're exactly identical by construction there's no reason to really include it since it's just adding that equal value to both sides.

The only other place other than talking about the golden ratio here that I've seen repeated fractions like this is this really interesting thing in knot theory where there's this isomorphism I guess it's called between tangles and fractions exactly like this. Except that knots are finite, so you can't really get an infinite repeating fraction this way but it's just another thing I'm considering.

Answer 23

I never studied knots/topology but that looks very interesting, it is easy to see that any finite fraction like this converges to a rational number.. idk if this is relevant but here it is :
any finite simple continued fraction converges to a rational number
any infinite simple continued fraction converges to a irrational number

Answer 24

Oh interesting I have never heard that before but it kind of makes sense since an infinite continued fraction will obey some sort of polynomial

Answer 25

While a finite one is clearly rational

Answer 26

im thinking about what happens to a "finite continued fraction" if we increase more than one of the even indexed positions.

earlier we observed increasing "exactly one even indexed position" increases the value.

Answer 27

Is it correct to say that the value increases if we increase "any number of even indexed positions" ?

one sec, im searching for counterexample..

Answer 28

let me put the question a bit more neat

Answer 29

Consider two finite simple continued fractions :\[a=[a_0;a_1,a_2,a_3,\ldots,a_n]\]
\[b=[b_0;b_1,b_2,b_3,\ldots,b_n]\]

such that \(a_i=b_i\) when \(i\) is odd
\(a_i\le b_i\) when \(i\) is even.

then is it true that \(a\le b\) ?

Answer 30

that seems to be true, if it is really true then we don't need to work in mod 3 !!

Answer 31

Hmmm I see what you're saying I think, let me play around with it a bit to figure out some stuff

Answer 32

Can u write a and b in fraction form ??

Answer 33

I don't get [] notations in a,b , so if it possible to rewrite it that I would understand :(

Answer 34

Yeah that seems completely right to me the way you've said it, they should only ever make the number larger. So if we have one base number such that for all i, \(1=a_i=b_i\) except at one place on each number maintains the parity \(a_{2i+k}=b_{2i+k+2}\) which one is larger? So a specific example is: \[ a =1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{1+\ddots}}}\]
and
\[ b= 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2+\ddots}}}\]

Answer 35

ikram :\[a=a_0+\dfrac{1}{a_1+\dfrac{1}{a_2+\dfrac{1}{a_3+\ddots+\dfrac{1}{a_n}}}}\]

\[b=b_0+\dfrac{1}{b_1+\dfrac{1}{b_2+\dfrac{1}{b_3+\ddots+\dfrac{1}{b_n}}}}\]

Answer 36

that looks more interesting, im still thinking..

Answer 37

Here's a little example of the notation hopefully this makes better sense.
\[n = [x;y,\overline{z}]=x+\dfrac{1}{y+\dfrac{1}{z+\dfrac{1}{z+\ddots}}}\]

so we have x before the semicolon cause it's just a number added to this big fraction thingy on its own, so it's sort of special. Then we just start numbering down, so next is y. Then next we have z with a bar on it, which means it repeats infinitely so it'll just be more z's all the way down. Hope this helps @ikram002p !

Answer 38

Yeah, got it :)

Answer 39

Hmm seems like combaring n-I is needed

Answer 40

I think the way I phrased it is sort of misleading, since there will be 4 different values at 2 spots... Hmmm

Answer 41

if we have one base number such that for all \(i\),
\(1=a_i=b_i\) except at one place on each number maintains the parity \(a_{2i+k}=b_{2i+k+2}\)

then \(b\gt a\) because the entire fraction to the right of \(b_2\) will be less than \(1\) :
\[a =1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{1+\ddots}}}\]

\[b= 1+\dfrac{1}{1+\color{blue}{\epsilon}}\]

Answer 42

\(\color{blue}{\epsilon}\lt 1\) is easy to see...

Answer 43

that means changing a number in the top affects the value more than changing the number in down...

