/sci/ - Science & Math » Thread #11296228

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Square Root of What number is -1?

Anonymous Sun 12 Jan 13:42:42 2020 No.11296228 View Reply Original Report

Quoted By: >>11296241 >>11296382 >>11296383 >>11296384 >>11296419 >>11296786 >>11297728

Well....

Anonymous

Anonymous Sun 12 Jan 2020 13:47:51 No.11296241 Report

Quoted By: >>11296365

>>11296228
No number. The root function is always defined for the positive range. -1=-root1.
This is also true for the imaginary. The root of minus 1 is i, not minus i

Anonymous

Anonymous Sun 12 Jan 2020 15:08:32 No.11296365 Report

Quoted By: >>11296442

>>11296241
Except that in order to make roots functions, they are undefined on negative numbers. Root of -1 is undefined, but I squared is -1.

Anonymous

Anonymous Sun 12 Jan 2020 15:29:45 No.11296382 Report

Quoted By: >>11296444

>>11296228
x = (-1)^2

Anonymous

Anonymous Sun 12 Jan 2020 15:30:18 No.11296383 Report

Quoted By: >>11296444

>>11296228
square root of one is both 1 and -1

Anonymous

Anonymous Sun 12 Jan 2020 15:31:41 No.11296384 Report

Quoted By: >>11296821

>>11296228
Define $\sqrt{ \cdot }$ .

Anonymous

Anonymous Sun 12 Jan 2020 16:01:57 No.11296419 Report

Quoted By:

>>11296228
U S E
P
H
A
S
O
R
S

Anonymous

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Anonymous Sun 12 Jan 2020 16:03:04 No.11296421 Report

Quoted By: >>11296791

Anonymous

Anonymous Sun 12 Jan 2020 16:10:51 No.11296442 Report

Quoted By: >>11296549

>>11296365
Thata what i said, but ok

Anonymous

Anonymous Sun 12 Jan 2020 16:12:05 No.11296444 Report

Quoted By:

>>11296382
>>11296383
This. $x = 1$ .

Anonymous

Anonymous Sun 12 Jan 2020 17:30:39 No.11296549 Report

Quoted By:

>>11296442
You literally said the root of -1 is i. That contradicts my statement.

Anonymous

Anonymous Sun 12 Jan 2020 20:01:12 No.11296786 Report

Quoted By:

>>11296228
I love you anime math poster. Please never leave this board

Anonymous

Anonymous Sun 12 Jan 2020 20:02:44 No.11296791 Report

Quoted By: >>11296843

>>11296421
I hate you pointless gif poster.
Please leave this board.

Anonymous

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Anonymous Sun 12 Jan 2020 20:07:52 No.11296806 Report

Quoted By:

The answer is "Kurisu is best girl"

Anonymous

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Anonymous Sun 12 Jan 2020 20:10:48 No.11296813 Report

Quoted By:

Nigga i don't exist!

Anonymous

Anonymous Sun 12 Jan 2020 20:14:11 No.11296816 Report

Quoted By: >>11296823

it´s x= i

Anonymous

Anonymous Sun 12 Jan 2020 20:17:48 No.11296821 Report

Quoted By: >>11296847

>>11296384
OP here,

$\sqrt{x}$ is the answer to this question:

Given a number $x$ , what number $y$ , when multiplied by itself, gives a product $x$ .

Anonymous

Anonymous Sun 12 Jan 2020 20:20:19 No.11296823 Report

Quoted By:

>>11296816
$\sqrt{i} = \frac{1}{\sqrt{2}} (1 + i)$

Anonymous

Anonymous Sun 12 Jan 2020 20:36:03 No.11296843 Report

Quoted By:

>>11296791
Ignore this fag

Anonymous

Anonymous Sun 12 Jan 2020 20:37:28 No.11296847 Report

Quoted By: >>11296855 >>11296861 >>11296898

>>11296821
That's not a well-defined operation then. You need to define it properly. What's the domain and co-domain?

Anonymous

Anonymous Sun 12 Jan 2020 20:40:53 No.11296855 Report

Quoted By:

>>11296847
>domain
-1 nigga he just toldyou in the OP

Anonymous

Anonymous Sun 12 Jan 2020 20:44:08 No.11296861 Report

Quoted By: >>11296911

>>11296847
OP here,

yeah. I see your point.

Anonymous

Anonymous Sun 12 Jan 2020 20:53:17 No.11296880 Report

Quoted By: >>11296911 >>11297202

OP here,

After thinking about it I just realized how fucking brainlet and dunning-kruger I actually am. Very sorry for this thread, hopefully the mods delete it soon.

Anonymous

Anonymous Sun 12 Jan 2020 21:01:35 No.11296898 Report

Quoted By: >>11296911

>>11296847
Holy shit this is actually quite good counterargument, and I havent seen this one before. Thanks anon.

Anonymous

Anonymous Sun 12 Jan 2020 21:07:05 No.11296911 Report

Quoted By:

>>11296861
>>11296880
>>11296898
samefag

Anonymous

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Anonymous Sun 12 Jan 2020 21:08:07 No.11296917 Report

Quoted By:

@11296911

Anonymous

Anonymous Sun 12 Jan 2020 21:25:07 No.11296959 Report

Quoted By:

The problem is that the square root operator is only defined on the first branch cut

$\sqrt{e^{i\theta}} = e^{\frac{i\theta}{2}}$
which means
$\frac{i\theta}{2} = \pi$
therefore $\theta = -2i\pi$

It looks like the answer here is just going to be "1", since $e^{-2i\pi} = 1[\math], but notice that we are on the next branch cut here, so it doesn't work.$

Anonymous

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Anonymous Sun 12 Jan 2020 21:39:24 No.11296997 Report

Quoted By:

Anonymous

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Anonymous Sun 12 Jan 2020 22:49:39 No.11297202 Report

Quoted By:

>>11296880
Ironically, I (the real OP) actually agree with that.

Anonymous

Anonymous Mon 13 Jan 2020 02:11:16 No.11297728 Report

Quoted By: >>11297732 >>11297736 >>11297741

>>11296228
isn't it just i^4 aka 1 or am I a brainlet

Anonymous

Anonymous Mon 13 Jan 2020 02:12:37 No.11297732 Report

Quoted By:

>>11297728
yes, $\sqrt{1} = \pom 1$

Anonymous

Anonymous Mon 13 Jan 2020 02:13:58 No.11297736 Report

Quoted By:

>>11297728
yes, \sqrt{1} = \pm 1

Anonymous

Anonymous Mon 13 Jan 2020 02:15:07 No.11297741 Report

Quoted By:

>>11297728
yes, $\sqrt{1} = \pm 1$

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