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SolomonZelman

  • one year ago

where are the properties faulty?

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  1. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle i=i^{1}=i^{4\times(1/4)}=\left(i^4\right)^{1/4}=\sqrt[4]{1}=1 }\)

  2. mathmate
    • one year ago
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    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, \(\pm 1~ and~ \pm i\).

  3. SolomonZelman
    • one year ago
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    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

  4. SolomonZelman
    • one year ago
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    but yeah it is a similar accomplishment

  5. SolomonZelman
    • one year ago
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    but even if i²=±1, then i still get i=1

  6. SolomonZelman
    • one year ago
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    i think i got what you are trying to say

  7. SolomonZelman
    • one year ago
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    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?

  8. anonymous
    • one year ago
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    \[\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}\] is true if \(a,b>0\) otherwise it is not

  9. mathmate
    • one year ago
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    There are 4 fourth-roots of 1, ±1 and ±i. +1 is one of them. The choice depends on the context.

  10. SolomonZelman
    • one year ago
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    Alright, yes, thanks:)

  11. mathmate
    • one year ago
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    You're welcome! :)

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