How is this possible?

I can proove that -2 = 2.

(-2) = 3root(-8) = (-8)^(1/3) = (-8)^(2/6) = 6root((-8)2) = 6root(64) = 2

Maybe you should watch the video and find out instead of whining about not knowing something that could easily be known.

>give me ad revenue
fuck OFF

1:50 "Any number whatsoever has a representation as a continued fraction" ... meh, even if we're only considering the definable reals, there a non-computables such as Caitlin's constant. The case doesn't seem so clear.

>not running an ad blocker but getting this mad about ads

lmao

You're all retarded, it's stated in the video, and should be obvious.

A rational number has a continued fraction if and only if it is finite

1 and 2 are rational numbers so they can't have infinite continued fraction representations

Alright, if that is the case then the continued fraction is equal to neither 1 nor 2. So what does it actually equal? What is the sequence of truncated fractions converge to? We know it must be an irrational number because it is an infinite continued fraction, but the question is can we express this irrational number simply using rational numbers like with sqrt(2).

The sequence of truncated fractions goes as follows: 2/3, 6/7, 14/15, 30/31, 62/63, 126/127, ...
So it appears to be getting closer to one, but we know this can't be the case because it's an infinite fraction. Does anyone recognize this sequence as converging toward some number? I'm not really a math wizard so I don't really have the intuition for this. But the fraction never exceeds 1, so it seems to me that at infinity this thing comes out to be .999... Which implies that .999... is different from 1 and also that .999... is an irrational number.

For any a,b, we'd formally get

[math] x = \dfrac{ (ab)^2 }{ 2 b + n x } = n\, b\, \left( \pm \sqrt{ 1 + n\,a^2 } - \sqrt{ 1 } \right) [/math]

and the one in the video is
[math] n = -1 [/math]
[math] b = 3/2 [/math]
[math] a = \sqrt{2}/b [/math]

Do all of those fail?

>You're all retarded
wtf is your problem?

A question that this raises is if there are theorems which tells us when a sequence of numbers, translated to a continued fraction like in the video, gives a converging sequence.

It might just not converge, like [math] \sum_{k=0}^\infty x^k [/math] doesn't converge for x = 9/8

It might also have to do with there being a negative sign, while in the rest of the video he's about representations with positive signs.

>We know it must be an irrational number because it is an infinite continued fraction
That's false. Only infinite continued fractions of a *certain type* result in only irrationals.

>It might just not converge
But the sequence obviously doesn't diverge, unlike (9/8)^k. It always remains below 1.
>It might also have to do with there being a negative sign
Possibly. Like I said I have no intuition for these things. I was just making an observation and it kinda makes sense because the concept of .999... is irrational (if you don't believe me just start a thread trying to rationalize it). And it's not really a number that you can converge to either (depending on your definition of converge), so you're right that the sequence doesn't converge in the classical sense.

What about the rest? They diverge I'm guessing? If I learned anything from the video it's that they definitely do not converge to rational numbers. So if it's not converging to a rational number, and by observation we see it's not diverging, what is it?

they aren't the same because the base that he is playing with is different.

the fraction isn't complete until he puts in the two or the one, therefore the infinite fraction does not equal two or one unless he places a 2 or 1 at the end of the series, it's just meaningless infinity.

[math] a_n := 7+(-2)^n [/math]
doesn't diverge either.

>awesome shirt dude

But the sequence does appear to converge to .999... I don't know why you're being difficult, I'm just stating what's obvious. 7+(-2)^n may not converge but it's also not a continued fraction so why are you even bringing it up? Are there other examples of continued fractions that don't diverge but don't converge but remain below a certain bound? Or can the continued fraction be written in the same form as you wrote in your post?

Don't overthink this problem; it's actually piss easy.

Notice that the sequence has a pattern. 2/3 to 6/7? 6/7 to 14/15? 14/15 to 30/31? The numerator is the sum of all numbers 2^k, where k is its order in the sequence. The denominator is just that sum plus one.

I'm sure you guys can figure this one out pretty quickly.

...

Nice! Dividing by zero! Good job user!

How is that delusional you fucking pseudo intellectual retard?

Which branch of Mathematics actually studies infinite continued fractions?

number theory

I could be wrong but it looks like it's wrong because the end results are different at the termination.

If you terminate the identity for 2 you get
2/(3-2) at the end.

If you terminate the identity for 1 at the end you get 2/(3-1).

This difference means the two can't be compared.

That's my basic analysis. The real answer might be as stupid as "Oh he broke some rule of math." But I'm inclined to believe the maker of the video intended for this to be a straightforward answer, rather than one that could be a subject of debate.

The 'mistake' is that the equality he presents is not true until you actually finish the fraction. If you just keep going on forever then you don't approach 1 nor 2.

So the mistake happens way before 1=2 because it is wrong the moment he said that 1= infinite fraction

>Its another let's trick retards with 11th grade math thread
Goodbye

>it literally is

just testing some shitty math here

x= 2/(3-x)
x(3-x)=2
3x-x^2-2=0
(x-2)(x-1)=0

x=2 OR x=1

help

what did it mean by this?

Not sure if serious, but as pointed out, this involves a division by zero. This lies in the first statement, a=b

If a = b, then (a - b) = 0

e.g. let a = 2 = b

2 - 2 is obviously zero, so from this, you CANNOT divide both sides by (a-b) as happens between lines 4 and 5 of that proof. That is the error of it, and how you are able to arrive at that conclusion.

I really hope you didn't already know that or something!

K=2/(3-K)
J=2/(3-J)

K=2/(3-K) is: K=2/(3-(2/(3-(2/(3-...K)))))
and
J=2/(3-J) is: J=2/(3-(2/(3-(2/(3-...J)))))
2/(3-(2/(3-(2/(3-...K))))) looks similar to 2/(3-(2/(3-(2/(3-...J))))) when not writing the J or K at the 'end'
K looks like J but not necessarily is J

LOL

I guess he asks why, by the current mathematical commonly accepted axioms, is that possible.

Answer would be that it is not, and probably they missed something on the comprobation of 2=1

You=faggot

Don't say you're not because this reply says you are.

You're wrong user.

6root(64) doesn't equals 2

Equals 2 and -2

IF 6root(64) would equal 2 OR -2 do'd be right, but sadly for you, equals 2 AND -2

AND and OR are different things