Here's the question you clicked on:
ParthKohli
@AravindG \(\sqrt{1} = \pm 1\) disprove.
Ok first of all Theory part
Eh, guise. This is a continuation of http://openstudy.com/study#/updates/51a721e1e4b08f9e59e230bf in some respects. Just so you know, strangers xD
Let's kick back and watch the greenies argue
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 \(-\sqrt{x}\)
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 \(n = 1\).
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 \(-\sqrt{1}\)
Notice that square root function always remains positive in the process
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.
Uh, actually, you are talking about \(-\sqrt{1}\) and \(+\sqrt{1}\). But what exactly \(\mathcal{is}\) \(\sqrt{1}\)?
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?
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.
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
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?
Guys............ http://en.wikipedia.org/wiki/Square_root
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?
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 \(1 \ne (-1)^2\)?
"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
@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
I was always taught that square rooting one made one. IDK though, cause I'm a n00b, just I learned it that way :)
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 :/
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!
@saifoo.khan had the same misconception But we convinced him in a post before .
@AravindG What was saif's question?
Look, Wolfram|Alpha agreed with @ParthKohli. I've used their API before, they're extremely accurate.
Maybe @.Sam. can assist me here
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.
What did Saifoo ask exactly?
I told you we never neglected the roots The are 2 roots : \(+\sqrt{1}\) and \(-\sqrt{1}\)
\(\sqrt{1}\) is like writing down \(\sin(\pi) \) for an answer. You actually want to write the numerical value.
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 \(-\sqrt{1}\)
WHY ARE YOU INCLUDING \(+\sqrt{1}\) AND \(-\sqrt1{}\)???????? I AM SIMPLY ASKING IF YOU CAN DISPROVE \(\sqrt1 = \pm 1 \)
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! :)
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
Awesome! @.Sam. is an awesome guy :)
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.
The match is now 2-1.
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. :)
I have gotta go. Please keep replying
\[\sqrt{x^2}=|x| \\ \\ +\sqrt{4}=+2 \\ \\ -\sqrt{4}=-2 \\ \\ \sqrt{4} \cancel{=}\pm2\] For situations with \(y^2=a\), then \[y=\pm \sqrt{a}\]
Notice the definition @ParthKohli
@.Sam. blame my math teacher! lol :P but yeah, i concur :)
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
I'm pretty sure for a function to be continuous it must be 1-to-1.