SolomonZelman
  • SolomonZelman
where are the properties faulty?
Mathematics
katieb
  • katieb
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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle i=i^{1}=i^{4\times(1/4)}=\left(i^4\right)^{1/4}=\sqrt[4]{1}=1 }\)
mathmate
  • mathmate
It'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 fourth-roots of 1, \(\pm 1~ and~ \pm i\).
SolomonZelman
  • SolomonZelman
Oh, 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

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SolomonZelman
  • SolomonZelman
but yeah it is a similar accomplishment
SolomonZelman
  • SolomonZelman
but even if i²=±1, then i still get i=1
SolomonZelman
  • SolomonZelman
i think i got what you are trying to say
SolomonZelman
  • SolomonZelman
the 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
  • anonymous
\[\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}\] is true if \(a,b>0\) otherwise it is not
mathmate
  • mathmate
There are 4 fourth-roots of 1, ±1 and ±i. +1 is one of them. The choice depends on the context.
SolomonZelman
  • SolomonZelman
Alright, yes, thanks:)
mathmate
  • mathmate
You're welcome! :)

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