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AravindGBest ResponseYou've already chosen the best response.0
Ok first of all Theory part
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
Eh, guise. This is a continuation of http://openstudy.com/study#/updates/51a721e1e4b08f9e59e230bf in some respects. Just so you know, strangers xD
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
Let's kick back and watch the greenies argue
 10 months ago

AravindGBest ResponseYou've already chosen the best response.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 \(\sqrt{x}\)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.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 \(n = 1\).
 10 months ago

AravindGBest ResponseYou've already chosen the best response.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 \(\sqrt{1}\)
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
Notice that square root function always remains positive in the process
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.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.
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Uh, actually, you are talking about \(\sqrt{1}\) and \(+\sqrt{1}\). But what exactly \(\mathcal{is}\) \(\sqrt{1}\)?
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
yes I talked abt them because they are the roots for x^2=1
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
And what is your point that you are making?
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
The square root function always remains positive
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Why does my use of the definition not convince you?
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.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.
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
Okay first I will provide you the standard definition \[\sqrt{x^2}=x\]
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
This is how square root function is defined
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
Notice the modulus sign on right ..That is why the range of it is positive real numbers
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
dw:1369910211377:dw
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
This is how it would look graphically
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
I was drawing a rough diagram for square root function @dumbsearch2
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Are you sure that's the definition of a square root?
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
Guys............ http://en.wikipedia.org/wiki/Square_root
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.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 nonnegative numbers. want proof? \(\huge \sqrt{4}=\pm2 \) \(\huge 2^2=4, \ and \ 2^2=4 \) everybody happy now?
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
In mathematics, a square root of a number \(a\) is a number \(y\) such that \(y^2 = a\) :P
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Let \(a = 1\) and \(y = 1\). Why is it that you think \(1 \ne (1)^2\)?
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.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
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
@ParthKohli Wolf agrees http://www.wolframalpha.com/input/?i=sqrt%7B1%7D
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
she or shi, u still got it :) it's a free world my friend xD
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
I was always taught that square rooting one made one. IDK though, cause I'm a n00b, just I learned it that way :)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
lol  scroll down and look at "all 2nd roots of 1"
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
Wow. Parth was so right, the whole while. WOW! :D
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
@dumbsearch2 no he isnt :/
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
@AravindG I know it's hard to admit :/
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.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!
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
@saifoo.khan had the same misconception But we convinced him in a post before .
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
@AravindG What was saif's question?
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
Look, WolframAlpha agreed with @ParthKohli. I've used their API before, they're extremely accurate.
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
Maybe @.Sam. can assist me here
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.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.
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
What did Saifoo ask exactly?
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
I told you we never neglected the roots The are 2 roots : \(+\sqrt{1}\) and \(\sqrt{1}\)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
\(\sqrt{1}\) is like writing down \(\sin(\pi) \) for an answer. You actually want to write the numerical value.
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
Gimme some time..let me think of how to explain it in a simpler way
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[x^2 = 1 \Rightarrow x^2  1 = 0 \Rightarrow (x + 1)(x  1) = 0 \Rightarrow x = \pm 1\]
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
yes completely right ^
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
and how we got it ? x=\(+\sqrt{1} \)and \(\sqrt{1}\)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
WHY ARE YOU INCLUDING \(+\sqrt{1}\) AND \(\sqrt1{}\)???????? I AM SIMPLY ASKING IF YOU CAN DISPROVE \(\sqrt1 = \pm 1 \)
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
Wow. Caps. Stay nice, buddies :P
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
lol, you got the opposite. The domain is \(x \ge 0\) and the range is any real number.
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
Wow! @experimentX is dropping his 2 cents! :)
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
The range is positive real numbers
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
That's nowhere in the definition of the square root relation.
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
ok I think Sam and Experimentx can make it clear
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.0
Awesome! @.Sam. is an awesome guy :)
 10 months ago

experimentXBest ResponseYou've already chosen the best response.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.
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
The match is now 21.
 10 months ago

dumbsearch2Best ResponseYou've already chosen the best response.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. :)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
I have gotta go. Please keep replying
 10 months ago

.Sam.Best ResponseYou've already chosen the best response.1
\[\sqrt{x^2}=x \\ \\ +\sqrt{4}=+2 \\ \\ \sqrt{4}=2 \\ \\ \sqrt{4} \cancel{=}\pm2\] For situations with \(y^2=a\), then \[y=\pm \sqrt{a}\]
 10 months ago

AravindGBest ResponseYou've already chosen the best response.0
Notice the definition @ParthKohli
 10 months ago

sasogeekBest ResponseYou've already chosen the best response.0
@.Sam. blame my math teacher! lol :P but yeah, i concur :)
 10 months ago

experimentXBest ResponseYou've already chosen the best response.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
 10 months ago

henpenBest ResponseYou've already chosen the best response.0
I'm pretty sure for a function to be continuous it must be 1to1.
 10 months ago
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