Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ParthKohli

  • one year ago

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

  • This Question is Closed
  1. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok first of all Theory part

  2. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Let's kick back and watch the greenies argue

  4. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Notice that square root function always remains positive in the process

  8. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes I talked abt them because they are the roots for x^2=1

  11. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And what is your point that you are making?

  12. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The square root function always remains positive

  13. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Why does my use of the definition not convince you?

  14. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay first I will provide you the standard definition \[\sqrt{x^2}=|x|\]

  16. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is how square root function is defined

  17. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Notice the modulus sign on right ..That is why the range of it is positive real numbers

  18. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1369910211377:dw|

  19. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is how it would look graphically

  20. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I was drawing a rough diagram for square root function @dumbsearch2

  21. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Are you sure that's the definition of a square root?

  22. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes 100%

  23. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Guys............ http://en.wikipedia.org/wiki/Square_root

  24. sasogeek
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    In mathematics, a square root of a number \(a\) is a number \(y\) such that \(y^2 = a\) :-P

  26. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  27. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    "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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ParthKohli Wolf agrees http://www.wolframalpha.com/input/?i=sqrt%7B1%7D

  29. sasogeek
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    she or shi, u still got it :) it's a free world my friend xD

  30. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol -- scroll down and look at "all 2nd roots of 1"

  32. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wow. Parth was so right, the whole while. WOW! :D

  33. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @dumbsearch2 no he isnt :/

  34. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @AravindG I know it's hard to admit :/

  35. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @saifoo.khan had the same misconception But we convinced him in a post before .

  37. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @AravindG What was saif's question?

  38. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No it didnt

  39. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Look, Wolfram|Alpha agreed with @ParthKohli. I've used their API before, they're extremely accurate.

  40. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Maybe @.Sam. can assist me here

  41. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @.Sam.

  43. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @experimentX

  44. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @dumbsearch2

  45. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :D

  46. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What did Saifoo ask exactly?

  47. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I told you we never neglected the roots The are 2 roots : \(+\sqrt{1}\) and \(-\sqrt{1}\)

  48. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  49. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Gimme some time..let me think of how to explain it in a simpler way

  50. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[x^2 = 1 \Rightarrow x^2 - 1 = 0 \Rightarrow (x + 1)(x - 1) = 0 \Rightarrow x = \pm 1\]

  51. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes completely right ^

  52. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and how we got it ? x=\(+\sqrt{1} \)and \(-\sqrt{1}\)

  53. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    WHY ARE YOU INCLUDING \(+\sqrt{1}\) AND \(-\sqrt1{}\)???????? I AM SIMPLY ASKING IF YOU CAN DISPROVE \(\sqrt1 = \pm 1 \)

  54. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wow. Caps. Stay nice, buddies :P

  55. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol, you got the opposite. The domain is \(x \ge 0\) and the range is any real number.

  56. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wow! @experimentX is dropping his 2 cents! :)

  57. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    neglect that reply

  58. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The range is positive real numbers

  59. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That's nowhere in the definition of the square root relation.

  60. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok I think Sam and Experimentx can make it clear

  61. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lets wait

  62. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Awesome! @.Sam. is an awesome guy :)

  63. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The match is now 2-1.

  65. dumbsearch2
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have gotta go. Please keep replying

  67. .Sam.
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\sqrt{x^2}=|x| \\ \\ +\sqrt{4}=+2 \\ \\ -\sqrt{4}=-2 \\ \\ \sqrt{4} \cancel{=}\pm2\] For situations with \(y^2=a\), then \[y=\pm \sqrt{a}\]

  68. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ^^

  69. AravindG
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Notice the definition @ParthKohli

  70. sasogeek
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @.Sam. blame my math teacher! lol :P but yeah, i concur :)

  71. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm pretty sure for a function to be continuous it must be 1-to-1.

  73. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.