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Spot the error. Surely there must be an error somewhere, 1 = -1 is absurd! so im having trouble with this 1 = -1 * -1 = sqrt (-1*-1) = sqrt (-1) * sqrt (-1) = i * i = i^2 = -1

Mathematics
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The problem is that \(\sqrt{1}=\pm1\).
no, sqrt(1) = 1
check your calculator

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By definition, \(\sqrt{x}=\{y\;|\;y^2=x\}\). When \(x=1\), both \(1,\) and \(-1\) are in the set. Alternatively, you can say that \(\sqrt{x}=|x|\), and for \(x=1\), then \(\sqrt{1}=1\).
The fallacy is that the rule √xy = √x√y is generally valid only if at least one of the two numbers x or y is positive.
If you use the second way, then \(\sqrt{-1}\cdot\sqrt{-1}=|-1|\cdot|-1|=1\cdot1=1\).
Although there are problems when you do what I just did.
\[\sqrt{-1}\sqrt{-1} \neq \sqrt{-1 \times -1}\]
but sqrt(-2) = sqrt(-1 * 2) = sqrt(-1) * sqrt(2)
"1 = -1 * -1 = sqrt (-1*-1) = sqrt (-1) * sqrt (-1) = i * i = i^2 = -1" This is just wrong (and it won't work if you try and use plus/minus either)
so sqrt(a*b) = sqrt(a)*sqrt(b) is only valid for a,b > 0 but what about sqrt(-1 *2) = sqrt(-1)*sqrt(2) ?
oh, at least one of a,b must be positive, ok
i sqrt 2 is fine
\[\sqrt{-2} = \sqrt{-1 * 2} = \sqrt{-1} * \sqrt{2}\] because one of them is positive so it is valid. √xy = √x√y is generally valid only if at least one of the two numbers x or y is positive.
micah has it !! The fallacy is that the rule √xy = √x√y is generally valid only if at least one of the two numbers x or y is positive.
convert to i wherever possible instead of messing with sqrt signs
sorry i can only give one medal
micah, then you have no choice but to convert two negative signs to positive in a product
king george, i see your response, its a bit complicated. the function sqrt(x) is not multivalued, but i guess it could be
but if we use king georges idea, thatn sqrt(-1) = + - i
because (-i)^2 = -1, (+i)^2 = -1
there are 2 roots of -1, i and -i
right, but what does sqrt(-1) equal to, is it multivalued or not?
i
in context
scroll above, and see king georges response
, king george, sqrt( x^2) = |x| , so i think you are wrong there
sqrt (-1*-1) = sqrt (-1) * sqrt (-1) --------------------------- In the set of Real Numbers, the square root of a product is the product of the square roots. That property does not hold true within the set of Complex Numbers as illustrated by the posted example. Product of Square Roots Theorem If m and n are not negative and are real numbers, then SQRT(m) * SQRT(n) = SQRT(mn).
ok so in the real numbers this is not valid
Ugh....i^2 = -1 and x^2 = -1 has 2 solutions Can you get abiguity, yes - so be careful
Oops. I totally was wrong there :(
directrix, this is not true what you said "In the set of Real Numbers, the square root of a product is the product of the square roots." if the numbers are both negative , or even one negative
directrix, this is very confusing, and i think wrong. points deducted // sqrt (-1*-1) = sqrt (-1) * sqrt (-1) --------------------------- In the set of Real Numbers, the square root of a product is the product of the square roots. That property does not hold true within the set of Complex Numbers as illustrated by the posted example. Product of Square Roots Theorem If m and n are not negative and are real numbers, then SQRT(m) * SQRT(n) = SQRT(mn). //
Correction: In the set of Real Numbers, the square root of a product is the product of the square roots *provided that the two numbers are both nonnegative*. In the complex numbers the square root of a product is the product of the square roots *provided that at least one is positive*
sqrt -1 is either principal sqrt (x>0) or principal branch of the complex square root. Don't mess with radicals, convert to i and to -1 pronto.
, what is principal branch of the complex square root
You don't want to know....
Edited correction: In the set of Real Numbers, the square root of a product xy is equal to the product sqrt(x)*sqrt(y) provided that both x and y are nonnegative. In the complex numbers the square root of xy is equal to the product sqrt(x)*sqrt(y) provided that at least one of x and y is positive.
estudier, i think i can handle it :) please send me a link
http://en.wikipedia.org/wiki/Branch_point
Edited correction 3: In the set of real numbers, sqrt(x*y) = sqrt(x)*sqrt(y) provided that both x and y are nonnegative. In the complex numbers we have sqrt(x*y) = sqrt(x)*sqrt(y), provided that at least one of x or y is positive.
Edited correction 3: In the set of real numbers, sqrt(x*y) = sqrt(x)*sqrt(y) provided that both x and y are nonnegative. In the complex numbers we have sqrt(x*y) = sqrt(x)*sqrt(y), provided that at least one of x or y is positive. Otherwise it is not equal. and in the reals we have sqrt(-2) is undefined, for example
Fallacy ----------- Sqrt(-1) * Sqrt(-1) = Sqrt ( -1 * -1) = Sqrt(1) = 1. ===== Sqrt(-1) * Sqrt(-1) = -1 by definition Consider the following assertion: (-1)^2 = (1)^2 log (-1)^2 = log (1)^2 2 log (-1) = 2 log (1) log(-1) = log(1) Therefore, -1 = 1 Where lies the error or misconception?

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