1 = -1, which part of this proof is wrong?

[Byeonggon Lee]

Of course 1 isn't same as -1.
This proof must be wrong but I can't find which part of this proof is wrong.
Could you help me with this problem?
(1)$$1 = \sqrt{1}$$
(2)$$= \sqrt{(-1)(-1)}$$
(3)$$= \sqrt{(-1)} \cdot i$$
(4)$$= i \cdot i$$
$$=-1$$

 

[symbolipoint]

Would you like better to start with $i^2+1=0$ ?

 

[Samy_A]

Of course 1 isn't same as -1.
This proof must be wrong but I can't find which part of this proof is wrong.
Could you help me with this problem?
(1)$$1 = \sqrt{1}$$
(2)$$= \sqrt{(-1)(-1)}$$
(3)$$= \sqrt{(-1)} \cdot i$$
(4)$$= i \cdot i$$
$$=-1$$

Read this insight article by @micromass : https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[HallsofIvy]

In the transition between the second line, $= \sqrt{(-1)(-1)}$, and the third line, $= \sqrt{-1}\cdot i$ you are assuming that $\sqrt{ab}= \sqrt{a}\cdot \sqrt{b}$. That is true for real numbers but not for general complex numbers.

 

[Deepak suwalka]

$1=\sqrt{1}$
$=\sqrt{(-1)(-1)}$
$=\sqrt{-1}\cdot\sqrt{-1}$
$=i^2$
$=-1$

 

[DrClaude]

$1=\sqrt{1}$
$=\sqrt{(-1)(-1)}$
$=\sqrt{-1}\cdot\sqrt{-1}$
$=i^2$
$=-1$

You are just repeating what others have pointed out to be wrong.

 

[LCKurtz]

In a thread that is 7 months old.

 

[phinds]

And for an OP who has not been here since posting the question.

 

[Deepak suwalka]

In a thread that is 7 months old.

sorry

 

[dkotschessaa]

sorry

It's ok.