What is the error in this proof?

[topu87]

please tell me where is the error here-
1 = $$\sqrt{}1$$ = $$\sqrt{}(-1) (-1)$$ =$$\sqrt{}-1$$ $$\sqrt{}-1$$ = i$$^{}2$$ = -1

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

 

[topu87]

sorry look at this form. This is more understandable than the previous one
1= $$\sqrt{1}$$ = $$\sqrt{(-1) (-1)}$$= $$\sqrt{-1}$$$$\sqrt{-1}$$ = i i = i$$^{2}$$ = -1

 

[arildno]

The error lies in the assumption that the square root operation when performed on negative (or in general complex) numbers follow exactly those laws valid when you restrict square rooting to positive numbers.

It just doesn't.

 

[mdnazmulh]

Did you want to say that we can't apply the laws for square-rooting of positive no. to this situation? Plz explain details. What is the exact logic to give in this situation?

 

[CRGreathouse]

Did you want to say that we can't apply the laws for square-rooting of positive no. to this situation? Plz explain details. What is the exact logic to give in this situation?

$$\sqrt{ab}=\sqrt a\sqrt b$$ is only valid for $a,b\ge0$.

 

[mdnazmulh]

Again it raises another question.
On the basis of which statement you can say the following statement is valid-

$$\sqrt{ab}=\sqrt a\sqrt b$$ is only valid for $a,b\ge0$.

 

[kamerling]

It's valid for a,b>=0 because there is a valid proof of it. this proof doesn't work if a or b can be negative. this proof uses as a definition of the square root that x is the square root of y, if x is positive and x*x = y. in that case there can be only one such number

1. :$$(\sqrt{ab})^2 = ab$$ by the definition of square root
2. : $$(\sqrt{a})^2 = a$$ by the definition of square root
3. : $$(\sqrt{b})^2 = b$$ by the definition of square root
4. : $$(\sqrt{ab})^2 = (\sqrt{a})^2 (\sqrt{b})^2$$
5. : $$(\sqrt{ab})^2 = (\sqrt{a}\sqrt{b})^2$$ associativity
6. : $$\sqrt{ab} = \sqrt{a}\sqrt{b}$$ definition of square root

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