Well, /sci/?

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

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

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Well, /sci/?

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>>7838630

1=/= sqrt 1

Dumbass

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>>7838630

1=/= sqrt 1

Dumbass

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>>7838630

1=/= sqrt 1

Dumbass

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>>7838630

1=/= sqrt 1

Dumbass

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>>7838649

sqrt(1) can be either 1 or -1.

It's not a bijective function.

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>>7838649

1 = 1^2

1 = +- 1

Well /sci?

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>>7838653

Ohhh look out OP, you've got a fucker who's taken some sort of elementary analysis! You're about to get wrecked!

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>>7838653

no. 1^(1/2) can be either 1 or -1. sqrt(1) is only 1

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>>7838653

No, √ yields the positive robot, not both, and is injective in R+

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Proof that sqrt(1) can be +/-1 just like any other positive integer

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>>7838659

But you retain the brackets through your whole proof, leaving you with |-1~=1 at the end.

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>>7838667

Why only injective? It should also be surjective.

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>>7838670

And that works, doesnt it?

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>>7838667

I've only ever seen and used sqrt and (1/2) exponent interchangeably.

Pretty sure you're wrong

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>>7838659

The correct definition of the square root is

√(x^2) = |x|

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>>7838686

\sqrt[3]{1} is certainly only 1

1^(1/3) in another matter, though.

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>>7838653

Finally a mathematically literate post.

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>>7838690

KEK

The root is a multiple valued function. Its outcome can be negative. Seems like a bad definition for abs( ), m80.

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>>7838675

god no, that would make it bijective.

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You guys still haven't found where the error lies. I thought better of you, /sci/

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>>7838675

No its not bijective

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>>7838630

CAN'T BREAK UP THE SQRT

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>>7838681

Yes, but it doesn't say -1=1

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>>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.

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>>7838719

You restricted the range to R+, Ya silly goose. Doesn't every non-negative real have a unique non-negative square root?

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>>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.

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1 is one solution to sqrt(1), one of two solutions, the first step is wrong

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> casually taking the log of negative values

OP confirmed for a sneaky faggot

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>>7838734

Yes, sqrt: R+ -> R+ is bijective.

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Why is /sci/ so bad at math? The third equality isn't justified for complex numbers.

>>7838728

This guy gets it.

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>>7838630

1 = -1

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>>7838630

sqrt(1) = +- 1

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Oh yeah? Well, let [math] n \in \mathbb{Z}^{+} [/math]

[math] \log { n } = \log { (1n) } = \log { 1 } + \log { n } =\log { (-1)^2 } + \log { n } =2 \log { (-1) } + \log { n } [/math]

[math]\therefore \log{(-1)}=0 [/math]

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>>7840035

Oops, [math] n \in \mathbb{R}^{+} [/math]

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>>7839995

You sure can split them up. You just can't take the solution to be over R+

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>>7838728

Sqrt 4 = sqrt 2 x sqrt 2 = +- 2

Sure can

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>>7838630

Squareroot of negative 1 is not defined.

In other words, you can't do that, you can't just split it out.

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>>7838728

winner

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>>7840086

It is defined as i

There is no problem doing this

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If we include imaginary numbers, then sqrt(1) = +/-1, similar to any other fucking positive integer.

Wow, so impressed, op, very profound.

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what is sqrt(i) ?

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>>7840099

sqrt(1/2) + sqrt(1/2) i.

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>>7840088

no it is not,

i^2 is defined as -1

i = sqrt(-1) is not defined.

Best regards

an engineer

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>>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

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>>7840111

But that's not exactly accurate, since |sqrt(-1)| |sqrt(-1)| = 1

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>>7840104

sqrt(2)x2?

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>>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.

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>>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

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>>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.

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>>7840951

When I divided (1) as (-1)*(-1), those two (-1) are not the same. One of them is [math]e^{i*pi}[/math] and the other one is [math]e^{-i*pi}[/math]

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