does the square root of 1, equal 1?
yes
yes
i feel stupid... thank you :)
or -1 :)
or^1/2
just 1 period :D
+/- 1 to be correct
actually : \[\sqrt(1) = |1|\] ^_^ and not 1 alone :)
don't feel stupid, alot of ppl missed such simple relation :)
\[\sqrt(x) = |x|\]
This is bullpellet. In real analysis the square root is only defined for non-negative numbers. And because it is a function, it can only assume one value for every argument. So the square root of 1 is that non-negative number which squared gives 1, so that is certainly 1. And you don't have to mess around with - or the absolute value function. \[x,y \geq 0 ⇒ \left( \sqrt{x} = y ⇔ y^2 = x\right)\]
>_< nowhereman watch your language lol
are you sure?
actually , that's what my prof said though, and I was shocked LOL
\[1 = \sqrt(1) = \sqrt(1)^2 = \sqrt(-1)^2 = ((-1))^{1/2} = |-1| = 1\] ._. maybe I have misunderstood him?
lol...wait a mind...I think I got you now ^_^"
min*
Yes, I'm absolutely sure. ( Only in complex analysis you can describe the square-root function as a riemannian surface with multiple value)
lol wut
And of course that definition leads to \[\sqrt{x^2} = |x|\]
spaceknight: what are you laughing about?
because i dont understand what you are talking about lol
I understood, thank you nowhereman ^_^
still studying DM :)
are 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.
Yes, 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.
Oh, this site is beautiful~ thank you spaceknight! ^_^
but not too much info though , still :)
why 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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