ParthKohli
  • ParthKohli
@AravindG \(\sqrt{1} = \pm 1\) disprove.
Mathematics
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AravindG
  • AravindG
Ok first of all Theory part
dumbsearch2
  • dumbsearch2
Eh, guise. This is a continuation of http://openstudy.com/study#/updates/51a721e1e4b08f9e59e230bf in some respects. Just so you know, strangers xD
dumbsearch2
  • dumbsearch2
Let's kick back and watch the greenies argue

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AravindG
  • AravindG
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 \(-\sqrt{x}\)
ParthKohli
  • ParthKohli
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 \(n = 1\).
AravindG
  • AravindG
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 \(-\sqrt{1}\)
AravindG
  • AravindG
Notice that square root function always remains positive in the process
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
Uh, actually, you are talking about \(-\sqrt{1}\) and \(+\sqrt{1}\). But what exactly \(\mathcal{is}\) \(\sqrt{1}\)?
AravindG
  • AravindG
yes I talked abt them because they are the roots for x^2=1
ParthKohli
  • ParthKohli
And what is your point that you are making?
AravindG
  • AravindG
The square root function always remains positive
ParthKohli
  • ParthKohli
Why does my use of the definition not convince you?
ParthKohli
  • ParthKohli
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.
AravindG
  • AravindG
Okay first I will provide you the standard definition \[\sqrt{x^2}=|x|\]
AravindG
  • AravindG
This is how square root function is defined
AravindG
  • AravindG
Notice the modulus sign on right ..That is why the range of it is positive real numbers
AravindG
  • AravindG
|dw:1369910211377:dw|
AravindG
  • AravindG
This is how it would look graphically
AravindG
  • AravindG
I was drawing a rough diagram for square root function @dumbsearch2
ParthKohli
  • ParthKohli
Are you sure that's the definition of a square root?
AravindG
  • AravindG
yes 100%
dumbsearch2
  • dumbsearch2
Guys............ http://en.wikipedia.org/wiki/Square_root
sasogeek
  • sasogeek
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?
ParthKohli
  • ParthKohli
In mathematics, a square root of a number \(a\) is a number \(y\) such that \(y^2 = a\) :-P
ParthKohli
  • ParthKohli
Let \(a = 1\) and \(y = -1\). Why is it that you think \(1 \ne (-1)^2\)?
dumbsearch2
  • dumbsearch2
"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
AravindG
  • AravindG
@ParthKohli Wolf agrees http://www.wolframalpha.com/input/?i=sqrt%7B1%7D
sasogeek
  • sasogeek
she or shi, u still got it :) it's a free world my friend xD
dumbsearch2
  • dumbsearch2
I was always taught that square rooting one made one. IDK though, cause I'm a n00b, just I learned it that way :)
ParthKohli
  • ParthKohli
lol -- scroll down and look at "all 2nd roots of 1"
dumbsearch2
  • dumbsearch2
Wow. Parth was so right, the whole while. WOW! :D
AravindG
  • AravindG
@dumbsearch2 no he isnt :/
dumbsearch2
  • dumbsearch2
@AravindG I know it's hard to admit :/
ParthKohli
  • ParthKohli
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!
AravindG
  • AravindG
@saifoo.khan had the same misconception But we convinced him in a post before .
ParthKohli
  • ParthKohli
@AravindG What was saif's question?
AravindG
  • AravindG
No it didnt
dumbsearch2
  • dumbsearch2
Look, Wolfram|Alpha agreed with @ParthKohli. I've used their API before, they're extremely accurate.
AravindG
  • AravindG
Maybe @.Sam. can assist me here
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
@.Sam.
ParthKohli
  • ParthKohli
@experimentX
dumbsearch2
  • dumbsearch2
@dumbsearch2
dumbsearch2
  • dumbsearch2
:D
ParthKohli
  • ParthKohli
What did Saifoo ask exactly?
AravindG
  • AravindG
I told you we never neglected the roots The are 2 roots : \(+\sqrt{1}\) and \(-\sqrt{1}\)
ParthKohli
  • ParthKohli
\(\sqrt{1}\) is like writing down \(\sin(\pi) \) for an answer. You actually want to write the numerical value.
AravindG
  • AravindG
Gimme some time..let me think of how to explain it in a simpler way
ParthKohli
  • ParthKohli
\[x^2 = 1 \Rightarrow x^2 - 1 = 0 \Rightarrow (x + 1)(x - 1) = 0 \Rightarrow x = \pm 1\]
AravindG
  • AravindG
yes completely right ^
AravindG
  • AravindG
and how we got it ? x=\(+\sqrt{1} \)and \(-\sqrt{1}\)
ParthKohli
  • ParthKohli
WHY ARE YOU INCLUDING \(+\sqrt{1}\) AND \(-\sqrt1{}\)???????? I AM SIMPLY ASKING IF YOU CAN DISPROVE \(\sqrt1 = \pm 1 \)
dumbsearch2
  • dumbsearch2
Wow. Caps. Stay nice, buddies :P
ParthKohli
  • ParthKohli
lol, you got the opposite. The domain is \(x \ge 0\) and the range is any real number.
dumbsearch2
  • dumbsearch2
Wow! @experimentX is dropping his 2 cents! :)
AravindG
  • AravindG
neglect that reply
AravindG
  • AravindG
The range is positive real numbers
ParthKohli
  • ParthKohli
That's nowhere in the definition of the square root relation.
AravindG
  • AravindG
ok I think Sam and Experimentx can make it clear
AravindG
  • AravindG
lets wait
dumbsearch2
  • dumbsearch2
Awesome! @.Sam. is an awesome guy :)
experimentX
  • experimentX
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.
ParthKohli
  • ParthKohli
The match is now 2-1.
dumbsearch2
  • dumbsearch2
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. :)
ParthKohli
  • ParthKohli
I have gotta go. Please keep replying
.Sam.
  • .Sam.
\[\sqrt{x^2}=|x| \\ \\ +\sqrt{4}=+2 \\ \\ -\sqrt{4}=-2 \\ \\ \sqrt{4} \cancel{=}\pm2\] For situations with \(y^2=a\), then \[y=\pm \sqrt{a}\]
AravindG
  • AravindG
^^
AravindG
  • AravindG
Notice the definition @ParthKohli
sasogeek
  • sasogeek
@.Sam. blame my math teacher! lol :P but yeah, i concur :)
experimentX
  • experimentX
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
anonymous
  • anonymous
I'm pretty sure for a function to be continuous it must be 1-to-1.

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