Proving 1 = -1: Help and Discussion

[IsotropicSpinManifol]

Ok
This is my first post HI everyone!

Whats wrong with this.

1 = sqrt(1) = sqrt (-1*-1) = sqrt(-1)*sqrt(-1) = i * i = i^2 = -1

therefore

1 = -1

0= -2,2

0 = Real Number Set

etc

Dammit I am right! And everyone in the history of maths is wrong!

AHHAHAHAA

ahum

 

[whozum]

Holy ****! You've gotten us!

You're mistake is in the second step, you need to consider the full definition, parameter-usage and application of the square root function before splitting it up like that.

 

[dextercioby]

Simply put:Sqrt always returns a real positive value.

$$\sqrt{ab}=\sqrt{|a|}\sqrt{|b|}$$ , $$ab\geq0$$.

Daniel.

 

[Galileo]

$\sqrt{-1 \cdot -1}\not=\sqrt{-1}\cdot \sqrt{-1}$.

 

[Zurtex]

Same thing but a bit more obvious, your step is like saying:

$$|1| = |-1|$$

Therefore:

$$1 = -1$$

 

[HallsofIvy]

What everyone else has said is true. And it is basically due to the fact that the complex numbers cannot be ordered. If we define i by "$$i= \sqrt{-1}$$ or even i²= -1, we cannot distinguish between "i" and "-i". (Since the complex numbers are not ordered, we can't say "the positive root" and "the negative root".)

More precise is to define the complex numbers as pairs of real numbers (a, b) and define addition by (a,b)+ (c,d)= (a+c, b+d) and multiplication by (a,b)(c,d)= (ac-bd, ad+ bc). (Then a+ bi is just a notation for (a,b).) Using that notation this problem disappears.

 

[toocool_sashi]

(ab)^1/2 = a^1/2*b^1/2 if and only if ATLEAST one of a, b is Non negative. If a and b are both non negative then (ab)^1/2 = - (a)^1/2*(b)1/2 ;) cheers!