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Well, /sci/?
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You are currently reading a thread in /sci/ - Science & Math

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Well, /sci/?
>>
>>7838630
1=/= sqrt 1

Dumbass
>>
>>7838630
1=/= sqrt 1

Dumbass
>>
>>7838630
1=/= sqrt 1

Dumbass
>>
>>7838642
>>7838645
>>7838647
>"1=/= sqrt 1"
Except it is.
>>
>>7838630
1=/= sqrt 1

Dumbass
>>
>>7838649
sqrt(1) can be either 1 or -1.
It's not a bijective function.
>>
>>7838642
>>7838645
>>7838647
Fuck off faggot
>>
>>7838651
>>7838647
>>7838645
>>7838642
I guess 1=|√1|
>>
>>7838649
1 = 1^2

1 = +- 1

Well /sci?
>>
>>7838653
Ohhh look out OP, you've got a fucker who's taken some sort of elementary analysis! You're about to get wrecked!
>>
>>7838653
no. 1^(1/2) can be either 1 or -1. sqrt(1) is only 1
>>
>>7838653
No, √ yields the positive robot, not both, and is injective in R+
>>
Proof that sqrt(1) can be +/-1 just like any other positive integer
>>
>>7838659
But you retain the brackets through your whole proof, leaving you with |-1~=1 at the end.
>>
>>7838667
Why only injective? It should also be surjective.
>>
>>7838670
And that works, doesnt it?
>>
>>7838667
I've only ever seen and used sqrt and (1/2) exponent interchangeably.

Pretty sure you're wrong
>>
>>7838659
The correct definition of the square root is
√(x^2) = |x|
>>
>>7838686
\sqrt[3]{1} is certainly only 1
1^(1/3) in another matter, though.
>>
>>7838653
Finally a mathematically literate post.
>>
>>7838690
KEK
The root is a multiple valued function. Its outcome can be negative. Seems like a bad definition for abs( ), m80.
>>
>>7838675
god no, that would make it bijective.
>>
You guys still haven't found where the error lies. I thought better of you, /sci/
>>
>>7838675
No its not bijective
>>
>>7838630
CAN'T BREAK UP THE SQRT
>>
>>7838681
Yes, but it doesn't say -1=1
>>
>>7838713
>The root is a multiple valued function.
Just tell where I went off in a different branch of the root other than the main one in the OP, then.
>>
>>7838719
You restricted the range to R+, Ya silly goose. Doesn't every non-negative real have a unique non-negative square root?
>>
>>7838713
>multiple valued
>function
Bruh a function is a triple, (M,N,G) such that
-G is a subset of MXN
-if (X1,Y) and (X2,Y) are members of G, then X1=X2.
>>
1 is one solution to sqrt(1), one of two solutions, the first step is wrong
>>
> casually taking the log of negative values
OP confirmed for a sneaky faggot
>>
>>7838734
Yes, sqrt: R+ -> R+ is bijective.
>>
Why is /sci/ so bad at math? The third equality isn't justified for complex numbers.

>>7838728
This guy gets it.
>>
>>7838630
1 = -1
>>
>>7838630
sqrt(1) = +- 1
>>
Oh yeah? Well, let $n \in \mathbb{Z}^{+}$
$\log { n } = \log { (1n) } = \log { 1 } + \log { n } =\log { (-1)^2 } + \log { n } =2 \log { (-1) } + \log { n }$
$\therefore \log{(-1)}=0$
>>
>>7840035
Oops, $n \in \mathbb{R}^{+}$
>>
>>7839995
You sure can split them up. You just can't take the solution to be over R+
>>
>>7838728
Sqrt 4 = sqrt 2 x sqrt 2 = +- 2

Sure can
>>
>>7838630
Squareroot of negative 1 is not defined.
In other words, you can't do that, you can't just split it out.
>>
>>7838728
winner
>>
>>7840086
It is defined as i

There is no problem doing this
>>
If we include imaginary numbers, then sqrt(1) = +/-1, similar to any other fucking positive integer.

Wow, so impressed, op, very profound.
>>
what is sqrt(i) ?
>>
>>7840099
sqrt(1/2) + sqrt(1/2) i.
>>
>>7840088
no it is not,

i^2 is defined as -1
i = sqrt(-1) is not defined.

Best regards
an engineer
>>
>>7840104
Me again,
i looked it up becuase I wanted to see if my examinator was an ass when he failed my exam during first year of undergrad.

Look what i found, the fucking pic in OP and it proved that I was right and that my examinator was not an ass.

Go back to your fucking pleb universites
Europe strong
>>
>>7840111
But that's not exactly accurate, since |sqrt(-1)| |sqrt(-1)| = 1
>>
>>7840104
sqrt(2)x2?
>>
>>7838630
The third equals sign is invalid. Square roots behave differently with imginary numbers.. Use the principle root argument and you will see. This was a question on my complex analysis assignment.
>>
>>7838685
Yes, √ and ^(1/2) are interchangeable
No, they cannot yield both positive and negative results, it's just something functions/operations dont do.
Result of √1 =/= solution of x^2=1
>>
>>7838630
I'm guessing the third equality is the problem but I don't know why. The fact that I've already passed a complex analysis course makes me feel stupid.
>>
>>7840171
>>7840951
Isn't it because $\sqrt{1}=\pm1$? So the first and last equality are incorrect.
>>
>>7840951
When I divided (1) as (-1)*(-1), those two (-1) are not the same. One of them is $e^{i*pi}$ and the other one is $e^{-i*pi}$