Continued Fraction of sqrt(2) (1 Viewer)

[vuhung]

Anyone please help me solve the part (d) of the problem below, rigorously.
(Cambridge math ext 1, part 2D)

 

[CirrusTheJet]

Will get to this after completing the other problem.

 

[alphxreturns]

proving 'rigorously' is the tough part i reckon

 

[vuhung]

proving 'rigorously' is the tough part i reckon

Can you explain why.
imo this one is a major problem how we teach math.

 

[alphxreturns]

Can you explain why.
imo this one is a major problem how we teach math.

wym why? because its just substitution over and over again, and idk how to 'rigorously' prove myself substituting the same thing infinitely

 

[vuhung]

wym why? because its just substitution over and over again, and idk how to 'rigorously' prove myself substituting the same thing infinitely

Spot on.

Conceptually, it is correct. But mathematically, we cannot apply an operation "infinity" times.

In this case, the sequence x_{n+1} = 2 + 1/x_n is tama-tama-luckily monotic and it converges. But in general case, apply "something" infinity times may lead to any values: a constant, infinity or does not converge at all.

 

[CirrusTheJet]

Mathematical induction? Still would be challenging.

 

[CirrusTheJet]

Maybe if use the formal definition of limits from MATH1131/41?

Note that I differed my course to term 2, so this may be incorrect.

 

[CirrusTheJet]

Is there an actual rigorous proof for this? Still working on this.

 

[vuhung]

> Is there an actual rigorous proof for this?
I think yes, consider the sequence x_{n+1} = 2 + 1/x_n

 

[RZ1234]

Maybe if use the formal definition of limits from MATH1131/41?

Note that I differed my course to term 2, so this may be incorrect.

you realise you have to pay the course fees still

 

[RZ1234]

math1a at UNSW shouldn't be hard at all, it's literally just the SAT (everything is done on a graphing calculator, even better, a algebra calc called maple)

 

[mathasanything]

Spot on.

Conceptually, it is correct. But mathematically, we cannot apply an operation "infinity" times.

In this case, the sequence x_{n+1} = 2 + 1/x_n is tama-tama-luckily monotic and it converges. But in general case, apply "something" infinity times may lead to any values: a constant, infinity or does not converge at all.

Not sure if I'm adding much to this thread, but perhaps the following may help in some way...

Consider how one would actually produce a continued fraction for sqrt(2) (spoiler, there is only 1 way since sqrt(2) is irrational).

We know that sqrt(2) is approx. 1.414213...

Therefore the fraction would start with 1 + 'something' since 1 is the whole number part of the value.

The next part of the continued fraction would need to consider the 0.414213..., as a fraction, this is 1/0.414213... which equals 2.414213... so now our fraction is 1 over (2 + 1 over...). But 2.414213... is actually 1 + sqrt(2)! We then repeat the process for the lower and lower fractions.

In doing so, you'll see that we have: 1 + sqrt(2) = 2.414213... so integral part is 2 again, and the denominator of the next fraction is 2.414213 - 2 = 0.414213. 1 divided by this value is again 2.414213... (which is 1 + sqrt(2))

It's a little difficult to demonstrate what I'm doing as I don't exactly know how to type out these continued fractions with correct formatting, but you can most likely see that the process of creating the continued fraction oscillates between working with sqrt(2) - 1 and sqrt(2) + 1 an infinite amount of times.

The overall value of the fraction, as you've pointed out, converges as the continued fraction grows:

a0 = 1
a1 = 1 + 1/2 = 1.5
a2 = 1 + 1/2 - 1/10 = 1.4
a3 = 1 + 1/2 - 1/10 + 1/60 = 1.416666666666...
a4 = 1 + 1/2 - 1/10 + 1/60 - 1/348 = 1.413793103
a5 = 1 + 1/2 - 1/10 + 1/60 - 1/348 + 1/2030 = 1.4142857143

FYI: a5 = 1 + fraction with 4 steps

a1 > a2
a2 < a3
a3 > a4
a4 < a5

and so on...

Ultimately, since sqrt(2) is irrational and therefore has an infinite number of decimal places, it follows that its continued fraction would continue infinitely. If you accept that a number with an infinite number of decimal places exists, then it follows that you should also accept that said number could be expressed as the sum of smaller and smaller fractions.

FYI - I did know a lecturer at UNSW who genuinely believed that any irrational number DOES NOT EXIST in the strict sense that the number 4 exists, specifically because said number could not be expressed as a fraction AND had an infinite number of decimal places.

I think I'm rambling a bit now...

 

[vuhung]

Introducing this problem in y12 ext 1, chapter Sequence is a quick fix.