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anonymous
 5 years ago
does the square root of 1, equal 1?
anonymous
 5 years ago
does the square root of 1, equal 1?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i feel stupid... thank you :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually : \[\sqrt(1) = 1\] ^_^ and not 1 alone :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't feel stupid, alot of ppl missed such simple relation :)

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0This is bullpellet. In real analysis the square root is only defined for nonnegative numbers. And because it is a function, it can only assume one value for every argument. So the square root of 1 is that nonnegative number which squared gives 1, so that is certainly 1. And you don't have to mess around with  or the absolute value function. \[x,y \geq 0 ⇒ \left( \sqrt{x} = y ⇔ y^2 = x\right)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0>_< nowhereman watch your language lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually , that's what my prof said though, and I was shocked LOL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1 = \sqrt(1) = \sqrt(1)^2 = \sqrt(1)^2 = ((1))^{1/2} = 1 = 1\] ._. maybe I have misunderstood him?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol...wait a mind...I think I got you now ^_^"

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I'm absolutely sure. ( Only in complex analysis you can describe the squareroot function as a riemannian surface with multiple value)

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0And of course that definition leads to \[\sqrt{x^2} = x\]

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0spaceknight: what are you laughing about?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because i dont understand what you are talking about lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I understood, thank you nowhereman ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you talking about the principal square root discussed in here http://mathworld.wolfram.com/PrincipalSquareRoot.html because it clearly states For example, the principal square root of 9 is 3, although both and 3 are square roots of 9.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, if you want to name it this way. In real analysis, if you are talking about _the_ square root, you mean the principal square root. Of course it is clear that the equation y^2 = x can have 0, 1 or 2 solutions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, this site is beautiful~ thank you spaceknight! ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but not too much info though , still :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why does it matter if you are talking about the principal square root or not? everytime i've had to sqrt anything in equations or whatever I've always been told to include the +/.
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