/sci/ - Science & Math » Thread #11288272

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Anonymous Thu 09 Jan 06:03:01 2020 No.11288272 View Reply Original Report

Quoted By: >>11288280 >>11288377 >>11288383 >>11289181 >>11289322 >>11289715 >>11289722 >>11289787 >>11289793 >>11290338 >>11290392 >>11290849 >>11290867 >>11290877 >>11294501 >>11294511 >>11294571 >>11294601

refute this

Anonymous

Anonymous Thu 09 Jan 2020 06:08:39 No.11288280 Report

Quoted By: >>11288349

>>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

Anonymous Thu 09 Jan 2020 06:23:02 No.11288310 Report

Quoted By: >>11288349

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

Anonymous

Anonymous Thu 09 Jan 2020 06:37:41 No.11288334 Report

Quoted By: >>11288349

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

Anonymous

Anonymous Thu 09 Jan 2020 06:51:56 No.11288349 Report

Quoted By: >>11288360 >>11292616

>>11288280
>>11288310
>>11288334
$4^{1/4} = (2^2)^{1/4} = 2^{2/4} = 2^{1/2}$

Anonymous

Anonymous Thu 09 Jan 2020 06:59:28 No.11288360 Report

Quoted By: >>11294494

>>11288349
Damn you got me there, please proceed

Anonymous

Anonymous Thu 09 Jan 2020 07:11:38 No.11288377 Report

Quoted By:

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

Anonymous

Anonymous Thu 09 Jan 2020 07:13:10 No.11288381 Report

Quoted By:

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

Anonymous

Anonymous Thu 09 Jan 2020 07:13:57 No.11288383 Report

Quoted By:

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

Anonymous

Anonymous Thu 09 Jan 2020 07:18:57 No.11288391 Report

Quoted By:

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

Anonymous Thu 09 Jan 2020 07:22:41 No.11288395 Report

Quoted By:

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

Anonymous

Anonymous Thu 09 Jan 2020 17:38:06 No.11289181 Report

Quoted By:

>>11288272
The equation has no solution.

Anonymous

Anonymous Thu 09 Jan 2020 19:03:35 No.11289322 Report

Quoted By: >>11289325 >>11289400

>>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

Anonymous Thu 09 Jan 2020 19:04:52 No.11289325 Report

Quoted By:

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

Anonymous

Anonymous Thu 09 Jan 2020 19:41:24 No.11289389 Report

Quoted By: >>11289392 >>11290127

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

Anonymous Thu 09 Jan 2020 19:42:31 No.11289392 Report

Quoted By:

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

Anonymous

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Anonymous Thu 09 Jan 2020 19:45:05 No.11289400 Report

Quoted By: >>11289728

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

Anonymous

Anonymous Thu 09 Jan 2020 19:53:21 No.11289416 Report

Quoted By: >>11289447 >>11291121

$1=\sqrt{1}=-1$ . Now adding $3$ to each side gives $4=2$

Anonymous

Anonymous Thu 09 Jan 2020 20:07:37 No.11289447 Report

Quoted By:

>>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

Anonymous Thu 09 Jan 2020 20:20:03 No.11289475 Report

Quoted By:

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

2 = 4

QED

Anonymous

Anonymous Thu 09 Jan 2020 22:17:24 No.11289715 Report

Quoted By:

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

Anonymous

Anonymous Thu 09 Jan 2020 22:22:51 No.11289722 Report

Quoted By:

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

Anonymous

Anonymous Thu 09 Jan 2020 22:26:00 No.11289728 Report

Quoted By: >>11289803

>>11289400
well what is it?

Anonymous

Anonymous Thu 09 Jan 2020 22:48:54 No.11289787 Report

Quoted By:

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

Anonymous

Anonymous Thu 09 Jan 2020 22:51:48 No.11289793 Report

Quoted By: >>11289831

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

Anonymous

Anonymous Thu 09 Jan 2020 22:54:52 No.11289803 Report

Quoted By: >>11289924

>>11289728
$\sqrt{2}<script type="math/tex">. The infinite power tower isn't a function, it has multiple values.$

Anonymous

Anonymous Thu 09 Jan 2020 23:06:18 No.11289831 Report

Quoted By:

>>11289793
huh?

