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ParthKohli
 3 years ago
@AravindG \(\sqrt{1} = \pm 1\) disprove.
ParthKohli
 3 years ago
@AravindG \(\sqrt{1} = \pm 1\) disprove.

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AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0Ok first of all Theory part

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0Eh, guise. This is a continuation of http://openstudy.com/study#/updates/51a721e1e4b08f9e59e230bf in some respects. Just so you know, strangers xD

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0Let's kick back and watch the greenies argue

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0We 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
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, 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
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0Notice that square root function always remains positive in the process

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

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0yes I talked abt them because they are the roots for x^2=1

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0And what is your point that you are making?

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0The square root function always remains positive

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Why does my use of the definition not convince you?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0You 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
 3 years ago
Best ResponseYou've already chosen the best response.0Okay first I will provide you the standard definition \[\sqrt{x^2}=x\]

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0This is how square root function is defined

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0Notice the modulus sign on right ..That is why the range of it is positive real numbers

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0This is how it would look graphically

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0I was drawing a rough diagram for square root function @dumbsearch2

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Are you sure that's the definition of a square root?

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0Guys............ http://en.wikipedia.org/wiki/Square_root

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0okay 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?

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

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

dumbsearch2
 3 years ago
Best 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

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

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

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0I was always taught that square rooting one made one. IDK though, cause I'm a n00b, just I learned it that way :)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0lol  scroll down and look at "all 2nd roots of 1"

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0Wow. Parth was so right, the whole while. WOW! :D

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0@dumbsearch2 no he isnt :/

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0@AravindG I know it's hard to admit :/

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Why 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
 3 years ago
Best ResponseYou've already chosen the best response.0@saifoo.khan had the same misconception But we convinced him in a post before .

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@AravindG What was saif's question?

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0Look, WolframAlpha agreed with @ParthKohli. I've used their API before, they're extremely accurate.

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0Maybe @.Sam. can assist me here

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0What 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
 3 years ago
Best ResponseYou've already chosen the best response.0What did Saifoo ask exactly?

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

ParthKohli
 3 years ago
Best 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.

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0Gimme some time..let me think of how to explain it in a simpler way

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0\[x^2 = 1 \Rightarrow x^2  1 = 0 \Rightarrow (x + 1)(x  1) = 0 \Rightarrow x = \pm 1\]

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0yes completely right ^

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0and how we got it ? x=\(+\sqrt{1} \)and \(\sqrt{1}\)

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

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0Wow. Caps. Stay nice, buddies :P

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

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0Wow! @experimentX is dropping his 2 cents! :)

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0The range is positive real numbers

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0That's nowhere in the definition of the square root relation.

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0ok I think Sam and Experimentx can make it clear

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0Awesome! @.Sam. is an awesome guy :)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0root 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
 3 years ago
Best ResponseYou've already chosen the best response.0The match is now 21.

dumbsearch2
 3 years ago
Best ResponseYou've already chosen the best response.0uh. 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
 3 years ago
Best ResponseYou've already chosen the best response.0I have gotta go. Please keep replying

.Sam.
 3 years ago
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 \[y=\pm \sqrt{a}\]

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0Notice the definition @ParthKohli

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0@.Sam. blame my math teacher! lol :P but yeah, i concur :)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0For 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
 3 years ago
Best ResponseYou've already chosen the best response.0I'm pretty sure for a function to be continuous it must be 1to1.
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