03453660
why i^2 = 1, where ' i ' is (iota).



This Question is Closed

surdawi
Best Response
You've already chosen the best response.
2
\[i = \sqrt{1}\]

03453660
Best Response
You've already chosen the best response.
0
yes then??

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

shubhamsrg
Best Response
You'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..

03453660
Best Response
You've already chosen the best response.
0
yes but what about this one

03453660
Best Response
You've already chosen the best response.
0
dw:1350311105247:dw

03453660
Best Response
You've already chosen the best response.
0
dw:1350311152497:dw

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

shubhamsrg
Best Response
You've already chosen the best response.
1
true,,i was about to write that only..

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

ZakaullahUET
Best Response
You've already chosen the best response.
0
hmm
dw:1350312680888:dw

shubhamsrg
Best Response
You've already chosen the best response.
1
sqrt(1) is only +1 and not 1 @zakaullahUET

shubhamsrg
Best Response
You've already chosen the best response.
1
sqrt is a +ve func..

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

shubhamsrg
Best Response
You'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..