@AravindG \(\sqrt{1} = \pm 1\) disprove.

- ParthKohli

@AravindG \(\sqrt{1} = \pm 1\) disprove.

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- katieb

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- AravindG

Ok first of all Theory part

- dumbsearch2

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

- dumbsearch2

Let's kick back
and watch the greenies argue

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## More answers

- 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

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

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

Notice that square root function always remains positive in the process

- 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

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

- AravindG

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

- ParthKohli

And what is your point that you are making?

- AravindG

The square root function always remains positive

- ParthKohli

Why does my use of the definition not convince you?

- 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

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

- AravindG

This is how square root function is defined

- AravindG

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

- AravindG

|dw:1369910211377:dw|

- AravindG

This is how it would look graphically

- AravindG

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

- ParthKohli

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

- AravindG

yes 100%

- dumbsearch2

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

- 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

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

- ParthKohli

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

- 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

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

- sasogeek

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

- 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

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

- dumbsearch2

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

- AravindG

@dumbsearch2 no he isnt :/

- dumbsearch2

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

- 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

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

- ParthKohli

@AravindG What was saif's question?

- AravindG

No it didnt

- dumbsearch2

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

- AravindG

Maybe @.Sam. can assist me here

- 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

@.Sam.

- ParthKohli

@experimentX

- dumbsearch2

@dumbsearch2

- dumbsearch2

:D

- ParthKohli

What did Saifoo ask exactly?

- AravindG

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

- ParthKohli

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

- AravindG

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

- ParthKohli

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

- AravindG

yes completely right ^

- AravindG

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

- ParthKohli

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

- dumbsearch2

Wow.
Caps.
Stay nice, buddies :P

- ParthKohli

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

- dumbsearch2

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

- AravindG

neglect that reply

- AravindG

The range is positive real numbers

- ParthKohli

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

- AravindG

ok I think Sam and Experimentx can make it clear

- AravindG

lets wait

- dumbsearch2

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

- 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

The match is now 2-1.

- 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

I have gotta go. Please keep replying

- .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

Notice the definition @ParthKohli

- sasogeek

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

- 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

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

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