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perl

  • 2 years ago

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

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  1. KingGeorge
    • 2 years ago
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    The problem is that \(\sqrt{1}=\pm1\).

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

  3. perl
    • 2 years ago
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    check your calculator

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

  5. micahwood50
    • 2 years ago
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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.

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

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

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

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

  10. estudier
    • 2 years ago
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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)

  11. perl
    • 2 years ago
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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) ?

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

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

  14. micahwood50
    • 2 years ago
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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.

  15. perl
    • 2 years ago
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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.

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

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

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

  19. perl
    • 2 years ago
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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

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

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

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

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

  24. estudier
    • 2 years ago
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    i

  25. estudier
    • 2 years ago
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    in context

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

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

  28. Directrix
    • 2 years ago
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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).

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

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

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

  32. perl
    • 2 years ago
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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

  33. perl
    • 2 years ago
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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). //

  34. perl
    • 2 years ago
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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*

  35. estudier
    • 2 years ago
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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.

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

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

  38. perl
    • 2 years ago
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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.

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

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

  41. perl
    • 2 years ago
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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.

  42. perl
    • 2 years ago
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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

  43. Directrix
    • 2 years ago
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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?

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