In class while messing with fractions and complex numbers I found this "paradox" $$ \sqrt{-1}=\sqrt{-1} $$ $$ \sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}} $$ $$ \frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}} $$ $$ \sqrt{-1}\cdot \sqrt{-1}=\sqrt{1}\cdot \sqrt{1} $$ $$ i\cdot i=\sqrt1 $$ $$ -1=1 $$ Could anybody explain me what is wrong with this passages?
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1This might be related: http://math.stackexchange.com/questions/1122038/what-is-wrong-with-my-proof-1-1?rq=1 – Wojowu Oct 29 '15 at 19:03
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notice that not only i^2 = $\sqrt{-1}$ , (-i)^2 =$\sqrt{-1}$ too – Oct 29 '15 at 19:03
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1https://en.wikipedia.org/wiki/Square_root Go down to "Notes". It explains why step 4 in your "proof" is incorrect. – qmd Oct 29 '15 at 19:04
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This states "The powers of negative real numbers are not always defined and are discontinuous even where defined. In fact, they are only defined when the exponent is a rational number with the denominator being an odd integer. When dealing with complex numbers the complex number operation is normally used instead." – Mark Viola Oct 29 '15 at 19:26
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$\sqrt{ab}\neq \sqrt{a}\sqrt{b}$ iff a or b <$0$
corcia candy
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By definition. And the definition is constructed so as to avoid exactly the contradiction that you bring up in your question. – Daniel R. Collins Oct 29 '15 at 22:21