- #1

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1 = [tex]\sqrt{}1[/tex] = [tex]\sqrt{}(-1) (-1)[/tex] =[tex]\sqrt{}-1[/tex] [tex]\sqrt{}-1[/tex] = i[tex]^{}2[/tex] = -1

so we get 1= -1. what is the error here?

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- Thread starter topu87
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- #1

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1 = [tex]\sqrt{}1[/tex] = [tex]\sqrt{}(-1) (-1)[/tex] =[tex]\sqrt{}-1[/tex] [tex]\sqrt{}-1[/tex] = i[tex]^{}2[/tex] = -1

so we get 1= -1. what is the error here?

- #2

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1= [tex]\sqrt{1}[/tex] = [tex]\sqrt{(-1) (-1)}[/tex]= [tex]\sqrt{-1}[/tex][tex]\sqrt{-1}[/tex] =

- #3

arildno

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It just doesn't.

- #4

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- #5

CRGreathouse

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[tex]\sqrt{ab}=\sqrt a\sqrt b[/tex] is only valid for [itex]a,b\ge0[/itex].

- #6

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On the basis of which statement you can say the following statement is valid-

[tex]\sqrt{ab}=\sqrt a\sqrt b[/tex] is only valid for [itex]a,b\ge0[/itex].

- #7

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- :[tex](\sqrt{ab})^2 = ab[/tex] by the definition of square root
- : [tex](\sqrt{a})^2 = a[/tex] by the definition of square root
- : [tex](\sqrt{b})^2 = b[/tex] by the definition of square root
- : [tex](\sqrt{ab})^2 = (\sqrt{a})^2 (\sqrt{b})^2 [/tex]
- : [tex](\sqrt{ab})^2 = (\sqrt{a}\sqrt{b})^2 [/tex] associativity
- : [tex]\sqrt{ab} = \sqrt{a}\sqrt{b} [/tex] definition of square root

the last step only works if [tex]\sqrt{ab}[/tex] and [tex]\sqrt{a}\sqrt{b}[/tex] must be positive. if this was not the case [tex]\sqrt{ab} = -\sqrt{a}\sqrt{b} [/tex] could also be possible

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