[bonanova]

Don't blink or you'll miss it.

1.  sqrt[-1] = sqrt[-1]


2.  sqrt[-1/1] = sqrt[1/-1]


3.  sqrt[-1]/sqrt[1] = sqrt[1]/sqrt[-1]


4.  sqrt[-1] x sqrt[-1] = sqrt[1] x sqrt[1]


5.  -1 = 1



[/code]

You know the drill; prove me wrong.


[/font]

[Guest]

sqrt[-1]/sqrt[1] = i

sqrt[1]/sqrt[-1] = -i

3. i = -i is false

5. -1 = 1 is therefore false

[Guest]

sqrt[-1]/sqrt[1] = i

sqrt[1]/sqrt[-1] = -i

3. i = -i is false

5. -1 = 1 is therefore false

It is wrong, because sqrt[1]/sqrt[-1]=1/i, not -i!!!

Edited November 6, 2010 by Balint Seregelyesi

[Guest]

In the set of C (complex numbers) the sqrt(1)=1,-i.

If you use -i as a root of 1 it will be clear that -1=-1, so the last step is wrong.

[bonanova]

sqrt[1]/sqrt[-1]=1/i, not -i!!!

Actually, they're the same.

Let x = 1/i

Multiply both sides by -i², which is 1.

x = -i

[bonanova]

In the set of C (complex numbers) the sqrt(1)=1,-i.

Not quite.

sqrt[1] = 1 or -1. It's always real.

Check your answer by squaring it. [-i]² is not 1.

[bonanova]

sqrt[-1]/sqrt[1] = i

sqrt[1]/sqrt[-1] = -i

3. i = -i is false

5. -1 = 1 is therefore false

If 3 is false, was it wrongly derived from 2?

Or is 2 false also?

Of the five statements, which is the last true one?

[Guest]

The whole equation is wrong becouse none negative nubmer has a square root and it is easily proofable:

Square root give us a number that if multiplied by itself, give us the number under the square root symbol

so if the number we are looking for is possitive: + * + = +

If the number is negative: - * - = +

Conclusion:Negative numbers don't have square roots sqrt[-1] = false

[Guest]

3 was wrongly derived from 2.

2. sqrt[-1/1] = sqrt[1/-1]

3. sqrt[-1]/sqrt[1] = sqrt[1]/sqrt[-1]

therefore, sqrt[1/-1] = sqrt[1]/sqrt[-1], but if you run the equation you get i = 1/i. In other words...

2. i = i

3. i = 1/i

then you can bring 3 to i*i = 1 and then -1 = 1.

As a side note to Tsopi...sqrt[-1] is the imaginary number i. The squareroot of any negative number can be represent with use of the number i.

[superprismatic]

Don't blink or you'll miss it.

1.  sqrt[-1] = sqrt[-1]


2.  sqrt[-1/1] = sqrt[1/-1]


3.  sqrt[-1]/sqrt[1] = sqrt[1]/sqrt[-1]


4.  sqrt[-1] x sqrt[-1] = sqrt[1] x sqrt[1]


5.  -1 = 1



[/code]


You know the drill; prove me wrong.



[/font]



I have a shorter version (don't even microblink):

[code]


1.  sqrt[-1] = sqrt[-1]


2.         i = -i

[araver]

1st and 2nd equations are valid as division is done before square rooting.

The rule used to get the 3rd equation is "broken". The rule only applies to positive real numbers (and except division by zero).

I have a shorter version (don't even microblink):

1.  sqrt[-1] = sqrt[-1]


2.         i = -i

Nice microblink! Since we're breaking rules, it could get even more "real" than that, no need to be "complex" about it

1.  sqrt[1] = sqrt[1]


2.         1 = -1

[Guest]

Don't blink or you'll miss it.

1.  sqrt[-1] = sqrt[-1]


2.  sqrt[-1/1] = sqrt[1/-1]


3.  sqrt[-1]/sqrt[1] = sqrt[1]/sqrt[-1]


4.  sqrt[-1] x sqrt[-1] = sqrt[1] x sqrt[1]


5.  -1 = 1

You know the drill; prove me wrong.

4-5

sqrt[1]=+-1; sqrt[-1]=+-i.

4.1 [+-i] x [+-i] = [+-1] x [+-1]

4.2 +-1 = +-1