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

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

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

03453660Best ResponseYou've already chosen the best response.0
dw:1350311105247:dw
 one year ago

03453660Best ResponseYou've already chosen the best response.0
dw:1350311152497:dw
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.1
true,,i was about to write that only..
 one year ago

zugzwangBest ResponseYou've already chosen the best response.2
I actually looked it up to be sure :P
 one year ago

ZakaullahUETBest ResponseYou've already chosen the best response.0
hmm dw:1350312680888:dw
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
sqrt(1) is only +1 and not 1 @zakaullahUET
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
sqrt is a +ve func..
 one year ago

ZakaullahUETBest ResponseYou've already chosen the best response.0
not necessary a function @shubhamsrg ? can b just a relation a*a=(a)*(a)
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
when 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..
 one year ago
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