On April 26, FromTheRafters wrote:
>>>>> Some initial geometric sketches:
>>>>> https://www.desmos.com/calculator/lr7o4i0uat
>>>> Here is a version that recursively decomposes a golden rectangle into
>>>> squares (use the M slider to control the level of recursion).
>>>> https://www.desmos.com/calculator/gf0wq6cwhc
>>> Decomposition into squares, but at the same time, constructing
>>> the sequence of rectangles.
>>>> Now the question remains how this recursive decomposition somehow
>>>> provides evidence that demonstrates geometrically that the golden ratio
>>>> is irrational.
>
>> I was thinking along the lines of whether there is some kind of feature that
>> differentiates between rational proportions, like 1/3 or 2/3 and irrational
>> proportions like the golden ratio.
>> Is it just that irrational proportions can only be subdivided into
>> successively smaller squares while rational proportions can be subdivided
>> into equal parts?
> >
>> The Fibonacci numbers can be used as successive rational approximations to
>> the golden ratio and from that you can kind of infer that there is no
>> ultimate rational proportion (given that there is no biggest Fibonacci
>> number).
>
> Perhaps Continued Fractional Expansions could help a little.

Very good! There's a mystical connection between Fibonacci
sequence and the golden ratio, and continued fractions provides insight.

An arithmetic approach:
r = 1/(1 + r)
Substitute for r, recursively:
r = 1/(1+ [1/(1 + r)])
= 1/(1+ [1/(1 + [etc.])])
= ....

Calculate the first five terms of this series, as rational numbers.
Stare at the pattern, what do you see?
Show that if one of these terms is p/q, the next is q/(p + q)

Next, we have to show that this sequence converges to r, hmmm...

Bonus credit:
Re-write the formula as:
r² = 1 - r
r = √(1 - r)

Then, compute recursively:
r = √(1 - [√(1 - r)])
= √(1 - [√(1 - √(1 - r))])
= ....

This method fails. Why? (consider the sequence of finite parts)

--
Rich

On Wednesday, April 27, 2022 at 8:27:18 PM UTC+2, RichD wrote:
> On April 26, Mike Terry wrote:
> >>>> Here is a version that recursively decomposes a golden rectangle into
> >>>> squares
> >>>> https://www.desmos.com/calculator/gf0wq6cwhc
> >>>> Now the question remains how this recursive decomposition somehow
> >>>> provides evidence that demonstrates geometrically that the golden ratio
> >>>> is irrational.
> >
> >>> The construction is not a proof, but evidence, a guide to a proof.
> >>> hint: Starting from the initial rectangle, with semi-arbitrary
> >>> dimensions, consider the sequence of calculated dimensions
> >>> of the ensuing rectangles.
> >
> >> I was thinking along the lines of whether there is some kind of feature that
> >> differentiates between rational proportions, like 1/3 or 2/3 and irrational
> >> proportions like the golden ratio.
> >> Is it just that irrational proportions can only be subdivided into successively smaller
> >> squares while rational proportions can be subdivided into equal parts?
> >
> > That's sort of heading in the right direction...
> > If we start with whole number lengths a,b with a>b, then a-b and b will have
> > the same greatest common divisor (GCD), and as the process continues
> > taking off more squares, the GCD always remains the same. Since the lengths
> > themselves keep decreasing, eventually the GCD itself will be reached and the
> > process can't continue because the new length will become zero.
> > (This is effectively applying Euclid's algorithm to calculate the GCD.)
> > The same applies with rational a,b because we can just change the units to
> > make a,b both whole number lengths.
> > So if, as with the golden ratio, the process continues indefinitely, that implies the ratio is
> > irrational.
> That works.
>
> The simplest explanation:
> Assume r = b/a is rational.
> Start with a rectangle of integer dimensions; like, a and a+b (duh)
> i) By construction, the rectangle dimensions continually shrink.
> ii) The dimensions are all positive.
> iii) By arithmetic, the dimensions of every rectangle are integers.
>
> Then, we see.... <drum roll>... an infinite descending sequence
> of positive integers! Which makes the baby Jesus cry.
>
> --
> Rich

But visually it seems there is a connection between numerical pattern in
the continued fraction and the visual pattern of squares decreasing in size.
For instance, for the square root of two, I would expect one big square and
then subtracting two smaller identical squares in every subsequent step from the
remaining area.

https://i.imgur.com/E6zgOSi.png

https://r-knott.surrey.ac.uk/Fibonacci/cfINTRO.html

On April 27, RichD wrote:
>>>>> Now the question remains how this recursive decomposition somehow
>>>>> provides evidence that demonstrates geometrically that the golden ratio
>>>>> is irrational.
>>> I was thinking along the lines of whether there is some kind of feature that
>>> differentiates between rational proportions, like 1/3 or 2/3 and irrational
>>> proportions like the golden ratio.
>>> Is it just that irrational proportions can only be subdivided into
>>> successively smaller squares while rational proportions can be subdivided
>>> into equal parts?
>
>>> The Fibonacci numbers can be used as successive rational approximations to
>>> the golden ratio and from that you can kind of infer that there is no
>>> ultimate rational proportion (given that there is no biggest Fibonacci
>>> number).
>
>> Perhaps Continued Fractional Expansions could help a little.
>
> There's a mystical connection between Fibonacci
> sequence and the golden ratio, and continued fractions provides insight.
>
> An arithmetic approach:
> r = 1/(1 + r)
> Substitute for r, recursively:
> r = 1/(1+ [1/(1 + r)])
> = 1/(1+ [1/(1 + [etc.])])
> = ....
> Calculate the first five terms of this series, as rational numbers.
> Stare at the pattern, what do you see?
> Show that if one of these terms is p/q, the next is q/(p + q)
> Next, we have to show that this sequence converges to r, hmmm...

The sequence successively oscillates around r, eventually trapping r.

Define:
u = p/q
v = q/(p + q)

uv + v = 1

And:
I) r² + r = 1
If u = v, they would satisfy (I), therefore u = v = r
But that's impossible, as r is irrational.
Hence, u ≠ v

Finally, it's an arithmetic exercise, to show that r lies
between them; replace the larger by the smaller, and
vice versa, plug into (I)

> Bonus credit:
> Re-write the formula as:
> r² = 1 - r
> r = √(1 - r)
> Then, compute recursively:
> r = √(1 - [√(1 - r)])
> = √(1 - [√(1 - √(1 - r))])
> = ....
> This method fails. Why? (consider the sequence of finite parts)

The sequence of finite parts:
√1, √(1 - √1), √(1 - √(1 - √1)), ...
doesn't converge to a limit.

However, define m = 1/r
Then:
m² = 1 + m
m = √(1 + m)

The recursive computation works for m.
Which is hard to explain -

--
Rich