[tex] \sqrt{x^2} = |x| [/tex] ??? Watch

lllllllllll
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#21
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#21
(Original post by morgan8002)
See http://www.thestudentroom.co.uk/show....php?t=3173187

A function cannot be multiple valued.

edit: Also, \sqrt{-1^2} = \sqrt{-1} = i.
what.... how does \sqrt{(-1^2)} = \sqrt{-1}
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Jammy Duel
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#22
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(Original post by Ano9901whichone)
Can you use your superior knowledge and explain my last post to me?
Having had 5 minutes to consider the matter getting ready for bed, the simplest way to consider the whole matter, supposing you have the correct tools, is to stop looking at it from the real line entirely and look at the problem in the complex plane, in particular using the complex exponential function and the problem goes away entirely
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Ano9901whichone
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#23
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#23
(Original post by Jammy Duel)
Having had 5 minutes to consider the matter getting ready for bed, the simplest way to consider the whole matter, supposing you have the correct tools, is to stop looking at it from the real line entirely and look at the problem in the complex plane, in particular using the complex exponential function and the problem goes away entirely
I wasn't attacking you or the other poster earlier, just in the name of banter.
Good night.
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SsEe
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#24
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#24
All you've discovered is that \sqrt{x^2} = {\sqrt{x}}^2 isn't always true.
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Notnek
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#25
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#25
(Original post by Ano9901whichone)
So we have  \sqrt{x^2}= |x| right?
'It makes sense try x=-1,' ok so  \sqrt{(-1)^2} = \sqrt 1 = 1 .
But wait a second let's try that again but square root first.
 \left ( \sqrt{-1} \right )^2 = i^2 = -1 . Is there some rule or convention that I do not know about.
If we consider exponentiation to be a single-valued function then index laws that hold for real numbers often don't hold for complex numbers.

E.g.

1 = \sqrt{-1 \times -1} = \sqrt{-1} \times \sqrt{-1} = -1.
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Notnek
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#26
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(Original post by Jammy Duel)
Except what he said was right, it is convention to take the positive root unless otherwise stated, but that does not mean the negative root does not exist as it is also a solution to the equation, in this case x^2=1
Ano9901whichone was correct.

Please read this.
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Ayman!
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#27
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#27
The imaginary number i by definition is \sqrt{-1} and thus cannot have two solutions.
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morgan8002
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#28
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#28
(Original post by lllllllllll)
what.... how does \sqrt{(-1^2)} = \sqrt{-1}
1^2= 1
-1^2=-(1^2)=-(1)=-1
\sqrt{(-1^2)}=\sqrt{-1^2}=\sqrt{-1}
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Zacken
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#29
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#29
(Original post by Jammy Duel)
...
You're the troll, in this case.
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Zacken
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#30
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#30
(Original post by Ano9901whichone)
But are they not the same?
 \sqrt{x^2}=(x^2)^{1/2} right?
 (\sqrt{x})^2 =(x^{1/2})^2
What's the difference?
\sqrt{ab} = \sqrt{a}\sqrt{b} only holds if a,b >0 and \left(x^{a}\right)^{1/b} = \left(x^{1/b}\right)^a only holds for x > 0.
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