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anonymous

  • 5 years ago

does the square root of 1, equal 1?

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  1. anonymous
    • 5 years ago
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    yes

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    i feel stupid... thank you :)

  4. anonymous
    • 5 years ago
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    or -1 :)

  5. anonymous
    • 5 years ago
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    or^1/2

  6. anonymous
    • 5 years ago
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    just 1 period :D

  7. anonymous
    • 5 years ago
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    +/- 1 to be correct

  8. anonymous
    • 5 years ago
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    actually : \[\sqrt(1) = |1|\] ^_^ and not 1 alone :)

  9. anonymous
    • 5 years ago
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    don't feel stupid, alot of ppl missed such simple relation :)

  10. anonymous
    • 5 years ago
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    \[\sqrt(x) = |x|\]

  11. nowhereman
    • 5 years ago
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    This is bullpellet. In real analysis the square root is only defined for non-negative numbers. And because it is a function, it can only assume one value for every argument. So the square root of 1 is that non-negative number which squared gives 1, so that is certainly 1. And you don't have to mess around with - or the absolute value function. \[x,y \geq 0 ⇒ \left( \sqrt{x} = y ⇔ y^2 = x\right)\]

  12. anonymous
    • 5 years ago
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    >_< nowhereman watch your language lol

  13. anonymous
    • 5 years ago
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    are you sure?

  14. anonymous
    • 5 years ago
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    actually , that's what my prof said though, and I was shocked LOL

  15. anonymous
    • 5 years ago
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    \[1 = \sqrt(1) = \sqrt(1)^2 = \sqrt(-1)^2 = ((-1))^{1/2} = |-1| = 1\] ._. maybe I have misunderstood him?

  16. anonymous
    • 5 years ago
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    lol...wait a mind...I think I got you now ^_^"

  17. anonymous
    • 5 years ago
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    min*

  18. nowhereman
    • 5 years ago
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    Yes, I'm absolutely sure. ( Only in complex analysis you can describe the square-root function as a riemannian surface with multiple value)

  19. anonymous
    • 5 years ago
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    lol wut

  20. nowhereman
    • 5 years ago
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    And of course that definition leads to \[\sqrt{x^2} = |x|\]

  21. nowhereman
    • 5 years ago
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    spaceknight: what are you laughing about?

  22. anonymous
    • 5 years ago
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    because i dont understand what you are talking about lol

  23. anonymous
    • 5 years ago
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    I understood, thank you nowhereman ^_^

  24. anonymous
    • 5 years ago
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    still studying DM :)

  25. anonymous
    • 5 years ago
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    are you talking about the principal square root discussed in here http://mathworld.wolfram.com/PrincipalSquareRoot.html because it clearly states For example, the principal square root of 9 is 3, although both and 3 are square roots of 9.

  26. nowhereman
    • 5 years ago
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    Yes, if you want to name it this way. In real analysis, if you are talking about _the_ square root, you mean the principal square root. Of course it is clear that the equation y^2 = x can have 0, 1 or 2 solutions.

  27. anonymous
    • 5 years ago
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    Oh, this site is beautiful~ thank you spaceknight! ^_^

  28. anonymous
    • 5 years ago
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    but not too much info though , still :)

  29. anonymous
    • 5 years ago
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    why does it matter if you are talking about the principal square root or not? everytime i've had to sqrt anything in equations or whatever I've always been told to include the +/-.

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