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SolomonZelman
 one year ago
where are the properties faulty?
SolomonZelman
 one year ago
where are the properties faulty?

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle i=i^{1}=i^{4\times(1/4)}=\left(i^4\right)^{1/4}=\sqrt[4]{1}=1 }\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3It's like \(1=\sqrt{(1)^2}=\sqrt 1= 1\), whenever we square an expression, we need to get rid of extraneous roots by back substitution. There are 4 fourthroots of 1, \(\pm 1~ and~ \pm i\).

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Oh, in the example you proposed, i think \(\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}\) doesn't work if √a and √b are both complex. that is what i heard

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1but yeah it is a similar accomplishment

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1but even if i²=±1, then i still get i=1

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1i think i got what you are trying to say

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1the even root of a constant is presumed to be positive, but that doesn't apply to i, so you get ±1. But, how would then (i^4)^(1/4)= 1 not be true?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}\] is true if \(a,b>0\) otherwise it is not

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3There are 4 fourthroots of 1, ±1 and ±i. +1 is one of them. The choice depends on the context.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Alright, yes, thanks:)
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