Answer 44

I meant to ask the more general question, for two fractions that are the same everywhere except at two spots that are 2 apart, then which one is more important to be greater? A specific case, but general enough to see what I mean(I hope) is: \[ a =n_0+\dfrac{1}{x_1+\dfrac{1}{n_2+\dfrac{1}{n_3+\ddots}}}\]
and
\[ b= n_0+\dfrac{1}{n_1+\dfrac{1}{n_2+\dfrac{1}{x_3+\ddots}}}\]

GIVEN: \[\LARGE n_1<x_1 \\ \LARGE n_3<x_3\] since we just found out what the answer is if x1 and n3 were both larger or or both smaller.

Answer 45

Wait ok nevermind I think you answered my question after all, I might not have understood it one sec...

Answer 46

Actually this result is kind of obvious too >_>

Answer 47

Yeah like also when ai=bi when I is odd and
ai<=bi.when I is even then a>b

Answer 48

yes! b > a is easy to see because we're assuming all are integers

Answer 49

1/(any integer) is less than 1

so \(n_1\lt x_1 \implies n_1+\color{blue}{\epsilon} \lt x_1+\color{blue}{\delta } \)

provided \(n_1, x_1\in \mathbb{N}\) and \(\color{blue}{0\lt \epsilon,\delta\lt 1}\)

Answer 50

Just a thought if you're interested in going this road, what if we can make some function or sequence there, are there any that simplify nicely, just a small example \[n = [1;2,3,4...]=1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\ddots}}}\] \[m = [f(x)]=f(1)+\dfrac{1}{f(2)+\dfrac{1}{f(3)+\dfrac{1}{f(4) +\ddots}}}\]

Answer 51

I'm noticing that if f(x) is a periodic function it will be exactly solvable. So one fun result is if the function is the constant function f(x)=1 then we get the golden ratio.

Answer 52

cool stuff xD i saw the first continued fraction before, it converges to
`0.697774657964007982006790592...`

http://oeis.org/A052119

Answer 53

+1

Answer 54

For any period of 1 (constant function) \[\large n^2=1+n f(x) \\ \large n= \phi \iff f(x)=1\] Cool I'll check this out interesting.

Answer 55

that also proves n is irrational i think
at least for period 1 case...

Answer 56

because the only possible rational roots of polynomial \(n^2=1+nf(x)\) are \(\pm 1\) and none of them satisfy the equation.

Answer 57

Nice

Answer 58

If the period of f is 2, in other words f(x)=f(x+2) then we get this equation for n (which simplifies down to the previous equation if f(1)=f(2) ). \[\Large n^2 f(2)-nf(1)f(2)-f(1)=0 \\ \Large \implies \\ \Large n= \frac{f_1f_2 \pm \sqrt{(f_1f_2)^2+4f_2f_1}}{2f_2}\]

Answer 59

nice :) and i think we can show \(m^2+4m\) is never perfect square...

Answer 60

I'm trying to understand if there's any significance to when f(1)=0 and f(2)=1 in the above equation since it will equal 0 if you plug that in.... hmmm

Answer 61

f(x) must always evaluate to a positive number

Answer 62

otherwise things will break..

Answer 63

Oh plugging in zero for either will always imply division by zero I see where it goes wrong in the algebra, interesting.

Answer 64

That's a cool comment, about how 4m+m^2 is never a perfect square. Interesting, that sorta just fell out of that. I am sorta barely brave enough to see what period 3 looks like, I wonder what we would get if we put like multiplicative functions or some weird number theory function there.

Answer 65

http://www.wolframalpha.com/input/?i=n%3Da%2B1%2F(b%2B1%2F(c%2B1%2Fn))&t=crmtb01

Or written out: \[n = \frac{f_1f_2f_3 +f_1-f_2+f_3 \pm \sqrt{(f_1f_2f_3+f_1-f_2+f_3)^2 +4(f_1f_2+1)(f_2f_3+1)}}{2(f_2f_3+1)}\] Hmmmm...