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Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0\[i=\sqrt{1}\] \[i^2 = (\sqrt{1})^2 = 1\]

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1its the definition of iota.. that i = sqrt(1) .. it'd be like asking why 2 is called two ? its the definition,,no theoritical reason..

03453660
 2 years ago
Best ResponseYou've already chosen the best response.0yes but what about this one

zugzwang
 2 years ago
Best ResponseYou've already chosen the best response.2I believe \[\sqrt{ab}= \sqrt{a}\sqrt{b}\] only applies if a and b are both nonnegative

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1true,,i was about to write that only..

zugzwang
 2 years ago
Best ResponseYou've already chosen the best response.2I actually looked it up to be sure :P

ZakaullahUET
 2 years ago
Best ResponseYou've already chosen the best response.0hmm dw:1350312680888:dw

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1sqrt(1) is only +1 and not 1 @zakaullahUET

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1sqrt is a +ve func..

ZakaullahUET
 2 years ago
Best ResponseYou've already chosen the best response.0not necessary a function @shubhamsrg ? can b just a relation a*a=(a)*(a)

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1when a no. comes out of sqrt, its in mod.. i.e. sqrt(a^2) = a see sqrt(x) graph,,it gives out only +ve values of y..
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