/sci/ - Science & Math

[Anonymous No.11288272]

refute this

[Anonymous No.11288280]

>>11288272
The fact that you assumed the number of powers of x are equal to the number of powers of y is retarded. Try harder next time

[Anonymous No.11288310]

Just because you put an equals sign in, doesn't make it true.

[Anonymous No.11288334]

Fourth square root of four is not equal to square root of 2

[Anonymous No.11288349]

>>11288280
>>11288310
>>11288334
[math]4^{1/4} = (2^2)^{1/4} = 2^{2/4} = 2^{1/2}[/math]

[Anonymous No.11288360]

>>11288349
Damn you got me there, please proceed

[Anonymous No.11288377]

>>11288272
You just showed that x^x^... is not equal to 2.

[Anonymous No.11288381]

You first have to say what you mean by x^x^x.... Since its not a common mathematical operation

[Anonymous No.11288383]

>>11288272
x^2=2
y^2=2
x=√2
y=-√2
x===y???!

[Anonymous No.11288391]

1. Assume something.
2. Figure out the consequences of that assumption.
3. Derive contradiction.

Gee, I guess the assumption that a solution exists is false.

[Anonymous No.11288395]

x^1/4 has more complex roots than x^1/2 so x and y aren't equal

[Anonymous No.11289181]

>>11288272
The equation has no solution.

[Anonymous No.11289322]

>>11288272
>x^x^x^x...
>...

Infinite powers, huh?

If x > 1 this turns into infinity, not 2 or 4.

If x < 1 then the result will be < 1 so not 2 or 4.

If x == 1 the result is 1, not 2 or 4.

There's no whole number we can substitute for x to make an infinite series of powers equal 2 or 4. If the input needs to be something other than an integer I'm not interested.

[Anonymous No.11289325]

>>11289322
P.S. I meant "real" number, not "whole" or integer.

[Anonymous No.11289389]

I have infinite balls. I decide to place all of them into Box A and Box B. I first place 10 balls into Box A, then I take one of those balls out of Box A and place it into Box B. So after the first dump, there are 9 balls in Box A and 1 ball in box B. I repeat the process until all of the balls are in the boxes. In the end, all of the balls will be in Box A. The balls were numbered, so I chose balls 1-10 for the first dump, placing ball 1 in box B, then in the next dumb, I placed ball 2 in Box B. So Box B contains 1,2,3... or every possible ball.

>tl;dr
Infinite sets are bullshit

[Anonymous No.11289392]

>>11289389
all of the balls will be in Box B*

[Anonymous No.11289400]

>>11289322
>There's no whole number we can substitute for x to make an infinite series of powers equal 2 or 4.

[Anonymous No.11289416]

[math] 1=\sqrt{1}=-1 [/math]. Now adding [math] 3 [/math] to each side gives [math] 4=2 [/math]

[Anonymous No.11289447]

>>11289416
square root is defined to be positive. The absolute value of a number is the same as the square root of its square

[Anonymous No.11289475]

Suppose x = 2, y = 4, and x = y.

2 = 4

QED

[Anonymous No.11289715]

>>11288272
You're assuming Infiniti = infuniti-1 and that either are numbers that have relevant consequence. I disagree

[Anonymous No.11289722]

>>11288272
>Refute this
Everything is ill defined you fucking retard.

[Anonymous No.11289728]

>>11289400
well what is it?

[Anonymous No.11289787]

>>11288272
Excuse me, but does not y^4=4=√4=2 and not 4^(1/4)? Or am I mistaken?

[Anonymous No.11289793]

>>11288272 (OP)
Excuse me, but does not y^4=4 imply y=√4 and not 4^(1/4)? Or am I mistaken?

[Anonymous No.11289803]

>>11289728
[math]\sqrt{2}[math]. The infinite power tower isn't a function, it has multiple values.

[Anonymous No.11289831]

>>11289793
huh?

[Anonymous No.11289903]

The power tower is 2 and 4 at sqrt of 2. It's a multivalued function.

[Anonymous No.11289924]

>>11289803
you're so dum

[Anonymous No.11290012]

>>11289903

How did you input the infinite power tower?

