Maths stuff

[Stuff No.6855807]

√(x)=±(x)
〖x〗^2=〖(-x)〗^2
∴x=(-x)???

How on earth does this work? 1 != -1 obviously but according to the relatively low level maths I am basing this on it does. Does anyone care to explain this to me.

[Anonymous No.6855820]

>>6855807
>√(x)=±(x)
Just die.

[Anonymous No.6855829]

principal square root, look it up dipshit.

[Anonymous No.6855838]

>>6855807
>Does anyone care to explain this to me.
>√(x)=±(x)
is this an equation? it's not an identity as one side is rooted and the other isn't. apart from the ± error

[Anonymous No.6855839]

>>6855829
1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1)*sqrt(-1) = -1

[Anonymous No.6855856]

>>6855839
http://en.wikipedia.org/wiki/Imaginary_unit#Proper_use

[Anonymous No.6855860]

>>6855856
>pointing out to me the point I was making

Can you be a little less faggy?

[Anonymous No.6855871]

>>6855839
>sqrt(-1 * -1) = sqrt(-1)*sqrt(-1)
No. If you are considering square roots of negative reals, then you are working within the complex number set, and <span class="math">\sqrt{ab}=\sqrt{a} \cdot \sqrt{b}[/spoiler] is NOT an identity there.

http://www.wolframalpha.com/input/?i=Sqrt[%28-1%29*%28-1%29]%3D%3DSqrt[-1]*Sqrt[-1]

[Anonymous No.6855886]

>>6855860
Given that the topic of the radical symbol and roots in general seems to have recently become a favorite of

- the badly educated (but convinced otherwise),
- the immature "I'll be contrarian towards definitions or conventions for the sake of being contrarian, because it is le edgy" dipshits, and
- the generic trolls,

it is difficult to tell what your intentions with such a post were, and you need to expect sarcasm not being recognized as such.

[Anonymous No.6855890]

>>6855829
This has nothing to do with the principal root and everything to do with >>6855871.

It's not an identity anymore.

[Anonymous No.6855903]

No, OP. √(x)=x. If you say that x2 = 4, that means that x is either 2 or -2.

[Anonymous No.6855904]

>>6855903
Shit, I meant √(x2)=x

[Anonymous No.6855907]

>>6855903
>>6855904
<span class="math">\sqrt{x^2} = |x| = (x \land x \geq 0) \lor (-x \land x < 0)[/spoiler].

<span class="math">\sqrt{x} < 0 \Leftrightarrow x \in \emptyset[/spoiler]

[Anonymous No.6855910]

>>6855890
It's not an identity because you are using principle roots. If you used solution sets the identity still works

[Anonymous No.6855912]

>>6855871
My post was made to point this out you fucking retard.

[Anonymous No.6855913]

Fuck this thread, it's just >>6852019 all over again.

Soon enough the three groups mentioned in >>6855886 will arrive and commence their usual storm of bullshit, insults and stupidity.

[Anonymous No.6855919]

>>6855912
You obviously haven't read >>6855886, have you?

[Anonymous No.6855991]

>>6855807
>How on earth does this work?

See >>6854657 and >>6855907.

[Anonymous No.6856115]

>>6855807
>√(x)=±(x)
No
>〖x〗^2=〖(-x)〗^2
Yes
>x=(-x)
No