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ParthKohli

@AravindG \(\sqrt{1} = \pm 1\) disprove.

  • 10 months ago
  • 10 months ago

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  1. AravindG
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    Ok first of all Theory part

    • 10 months ago
  2. dumbsearch2
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    Eh, guise. This is a continuation of http://openstudy.com/study#/updates/51a721e1e4b08f9e59e230bf in some respects. Just so you know, strangers xD

    • 10 months ago
  3. dumbsearch2
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    Let's kick back and watch the greenies argue

    • 10 months ago
  4. AravindG
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    We have 2 types of square roots ..Positive and negative .Unless stated explicitly we always take the positive square root.The square root function ALWAYS has a set of real positive numbers as range .Hence we denote the positive and negative square roots as : \(+\sqrt{x}\) and \(-\sqrt{x}\)

    • 10 months ago
  5. ParthKohli
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    Actually, we need to get back to the definition of a square root. If \(\sqrt{x} = n\), then the equation \(x = n^2 \) is satisfied. That is the definition. Does \(1 = (-1)^2\) satisfy? In that case, \(-1 = \sqrt{1}\). So is the case with \(n = 1\).

    • 10 months ago
  6. AravindG
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    The solutions for the equality \[x^2=1\] are 1 and -1 But notice how we get that : First root which is positive is \(+\sqrt{1}=1\) and the other negative root is \(-\sqrt{1}\)

    • 10 months ago
  7. AravindG
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    Notice that square root function always remains positive in the process

    • 10 months ago
  8. ParthKohli
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    Also, my question clearly on the top does not mention \(+\sqrt{1} = 1\) or \(-\sqrt{1} = 1\), but says \(\sqrt{1} = \pm1\) without any sign.

    • 10 months ago
  9. ParthKohli
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    Uh, actually, you are talking about \(-\sqrt{1}\) and \(+\sqrt{1}\). But what exactly \(\mathcal{is}\) \(\sqrt{1}\)?

    • 10 months ago
  10. AravindG
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    yes I talked abt them because they are the roots for x^2=1

    • 10 months ago
  11. ParthKohli
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    And what is your point that you are making?

    • 10 months ago
  12. AravindG
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    The square root function always remains positive

    • 10 months ago
  13. ParthKohli
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    Why does my use of the definition not convince you?

    • 10 months ago
  14. ParthKohli
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    You may be talking about the principal square root, but that has nothing to do with square root. They both are similar but different functions.

    • 10 months ago
  15. AravindG
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    Okay first I will provide you the standard definition \[\sqrt{x^2}=|x|\]

    • 10 months ago
  16. AravindG
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    This is how square root function is defined

    • 10 months ago
  17. AravindG
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    Notice the modulus sign on right ..That is why the range of it is positive real numbers

    • 10 months ago
  18. AravindG
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    |dw:1369910211377:dw|

    • 10 months ago
  19. AravindG
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    This is how it would look graphically

    • 10 months ago
  20. AravindG
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    I was drawing a rough diagram for square root function @dumbsearch2

    • 10 months ago
  21. ParthKohli
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    Are you sure that's the definition of a square root?

    • 10 months ago
  22. AravindG
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    yes 100%

    • 10 months ago
  23. dumbsearch2
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    Guys............ http://en.wikipedia.org/wiki/Square_root

    • 10 months ago
  24. sasogeek
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    okay to end all this shinanigans going on, let's just say that \(\huge \sqrt{x^2}=\pm x \) this is because, \(\huge +x^2=-x^2\) and both are always non-negative numbers. want proof? \(\huge \sqrt{4}=\pm2 \) \(\huge -2^2=4, \ and \ 2^2=4 \) everybody happy now?

    • 10 months ago
  25. ParthKohli
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    In mathematics, a square root of a number \(a\) is a number \(y\) such that \(y^2 = a\) :-P

    • 10 months ago
  26. ParthKohli
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    Let \(a = 1\) and \(y = -1\). Why is it that you think \(1 \ne (-1)^2\)?

    • 10 months ago
  27. dumbsearch2
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    "shinanigans" @sasogeek LOL! 1. shinanigans is spelled wrong. 2. Their debate is hardly shenanigans, it's just a real interesting discussion that fascinates our curious minds :D

    • 10 months ago
  28. AravindG
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    @ParthKohli Wolf agrees http://www.wolframalpha.com/input/?i=sqrt%7B1%7D

    • 10 months ago
  29. sasogeek
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    she or shi, u still got it :) it's a free world my friend xD

    • 10 months ago
  30. dumbsearch2
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    I was always taught that square rooting one made one. IDK though, cause I'm a n00b, just I learned it that way :)

    • 10 months ago
  31. ParthKohli
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    lol -- scroll down and look at "all 2nd roots of 1"

    • 10 months ago
  32. dumbsearch2
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    Wow. Parth was so right, the whole while. WOW! :D

    • 10 months ago
  33. AravindG
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    @dumbsearch2 no he isnt :/

    • 10 months ago
  34. dumbsearch2
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    @AravindG I know it's hard to admit :/

    • 10 months ago
  35. ParthKohli
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    Why are you dragging in the principal root all the time? What you said is just a property of square roots. It's not the definition!

