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/sci/ - Science & Math

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10730895 No.10730895 [Reply] [Original]

What precisely is wrong with the following logic? What definitions am I violating?

i^2 = -1
i^4 = 1
(i^4)^0.5 = 1^0.5
i^2 = 1

>> No.10730900

>>10730895
i isn't real

>> No.10730906

>>10730895
1^0.5 = 1 or -1
Looks fine to me until the last line

>> No.10730909

>>10730906
Oh yeah you're right, can't believe I neglected that. Thanks

>> No.10730918

>>10730895
that image xD

>> No.10730920

>>10730906
Brainlet

>>10730895
(i^4)^0.5 =/= i^2

https://en.wikipedia.org/wiki/Branch_point

>> No.10730939

>>10730920
That looks pretty amazing desu

>> No.10730955

>>10730920
For all intents and purposes, it absolutely fucking does. I don't believe complex analysis is relevant to OP's homework question.

On the complex plane, the square root of an imaginary is equivalent to halving the angle from 0. In this case, theta can equal either 2π, or 4π (or an infinite trivial other solutions) and when 2π or 4π are divided by 2, π and 2π give -1 (i^2) and 1 (i^4), respectively.

>> No.10730958

>>10730955
>For all intents and purposes, it absolutely fucking does.
I can already tell I'm dealing with a special kind of retard so here you go:

https://www.wolframalpha.com/input/?i=(i%5E4)%5E0.5+%3D+i%5E2

>> No.10730964

>>10730958
Wolfram alpha will also tell you sqrt(4) = -2 is false. That does not mean in the context of a problem square rooting both sides an answer cannot be -2.

x^2=4, you're telling me the only solution is 2?

>> No.10730977

>>10730964
You claimed (i^4)^0.5 = i^2 "for all intents and purposes." Now you are admitting it's not and projecting your claim about all contexts onto me.

(i^4)^0.5 =/= i^2 when you are taking the positive branching path as OP does. It's standard in ambiguous situations to take the positive branching path of the square root, effectively turning the square root into a function.

>> No.10730980

>>10730895
bruh you gotta consider both roots

>> No.10730999

>>10730977
Here's a source that corroborates my claim:
http://mathforum.org/library/drmath/view/64430.html

>1^2 = -1
>1^4 = 1
>(i^4)^0.5 = 1^0.5
>i^2 = -1
OP took the square root of both sides, giving both i^2 = -1 and i^4 = 1, but not noticing that there are two different answers when taking a root.

The positive branching path wasn't taken, wolfram brainlet.

>> No.10731005

i and -i are interchangeable because the square root can be negative or positive. any equation with one can be switched to the conjugate and it will work out

>> No.10731034

>>10730999
>Here's a source that corroborates my claim:
Again you appear to be projecting, this disproves your claim that (i^4)^0.5 = i^2 in all contexts.

>The positive branching path wasn't taken
>1^0.5 = 1
Boy you sure are dumb.

>> No.10731041

>>10731034
I never claimed that. I claimed in this case, for all intents and purposes, we can neglect the "principle square root."

>> No.10731060

ITT: /sci/ begins to realize that [math]\sqrt{x^{2}}=|x|[/math]

>> No.10731687

>>10731041
Then 1^0.5 =/= 1

Instead of saying the answer is 1 or -1 OP should have chosen the same path for both sides and then the equality holds without caveats.

>> No.10731698

>>10730920
What's your problem? 1^4=1

>> No.10731899

>>10731698
And?

>> No.10731905

>>10730895
fundamental theorem of algebra.

>> No.10731912

>>10731060
wrong. sqrt(x * conj(x))