Question

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

Answer 1

Ok first of all Theory part

Answer 2

Eh, guise.
This is a continuation of http://openstudy.com/study#/updates/51a721e1e4b08f9e59e230bf in some respects.

Just so you know, strangers xD

Answer 3

Let's kick back
and watch the greenies argue

Answer 4

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 \(-\sqrt{x}\)

Answer 5

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 \(n = 1\).

Answer 6

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 \(-\sqrt{1}\)

Answer 7

Notice that square root function always remains positive in the process

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

And what is your point that you are making?

Answer 12

The square root function always remains positive

Answer 13

Why does my use of the definition not convince you?

Answer 14

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.

Answer 15

Okay first I will provide you the standard definition

\[\sqrt{x^2}=|x|\]

Answer 16

This is how square root function is defined

Answer 17

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

Answer 18

This is how it would look graphically

Answer 19

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

Answer 20

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

Answer 21

yes 100%

Answer 22

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

Answer 23

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?

Answer 24

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

Answer 25

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

Answer 26

"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

Answer 27

@ParthKohli Wolf agrees

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

Answer 28

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

Answer 29

I was always taught that square rooting one made one.
IDK though, cause I'm a n00b, just I learned it that way :)

Answer 30

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

Answer 31

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

Answer 32

@dumbsearch2 no he isnt :/

Answer 33

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

Answer 34

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!

Answer 35

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

Answer 36

@AravindG What was saif's question?

Answer 37

No it didnt

Answer 38

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

Answer 39

Maybe @.Sam. can assist me here

Answer 40

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.

Answer 41

@.Sam.

Answer 42

@experimentX

Answer 43

@dumbsearch2

Answer 44

:D

Answer 45

What did Saifoo ask exactly?

Answer 46

I told you we never neglected the roots

The are 2 roots :

\(+\sqrt{1}\) and \(-\sqrt{1}\)

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

yes completely right ^

Answer 51

and how we got it ?

x=\(+\sqrt{1} \)and \(-\sqrt{1}\)

Answer 52

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

Answer 53

Wow.
Caps.
Stay nice, buddies :P

Answer 54

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

Answer 55

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