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Ok first of all Theory part

Let's kick back
and watch the greenies argue

Notice that square root function always remains positive in the process

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

And what is your point that you are making?

The square root function always remains positive

Why does my use of the definition not convince you?

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

This is how square root function is defined

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

|dw:1369910211377:dw|

This is how it would look graphically

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

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

yes 100%

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

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

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

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

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

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

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

@dumbsearch2 no he isnt :/

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

@AravindG What was saif's question?

No it didnt

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

Maybe @.Sam. can assist me here

@.Sam.

:D

What did Saifoo ask exactly?

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

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

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

yes completely right ^

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

Wow.
Caps.
Stay nice, buddies :P

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

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

neglect that reply

The range is positive real numbers

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

ok I think Sam and Experimentx can make it clear

lets wait

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

The match is now 2-1.

I have gotta go. Please keep replying

^^

Notice the definition @ParthKohli

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

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