What is this infinite fraction equal to? Is it undefined? You should be able to solve this

To do this rigorously, you need to project onto a triadic semi-complete Chomper space, then perform the universal recomplication in the obvious manner to reach the result of unity.

Theta = 1/(0+Theta)

Theta = 1/Theta

Theta^2 = 1

Theta = 1

If you guys didn't get this you're retarded and should probably go back to /b/

good meme

So theta can also be -1?

Let's say
x = 1
= 1 / 1
= 1 / x
so x^2 = 1
so x = -1 or 1

this is how dumb you are lol

Theta = 1/ (0+1/(0+theta)) = 1/(1/theta)=theta

Theta has infinite solutions.

You're retarded

Another thread called me a retard too. Guess I am a retard.

that shit is childish

x = 1 / (0 + x)
x = 1

Well that's pretty honest from you.

brainlet

for people that actually don't know how to do this

[math]\theta = \frac{1}{0+\frac{1}{0+\frac{1}{0+\frac{1}{0+...}}}}
=\frac{1}{\frac{1}{\frac{1}{\frac{1}{...}}}}= \frac{1}{\theta}[/math]
note that:
[math]\theta > 0[/math]
we obtain:
[math]\theta = \frac{1}{\theta} \Leftrightarrow \theta^2 = 1 \Leftrightarrow \theta = 1 \quad \quad \quad \quad \square[/math]

qed

en.wikipedia.org/wiki/Continued_fraction

Mind blown

Rewrite it as [math] \theta = \frac{1}{\theta} [/math] and it's clear that this has two solutions, 1 and -1. Pretty interesting.

>θ2=1⇔θ=1
"no"

1/infinity

thus 1 / (-1/12)

turns into

1 * (-12/1)

thus, -12

QED

i don't know what is the biggest bait: the original post or -1/12 replies

Kek.

Let [math]\theta \,\in\, \mathbf R[/math].
[eqn]\theta \,=\, \frac{1}{0 \,+\, \theta} \,=\, \frac{1}{\theta} \quad\text{iff}\quad \theta^2 \,=\, 1 \quad\text{iff}\quad \theta \,\in\, \left\{-1,\,1 \right\}[/eqn]
In other words, your shit is undefined. As always, OP is a faggot.

how do you assign a value to this shit when all of the truncated fractions are 1/0

...

x = 1/(0 + x)
x = 1/x iff
x^2 = 1 iff
x = 1 or x = -1

Consider the partial fractions to discard -1 as an option.

x=1

No, you fucking pleblet.

Don't tell me none of you "high IQ people" ever saw infinite fractions in number theory? It is a trivial procedure to compute the partial fractions and see why -1 cannot be a solution.

\square

Please do compute the partial fractions.

VERY IMPORTANT POST

Everyone who said the answer is 1 or -1 is wrong. What you have shown is that IF AN ANSWER EXISTS, then it must be 1 or -1.

However, every infinite fraction is irrational (google it). This means that theta is not well-defined (because if it were defined, it would have to be 1 or -1, a contradiction).

Alternatively, an infinite fraction is BY DEFINITION the limit of the sequence of finite continued fractions obtained by truncating after n levels. Since each such truncation has a zero in the denominator, every element of this sequence is undefined, so the limit is also undefined.

no he's not. He's wrong, but not any more wrong than the people who say the answer is 1.

>infinite fraction
Is that right?

kind of lol at 1 or -1

it will oscillate as x = 1/x for x not = 1

Thank god. I was getting to the end of the thread and seriously feared no one had given a decent answer.

s1 = 1
s2 = 1
s3 = 1

If you want to be autistic then you can get the weird looking general form:

sn = 1/(0 + 1)^(n-1)

This looks weird but you need to re arrange

0 + 1/x = (0 + 1)/x

And then take the limit as n approaches infinity which is 1/1^infinity = 1

That algebraic solution you've maybe seen in numberphile videos is a good tool to find possible answers. Maybe you have a series of fractions that you can prove converges but cannot truly find what it converges to by taking the limit, so you prove it converges and use the algebraic method to find possible solutions. But alone it does not say anything.

It, if anything, says that if your infinite fraction converges, then it must converge to one of those values.

>every infinite fraction is irrational
so undefined is an irrational number now?

given that θ > 0 the statement is valid.

brainlet

wrong
right