[Anonymous No.11290034]

>>11290012
just graph y=x^y

[Anonymous No.11290127]

>>11289389
It isn't possible to do something an infinite amount of times. The best we can do is take a limit. Then you have to carefully define what you mean by the limit set for box a.

Pic semi-related

[Anonymous No.11290338]

>>11288272
Tricky. But not a proof for [math]4 = 2[/math]. We've only shown that [math]x^{x^{x^{.^{.^{.}}}}}[/math] associates [math]\sqrt{2}[/math] to both 2 and 4.

[Anonymous No.11290392]

>>11288272
Dont mess with infinity unless you know what those 3 dots mean.

Dont do a numberphile-like trick.

[Anonymous No.11290634]

>let p = 2. Then let x^p = 2, so x = sqrt(2)
>let q = 4. Then let y^q = 4, so y = sqrt(2)
The infinite powers seem just to be a distraction. Not sure if this is supposed to be acceptable in a math theorem, infinite values or not, but it looks like the pic just goes equivocates x and then y, i.e. x is said to equal 2 and then the a new x to the power of the old x is said to equal 2. I guess this is supposed to create some kind of illusion that this follows from the original statement. I've changed the variables such that the meaning of the theorem is the same, but this equivocation is more apparent.

[Anonymous No.11290641]

>let p = 2. Then let x^p = 2, so x = sqrt(2)
>let q = 4. Then let y^q = 4, so y = sqrt(2)
The infinite powers seem just to be a distraction. Not sure if this is supposed to be acceptable in a math theorem, infinite values or not, but it looks like the pic just goes and equivocates x, and then y, i.e. x is said to equal 2 and then the a new x to the power of the old x is said to equal 2. I guess this is supposed to create some kind of illusion that this follows from the original statement. I've changed the variables such that the meaning of the theorem is the same, but this equivocation is more apparent.

[Anonymous No.11290676]

>>11290127
Wouldn't you end up with an octagon that way?

[Anonymous No.11290693]

>>11290676
no

[Anonymous No.11290716]

>>11290693
Every new fold can only be as long as the last fold, so yes.

[Anonymous No.11290822]

>>11290716
are you retarded

[Anonymous No.11290849]

>>11288272
Has anyone actually tried computing it, it seems to tend to 2. Not sure how you'd get 4 to make sense.

[Anonymous No.11290867]

>>11288272
>infinite anything
>expecting it to obey normal rules
kek, take analysis.

Fact: ...9999.0 = -1, ...9999.9999... = 0, 0.9999... = 1

[Anonymous No.11290868]

>>11290127
pi is 4 in the L1 metric.

[Anonymous No.11290877]

>>11288272
So there's two points of intersection y=x^y with x=√2, then it's true, but it isn't Identity.

[Anonymous No.11291121]

>>11289416
[math]4 = i + 3[/math]

[Anonymous No.11291698]

>>11290822
Yes, but what does that make you? Each new fold can only be as long as where the last fold ended. Therefore, octagon. It wouldn't be able to just magically keep folding to cover every nook of the circle.

[Anonymous No.11291702]

>>11290867
None of that is true, you idiot. unquestionably believing what you're told < thinking

[Anonymous No.11292604]

>>11291698

[Anonymous No.11292616]

>>11288349
fuck this

[Anonymous No.11292627]

>refute this
>4=2
refute THIS

[Anonymous No.11292711]

this is what happens when you dont learn math to understand it but instead memorize concepts by learning to shuffle letters around

[Anonymous No.11294494]

>>11288360
read again anon

[Anonymous No.11294499]

>>11292711
And shuffling letters around incorrectly does not help much either.

[Anonymous No.11294501]

>>11288272

[Anonymous No.11294511]

>>11288272
Sure, but first prove that the first step of your proof is possible

[Anonymous No.11294571]

>>11288272
x has no convergent value, you cant just assume that the power tower converges to a finite value.

[Anonymous No.11294591]

>>11294571
>power tower
zozzle

[Anonymous No.11294601]

>>11288272
When x or y is sqrt(2), the power tower sequence is monotonically increasing, and bounded from above by 2, so the sequence will never converge to 4.