## ParthKohli Group Title @AravindG $$\sqrt{1} = \pm 1$$ disprove. one year ago one year ago

1. AravindG Group Title

Ok first of all Theory part

2. dumbsearch2 Group Title

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

3. dumbsearch2 Group Title

Let's kick back and watch the greenies argue

4. AravindG Group Title

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 Group Title

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 Group Title

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 Group Title

Notice that square root function always remains positive in the process

8. ParthKohli Group Title

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 Group Title

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

10. AravindG Group Title

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

11. ParthKohli Group Title

And what is your point that you are making?

12. AravindG Group Title

The square root function always remains positive

13. ParthKohli Group Title

Why does my use of the definition not convince you?

14. ParthKohli Group Title

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 Group Title

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

16. AravindG Group Title

This is how square root function is defined

17. AravindG Group Title

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

18. AravindG Group Title

|dw:1369910211377:dw|

19. AravindG Group Title

This is how it would look graphically

20. AravindG Group Title

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

21. ParthKohli Group Title

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

22. AravindG Group Title

yes 100%

23. dumbsearch2 Group Title

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

24. sasogeek Group Title

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 Group Title

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

26. ParthKohli Group Title

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

27. dumbsearch2 Group Title

"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 Group Title

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

29. sasogeek Group Title

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

30. dumbsearch2 Group Title

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 Group Title

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

32. dumbsearch2 Group Title

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

33. AravindG Group Title

@dumbsearch2 no he isnt :/

34. dumbsearch2 Group Title

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

35. ParthKohli Group Title

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 Group Title

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

37. ParthKohli Group Title

@AravindG What was saif's question?

38. AravindG Group Title

No it didnt

39. dumbsearch2 Group Title

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

40. AravindG Group Title

Maybe @.Sam. can assist me here

41. ParthKohli Group Title

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 Group Title

@.Sam.

43. ParthKohli Group Title

@experimentX

44. dumbsearch2 Group Title

@dumbsearch2

45. dumbsearch2 Group Title

:D

46. ParthKohli Group Title

47. AravindG Group Title

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

48. ParthKohli Group Title

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

49. AravindG Group Title

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

50. ParthKohli Group Title

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

51. AravindG Group Title

yes completely right ^

52. AravindG Group Title

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

53. ParthKohli Group Title

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

54. dumbsearch2 Group Title

Wow. Caps. Stay nice, buddies :P

55. ParthKohli Group Title

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

56. dumbsearch2 Group Title

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

57. AravindG Group Title

58. AravindG Group Title

The range is positive real numbers

59. ParthKohli Group Title

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

60. AravindG Group Title

ok I think Sam and Experimentx can make it clear

61. AravindG Group Title

lets wait

62. dumbsearch2 Group Title

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

63. experimentX Group Title

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 Group Title

The match is now 2-1.

65. dumbsearch2 Group Title

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 Group Title

67. .Sam. Group Title

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

68. AravindG Group Title

^^

69. AravindG Group Title

Notice the definition @ParthKohli

70. sasogeek Group Title

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

71. experimentX Group Title

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 Group Title

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