View Full Version : 1 = -1
IsotropicSpinManifol
Jun17-05, 02:15 AM
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 im right! And everyone in the history of maths is wrong!
AHHAHAHAA
ahum
Holy ****! You've gotten us! :rolleyes:
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
Jun17-05, 03:03 AM
Simply put:Sqrt always returns a real positive value.
\sqrt{ab}=\sqrt{|a|}\sqrt{|b|} , ab\geq0.
Daniel.
Galileo
Jun17-05, 04:24 AM
\sqrt{-1 \cdot -1}\not=\sqrt{-1}\cdot \sqrt{-1}.
Same thing but a bit more obvious, your step is like saying:
|1| = |-1|
Therefore:
1 = -1
HallsofIvy
Jun17-05, 11:42 AM
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
Jul13-05, 10:43 PM
(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!!
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