Anonymous

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Anonymous Thu 09 Jan 2020 23:50:03 No.11289903 Report

Quoted By: >>11290012

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

Anonymous

Anonymous Fri 10 Jan 2020 00:05:20 No.11289924 Report

Quoted By:

>>11289803
you're so dum

Anonymous

Anonymous Fri 10 Jan 2020 00:38:12 No.11290012 Report

Quoted By: >>11290034

>>11289903

How did you input the infinite power tower?

Anonymous

Anonymous Fri 10 Jan 2020 00:46:49 No.11290034 Report

Quoted By:

>>11290012
just graph y=x^y

Anonymous

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Anonymous Fri 10 Jan 2020 01:26:03 No.11290127 Report

Quoted By: >>11290676 >>11290868

>>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

Anonymous Fri 10 Jan 2020 02:52:33 No.11290338 Report

Quoted By:

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

Anonymous

Anonymous Fri 10 Jan 2020 03:18:17 No.11290392 Report

Quoted By:

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

Dont do a numberphile-like trick.

Anonymous

Anonymous Fri 10 Jan 2020 05:54:48 No.11290634 Report

Quoted By:

>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

Anonymous Fri 10 Jan 2020 05:59:50 No.11290641 Report

Quoted By:

>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

Anonymous Fri 10 Jan 2020 06:17:34 No.11290676 Report

Quoted By: >>11290693

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

Anonymous

Anonymous Fri 10 Jan 2020 06:25:11 No.11290693 Report

Quoted By: >>11290716

>>11290676
no

Anonymous

Anonymous Fri 10 Jan 2020 06:38:48 No.11290716 Report

Quoted By: >>11290822

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

Anonymous

Anonymous Fri 10 Jan 2020 07:51:05 No.11290822 Report

Quoted By: >>11291698

>>11290716
are you retarded

Anonymous

Anonymous Fri 10 Jan 2020 08:24:35 No.11290849 Report

Quoted By:

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

Anonymous

Anonymous Fri 10 Jan 2020 08:44:26 No.11290867 Report

Quoted By: >>11291702

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

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

Anonymous

Anonymous Fri 10 Jan 2020 08:45:27 No.11290868 Report

Quoted By:

>>11290127
pi is 4 in the L1 metric.

Anonymous

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Anonymous Fri 10 Jan 2020 09:02:24 No.11290877 Report

Quoted By:

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

Anonymous

Anonymous Fri 10 Jan 2020 12:40:02 No.11291121 Report

Quoted By:

>>11289416
$4 = i + 3$

Anonymous

Anonymous Fri 10 Jan 2020 17:31:14 No.11291698 Report

Quoted By: >>11292604

>>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

Anonymous Fri 10 Jan 2020 17:33:40 No.11291702 Report

Quoted By:

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

Anonymous

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Anonymous Sat 11 Jan 2020 00:18:31 No.11292604 Report

Quoted By:

>>11291698

Anonymous

Anonymous Sat 11 Jan 2020 00:26:21 No.11292616 Report

Quoted By:

>>11288349
fuck this

Anonymous

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Anonymous Sat 11 Jan 2020 00:31:58 No.11292627 Report

Quoted By:

>refute this
>4=2
refute THIS

Anonymous

Anonymous Sat 11 Jan 2020 01:07:20 No.11292711 Report

Quoted By: >>11294499

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

Anonymous

Anonymous Sat 11 Jan 2020 19:42:29 No.11294494 Report

Quoted By:

>>11288360
read again anon

Anonymous

Anonymous Sat 11 Jan 2020 19:45:29 No.11294499 Report

Quoted By:

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

Anonymous

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Anonymous Sat 11 Jan 2020 19:48:37 No.11294501 Report

Quoted By:

>>11288272

Anonymous

Anonymous Sat 11 Jan 2020 19:51:53 No.11294511 Report

Quoted By:

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

Anonymous

Anonymous Sat 11 Jan 2020 20:31:41 No.11294571 Report

Quoted By: >>11294591

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

Anonymous

Anonymous Sat 11 Jan 2020 20:40:40 No.11294591 Report

Quoted By:

>>11294571
>power tower
zozzle

Anonymous

Anonymous Sat 11 Jan 2020 20:44:53 No.11294601 Report

Quoted By:

>>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.

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