    • 10 months ago
  36. AravindG
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    @saifoo.khan had the same misconception But we convinced him in a post before .

    • 10 months ago
  37. ParthKohli
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    @AravindG What was saif's question?

    • 10 months ago
  38. AravindG
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    No it didnt

    • 10 months ago
  39. dumbsearch2
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    Look, Wolfram|Alpha agreed with @ParthKohli. I've used their API before, they're extremely accurate.

    • 10 months ago
  40. AravindG
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    Maybe @.Sam. can assist me here

    • 10 months ago
  41. ParthKohli
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    What you're doing is actually neglecting a root. \(x^2 - x = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1\) is an example of neglecting a root.

    • 10 months ago
  42. ParthKohli
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    @.Sam.

    • 10 months ago
  43. ParthKohli
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    @experimentX

    • 10 months ago
  44. dumbsearch2
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    @dumbsearch2

    • 10 months ago
  45. dumbsearch2
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    :D

    • 10 months ago
  46. ParthKohli
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    What did Saifoo ask exactly?

    • 10 months ago
  47. AravindG
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    I told you we never neglected the roots The are 2 roots : \(+\sqrt{1}\) and \(-\sqrt{1}\)

    • 10 months ago
  48. ParthKohli
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    \(\sqrt{1}\) is like writing down \(\sin(\pi) \) for an answer. You actually want to write the numerical value.

    • 10 months ago
  49. AravindG
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    Gimme some time..let me think of how to explain it in a simpler way

    • 10 months ago
  50. ParthKohli
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    \[x^2 = 1 \Rightarrow x^2 - 1 = 0 \Rightarrow (x + 1)(x - 1) = 0 \Rightarrow x = \pm 1\]

    • 10 months ago
  51. AravindG
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    yes completely right ^

    • 10 months ago
  52. AravindG
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    and how we got it ? x=\(+\sqrt{1} \)and \(-\sqrt{1}\)

    • 10 months ago
  53. ParthKohli
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    WHY ARE YOU INCLUDING \(+\sqrt{1}\) AND \(-\sqrt1{}\)???????? I AM SIMPLY ASKING IF YOU CAN DISPROVE \(\sqrt1 = \pm 1 \)

    • 10 months ago
  54. dumbsearch2
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    Wow. Caps. Stay nice, buddies :P

    • 10 months ago
  55. ParthKohli
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    lol, you got the opposite. The domain is \(x \ge 0\) and the range is any real number.

    • 10 months ago
  56. dumbsearch2
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    Wow! @experimentX is dropping his 2 cents! :)

    • 10 months ago
  57. AravindG
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    neglect that reply

    • 10 months ago
  58. AravindG
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    The range is positive real numbers

    • 10 months ago
  59. ParthKohli
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    That's nowhere in the definition of the square root relation.

    • 10 months ago
  60. AravindG
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    ok I think Sam and Experimentx can make it clear

    • 10 months ago
  61. AravindG
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    lets wait

    • 10 months ago
  62. dumbsearch2
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    Awesome! @.Sam. is an awesome guy :)

    • 10 months ago
  63. experimentX
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    root is a multivalued function, function by definition must be single valued. +1 and -1 are different values (or branches) we restrict only to one particular branch. sqrt(1) = +-1, this is untrue. by taking sqrt(1), we are explicitly choosing sqrt(1) as solution of x^2 = 1.

    • 10 months ago
  64. ParthKohli
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    The match is now 2-1.

    • 10 months ago
  65. dumbsearch2
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    uh. Very frankly, I need help with my question *sigh* it's 3:57 A.M. here, and maybe you can debate a lil' later, there's no one else on this math topic ready to help right now, at least it doesn't seem like it, so when you're done please try to help me a little bitt haha. :)

    • 10 months ago
  66. ParthKohli
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    I have gotta go. Please keep replying

    • 10 months ago
  67. .Sam.
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    \[\sqrt{x^2}=|x| \\ \\ +\sqrt{4}=+2 \\ \\ -\sqrt{4}=-2 \\ \\ \sqrt{4} \cancel{=}\pm2\] For situations with \(y^2=a\), then \[y=\pm \sqrt{a}\]

    • 10 months ago
  68. AravindG
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    ^^

    • 10 months ago
  69. AravindG
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    Notice the definition @ParthKohli

    • 10 months ago
  70. sasogeek
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    @.Sam. blame my math teacher! lol :P but yeah, i concur :)

    • 10 months ago
  71. experimentX
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    For situations with y^2=a, then, +- sqrt(a) are roots not values of y, just possible candidates of 'y'. when we arrive at problem when we have to determine the value of 'y' to work out, we restrict with one value considering different cases. this is same are inverse trig functions, but we choose -pi/2 to pi/2 for sin, 0 to pi for cosine and -pi/2 to pi/2 for tan

    • 10 months ago
  72. henpen
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    I'm pretty sure for a function to be continuous it must be 1-to-1.

    • 10 months ago
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