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ParthKohli Group Title

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

  • one year ago
  • one year ago

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  1. AravindG Group Title
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    Ok first of all Theory part

    • one year ago
  2. dumbsearch2 Group Title
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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

    • one year ago
  3. dumbsearch2 Group Title
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    Let's kick back and watch the greenies argue

    • one year ago
  4. AravindG Group Title
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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 \(-\sqrt{x}\)

    • one year ago
  5. ParthKohli Group Title
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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 \(n = 1\).

    • one year ago
  6. AravindG Group Title
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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 \(-\sqrt{1}\)

    • one year ago
  7. AravindG Group Title
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    Notice that square root function always remains positive in the process

    • one year ago
  8. ParthKohli Group Title
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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.

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

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

    • one year ago
  11. ParthKohli Group Title
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    And what is your point that you are making?

    • one year ago
  12. AravindG Group Title
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    The square root function always remains positive

    • one year ago
  13. ParthKohli Group Title
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    Why does my use of the definition not convince you?

    • one year ago
  14. ParthKohli Group Title
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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.

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

    • one year ago
  16. AravindG Group Title
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    This is how square root function is defined

    • one year ago
  17. AravindG Group Title
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    Notice the modulus sign on right ..That is why the range of it is positive real numbers

    • one year ago
  18. AravindG Group Title
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    |dw:1369910211377:dw|

    • one year ago
  19. AravindG Group Title
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    This is how it would look graphically

    • one year ago
  20. AravindG Group Title
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    I was drawing a rough diagram for square root function @dumbsearch2

    • one year ago
  21. ParthKohli Group Title
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    Are you sure that's the definition of a square root?

    • one year ago
  22. AravindG Group Title
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    yes 100%

    • one year ago
  23. dumbsearch2 Group Title
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    Guys............ http://en.wikipedia.org/wiki/Square_root

    • one year ago
  24. sasogeek Group Title
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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?

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

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

    • one year ago
  27. dumbsearch2 Group Title
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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

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

    • one year ago
  29. sasogeek Group Title
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    she or shi, u still got it :) it's a free world my friend xD

    • one year ago
  30. dumbsearch2 Group Title
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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 :)

    • one year ago
  31. ParthKohli Group Title
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    lol -- scroll down and look at "all 2nd roots of 1"

    • one year ago
  32. dumbsearch2 Group Title
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    Wow. Parth was so right, the whole while. WOW! :D

    • one year ago
  33. AravindG Group Title
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    @dumbsearch2 no he isnt :/

    • one year ago
  34. dumbsearch2 Group Title
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    @AravindG I know it's hard to admit :/

    • one year ago
  35. ParthKohli Group Title
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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!

    • one year ago
  36. AravindG Group Title
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    @saifoo.khan had the same misconception But we convinced him in a post before .

    • one year ago
  37. ParthKohli Group Title
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    @AravindG What was saif's question?

    • one year ago
  38. AravindG Group Title
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    No it didnt

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

    • one year ago
  40. AravindG Group Title
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    Maybe @.Sam. can assist me here

    • one year ago
  41. ParthKohli Group Title
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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.

    • one year ago
  42. ParthKohli Group Title
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    @.Sam.

    • one year ago
  43. ParthKohli Group Title
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    @experimentX

    • one year ago
  44. dumbsearch2 Group Title
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    @dumbsearch2

    • one year ago
  45. dumbsearch2 Group Title
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    :D

    • one year ago
  46. ParthKohli Group Title
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    What did Saifoo ask exactly?

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

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

    • one year ago
  49. AravindG Group Title
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    Gimme some time..let me think of how to explain it in a simpler way

    • one year ago
  50. ParthKohli Group Title
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    \[x^2 = 1 \Rightarrow x^2 - 1 = 0 \Rightarrow (x + 1)(x - 1) = 0 \Rightarrow x = \pm 1\]

    • one year ago
  51. AravindG Group Title
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    yes completely right ^

    • one year ago
  52. AravindG Group Title
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    and how we got it ? x=\(+\sqrt{1} \)and \(-\sqrt{1}\)

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

    • one year ago
  54. dumbsearch2 Group Title
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    Wow. Caps. Stay nice, buddies :P

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

    • one year ago
  56. dumbsearch2 Group Title
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    Wow! @experimentX is dropping his 2 cents! :)

    • one year ago
  57. AravindG Group Title
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    neglect that reply

    • one year ago
  58. AravindG Group Title
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    The range is positive real numbers

    • one year ago
  59. ParthKohli Group Title
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    That's nowhere in the definition of the square root relation.

    • one year ago
  60. AravindG Group Title
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    ok I think Sam and Experimentx can make it clear

    • one year ago
  61. AravindG Group Title
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    lets wait

    • one year ago
  62. dumbsearch2 Group Title
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    Awesome! @.Sam. is an awesome guy :)

    • one year ago
  63. experimentX Group Title
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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.

    • one year ago
  64. ParthKohli Group Title
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    The match is now 2-1.

    • one year ago
  65. dumbsearch2 Group Title
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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. :)

    • one year ago
  66. ParthKohli Group Title
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    I have gotta go. Please keep replying

    • one year ago
  67. .Sam. Group Title
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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 \[y=\pm \sqrt{a}\]

    • one year ago
  68. AravindG Group Title
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    ^^

    • one year ago
  69. AravindG Group Title
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    Notice the definition @ParthKohli

    • one year ago
  70. sasogeek Group Title
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    @.Sam. blame my math teacher! lol :P but yeah, i concur :)

    • one year ago
  71. experimentX Group Title
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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

    • one year ago
  72. henpen Group Title
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    I'm pretty sure for a function to be continuous it must be 1-to-1.

    • one year ago
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