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ParthKohli

  • 2 years ago

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

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  1. AravindG
    • 2 years ago
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    Ok first of all Theory part

  2. dumbsearch2
    • 2 years ago
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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

  3. dumbsearch2
    • 2 years ago
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    Let's kick back and watch the greenies argue

  4. AravindG
    • 2 years ago
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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 \(-\sqrt{x}\)

  5. ParthKohli
    • 2 years ago
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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 \(n = 1\).

  6. AravindG
    • 2 years ago
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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 \(-\sqrt{1}\)

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

  8. ParthKohli
    • 2 years ago
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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.

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

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

  11. ParthKohli
    • 2 years ago
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    And what is your point that you are making?

  12. AravindG
    • 2 years ago
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    The square root function always remains positive

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

  14. ParthKohli
    • 2 years ago
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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.

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

  16. AravindG
    • 2 years ago
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    This is how square root function is defined

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

  18. AravindG
    • 2 years ago
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    |dw:1369910211377:dw|

  19. AravindG
    • 2 years ago
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    This is how it would look graphically

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

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

  22. AravindG
    • 2 years ago
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    yes 100%

  23. dumbsearch2
    • 2 years ago
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    Guys............ http://en.wikipedia.org/wiki/Square_root

  24. sasogeek
    • 2 years ago
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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?

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

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

  27. dumbsearch2
    • 2 years ago
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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

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

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

  30. dumbsearch2
    • 2 years ago
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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 :)

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

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

  33. AravindG
    • 2 years ago
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    @dumbsearch2 no he isnt :/

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

  35. ParthKohli
    • 2 years ago
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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!

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

  37. ParthKohli
    • 2 years ago
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    @AravindG What was saif's question?

  38. AravindG
    • 2 years ago
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    No it didnt

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

  40. AravindG
    • 2 years ago
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    Maybe @.Sam. can assist me here

  41. ParthKohli
    • 2 years ago
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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.

  42. ParthKohli
    • 2 years ago
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    @.Sam.

  43. ParthKohli
    • 2 years ago
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    @experimentX

  44. dumbsearch2
    • 2 years ago
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    @dumbsearch2

  45. dumbsearch2
    • 2 years ago
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    :D

  46. ParthKohli
    • 2 years ago
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    What did Saifoo ask exactly?

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

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

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

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

  51. AravindG
    • 2 years ago
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    yes completely right ^

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

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

  54. dumbsearch2
    • 2 years ago
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    Wow. Caps. Stay nice, buddies :P

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

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

  57. AravindG
    • 2 years ago
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    neglect that reply

  58. AravindG
    • 2 years ago
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    The range is positive real numbers

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

  60. AravindG
    • 2 years ago
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    ok I think Sam and Experimentx can make it clear

  61. AravindG
    • 2 years ago
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    lets wait

  62. dumbsearch2
    • 2 years ago
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    Awesome! @.Sam. is an awesome guy :)

  63. experimentX
    • 2 years ago
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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.

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

  65. dumbsearch2
    • 2 years ago
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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. :)

  66. ParthKohli
    • 2 years ago
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    I have gotta go. Please keep replying

  67. .Sam.
    • 2 years ago
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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 \[y=\pm \sqrt{a}\]

  68. AravindG
    • 2 years ago
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    ^^

  69. AravindG
    • 2 years ago
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    Notice the definition @ParthKohli

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

  71. experimentX
    • 2 years ago
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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

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

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