perl
  • perl
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
KingGeorge
  • KingGeorge
The problem is that \(\sqrt{1}=\pm1\).
perl
  • perl
no, sqrt(1) = 1
perl
  • perl
check your calculator

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KingGeorge
  • KingGeorge
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 \(\sqrt{1}=1\).
anonymous
  • anonymous
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.
KingGeorge
  • KingGeorge
If you use the second way, then \(\sqrt{-1}\cdot\sqrt{-1}=|-1|\cdot|-1|=1\cdot1=1\).
KingGeorge
  • KingGeorge
Although there are problems when you do what I just did.
anonymous
  • anonymous
\[\sqrt{-1}\sqrt{-1} \neq \sqrt{-1 \times -1}\]
perl
  • perl
but sqrt(-2) = sqrt(-1 * 2) = sqrt(-1) * sqrt(2)
anonymous
  • anonymous
"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)
perl
  • perl
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) ?
perl
  • perl
oh, at least one of a,b must be positive, ok
anonymous
  • anonymous
i sqrt 2 is fine
anonymous
  • anonymous
\[\sqrt{-2} = \sqrt{-1 * 2} = \sqrt{-1} * \sqrt{2}\] 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.
perl
  • perl
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.
anonymous
  • anonymous
convert to i wherever possible instead of messing with sqrt signs
perl
  • perl
sorry i can only give one medal
perl
  • perl
micah, then you have no choice but to convert two negative signs to positive in a product
perl
  • perl
king george, i see your response, its a bit complicated. the function sqrt(x) is not multivalued, but i guess it could be
perl
  • perl
but if we use king georges idea, thatn sqrt(-1) = + - i
perl
  • perl
because (-i)^2 = -1, (+i)^2 = -1
anonymous
  • anonymous
there are 2 roots of -1, i and -i
perl
  • perl
right, but what does sqrt(-1) equal to, is it multivalued or not?
anonymous
  • anonymous
i
anonymous
  • anonymous
in context
perl
  • perl
scroll above, and see king georges response
perl
  • perl
, king george, sqrt( x^2) = |x| , so i think you are wrong there
Directrix
  • Directrix
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).
perl
  • perl
ok so in the real numbers this is not valid
anonymous
  • anonymous
Ugh....i^2 = -1 and x^2 = -1 has 2 solutions Can you get abiguity, yes - so be careful
KingGeorge
  • KingGeorge
Oops. I totally was wrong there :(
perl
  • perl
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
perl
  • perl
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). //
perl
  • perl
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*
anonymous
  • anonymous
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.
perl
  • perl
, what is principal branch of the complex square root
anonymous
  • anonymous
You don't want to know....
perl
  • perl
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.
perl
  • perl
estudier, i think i can handle it :) please send me a link
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Branch_point
perl
  • perl
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.
perl
  • perl
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
Directrix
  • Directrix
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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