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perl Group Title

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

  • 2 years ago
  • 2 years ago

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  1. KingGeorge Group Title
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    The problem is that \(\sqrt{1}=\pm1\).

    • 2 years ago
  2. perl Group Title
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    no, sqrt(1) = 1

    • 2 years ago
  3. perl Group Title
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    check your calculator

    • 2 years ago
  4. KingGeorge Group Title
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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 \(\sqrt{1}=1\).

    • 2 years ago
  5. micahwood50 Group Title
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    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.

    • 2 years ago
  6. KingGeorge Group Title
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    If you use the second way, then \(\sqrt{-1}\cdot\sqrt{-1}=|-1|\cdot|-1|=1\cdot1=1\).

    • 2 years ago
  7. KingGeorge Group Title
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    Although there are problems when you do what I just did.

    • 2 years ago
  8. micahwood50 Group Title
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    \[\sqrt{-1}\sqrt{-1} \neq \sqrt{-1 \times -1}\]

    • 2 years ago
  9. perl Group Title
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    but sqrt(-2) = sqrt(-1 * 2) = sqrt(-1) * sqrt(2)

    • 2 years ago
  10. estudier Group Title
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    "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)

    • 2 years ago
  11. perl Group Title
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    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) ?

    • 2 years ago
  12. perl Group Title
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    oh, at least one of a,b must be positive, ok

    • 2 years ago
  13. estudier Group Title
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    i sqrt 2 is fine

    • 2 years ago
  14. micahwood50 Group Title
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    \[\sqrt{-2} = \sqrt{-1 * 2} = \sqrt{-1} * \sqrt{2}\] 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.

    • 2 years ago
  15. perl Group Title
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    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.

    • 2 years ago
  16. estudier Group Title
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    convert to i wherever possible instead of messing with sqrt signs

    • 2 years ago
  17. perl Group Title
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    sorry i can only give one medal

    • 2 years ago
  18. perl Group Title
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    micah, then you have no choice but to convert two negative signs to positive in a product

    • 2 years ago
  19. perl Group Title
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    king george, i see your response, its a bit complicated. the function sqrt(x) is not multivalued, but i guess it could be

    • 2 years ago
  20. perl Group Title
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    but if we use king georges idea, thatn sqrt(-1) = + - i

    • 2 years ago
  21. perl Group Title
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    because (-i)^2 = -1, (+i)^2 = -1

    • 2 years ago
  22. estudier Group Title
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    there are 2 roots of -1, i and -i

    • 2 years ago
  23. perl Group Title
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    right, but what does sqrt(-1) equal to, is it multivalued or not?

    • 2 years ago
  24. estudier Group Title
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    i

    • 2 years ago
  25. estudier Group Title
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    in context

    • 2 years ago
  26. perl Group Title
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    scroll above, and see king georges response

    • 2 years ago
  27. perl Group Title
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    , king george, sqrt( x^2) = |x| , so i think you are wrong there

    • 2 years ago
  28. Directrix Group Title
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    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).

    • 2 years ago
  29. perl Group Title
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    ok so in the real numbers this is not valid

    • 2 years ago
  30. estudier Group Title
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    Ugh....i^2 = -1 and x^2 = -1 has 2 solutions Can you get abiguity, yes - so be careful

    • 2 years ago
  31. KingGeorge Group Title
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    Oops. I totally was wrong there :(

    • 2 years ago
  32. perl Group Title
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    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

    • 2 years ago
  33. perl Group Title
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    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). //

    • 2 years ago
  34. perl Group Title
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    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*

    • 2 years ago
  35. estudier Group Title
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    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.

    • 2 years ago
  36. perl Group Title
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    , what is principal branch of the complex square root

    • 2 years ago
  37. estudier Group Title
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    You don't want to know....

    • 2 years ago
  38. perl Group Title
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    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.

    • 2 years ago
  39. perl Group Title
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    estudier, i think i can handle it :) please send me a link

    • 2 years ago
  40. estudier Group Title
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    http://en.wikipedia.org/wiki/Branch_point

    • 2 years ago
  41. perl Group Title
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    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.

    • 2 years ago
  42. perl Group Title
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    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

    • 2 years ago
  43. Directrix Group Title
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    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?

    • 2 years ago
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