$i=(-1)^{\frac{1}{2}} $ So
$$=[(-1)^3]^{\frac{1}{2}} =(-1)^{\frac{3}{2}} =[(-1)^{\frac{1}{2}}]^3 =i^3 =-i $$
If I solve
$$=i =-i =-i =-i =i? $$
This step its ok?
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2What's the question? – Nathaniel Bubis Jun 28 '17 at 18:22
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2Exponents don't work the same way on negative bases. See this post for a similar fallacy: https://math.stackexchange.com/questions/2303412/where-is-the-fallacy-i-1 – Franklin Pezzuti Dyer Jun 28 '17 at 18:23
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Instead of Nilknarf's link I recommend you start here. This and related questions have been asked umpteen times here. The umbrella tag [tag:fake-proofs] is IMHO appropriate here :-) – Jyrki Lahtonen Jun 29 '17 at 04:50
1 Answers
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Property of exponents can not be applied to non-real and negative bases.
$a^{mn}=(a^m)^n$ if and only if $a>0$ or $m,n \in \mathbb Z$. At least one of these conditions is must.
Jaideep Khare
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@Typhon IMO, question is clear. The OP tried to prove $i=-i$, and thus concluded that $i=0$, and hence is confused that what is the value of $i$. – Jaideep Khare Jun 28 '17 at 19:22