- #1
kntsy
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Is square root a function in this way?
[tex]f:\mathbb C\rightarrow\mathbb R^+[/tex]
However contradiction can be drawn:[tex]\sqrt{x^2}=|x|\text{ and } i^2=-1[/tex]
[tex]\sqrt{-1}=\sqrt{i^2}=|i|=1[/tex]
[tex]\sqrt{-1}=1[/tex] ??What is the problem in the definition or the deduction?
[tex]f:\mathbb C\rightarrow\mathbb R^+[/tex]
However contradiction can be drawn:[tex]\sqrt{x^2}=|x|\text{ and } i^2=-1[/tex]
[tex]\sqrt{-1}=\sqrt{i^2}=|i|=1[/tex]
[tex]\sqrt{-1}=1[/tex] ??What is the problem in the definition or the deduction?
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