When I was researching on imaginary numbers, I came across this proof -

\(i=\sqrt { -1 } =\sqrt { -\frac { 1 }{ 1 } } =\frac { \sqrt { 1 } }{ \sqrt { -1 } } =\frac { 1 }{ i } =-i\)

which is clearly wrong. But even after analyzing it carefully, I couldn't find the error in it.

Where exactly is the error in the proof?

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TopNewestwait how is 1/i equal to negative i? – Jonathan Hsu · 2 years, 2 months ago

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– Aryan Gaikwad · 1 year, 11 months ago

Multiple 1/i by i/iLog in to reply

root(a/b) = roota/rootb

only for positive numbers – Dev Sharma · 1 year, 10 months ago

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For complex numbers (unlike non-negative real numbers) the square root is defined as a multi-values function: \(\sqrt{-1}=\pm{i}\). If \(z\) is any non-zero complex number, then \(\sqrt{z}\) has two values.

Does that help? – Otto Bretscher · 2 years, 2 months ago

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– Jonathan Hsu · 2 years, 2 months ago

wait so if you take -1 and raise it to the 3/2 power, it would be i and -i? because -1^3/2 = (/sqrt{-1})^3/2 = /sqrt{-1} * /sqrt{-1} * /sqrt{-1} = -1 * /sqrt{-1} = -i, but it can't be i...Log in to reply

– Otto Bretscher · 2 years, 2 months ago

Again, \((-1)^{3/2}=\pm{i}\) , with the "principal value" being \(-i\).Log in to reply