\(-1 \times -1\) = \(1 \times 1\)

By cross multiplication

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

Taking square root on both sides

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

We know that \(i\) = \(\sqrt{-1}\)

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

\(i \times i\) = \(\sqrt{1} \times \sqrt{1}\)

\(i^{2}\) = \(1\)

Also we know \(i^{2}\) = \(-1\)

This implies \(-1\) = \(1\)

Taking log on both sides we get \(log -1\) = \(log 1\)

Also \(1\) =\(-1^{2}\)

\(log -1^{3}\) = \(log -1^{2}\)

\(3 log -1\) = \( 2 log -1\)

Since \(log -1\) is not zero, we can divide by \(log -1\) on both sides we get

\(3\) = \(2\)

\(3\) = \(1\) + \(1\)

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## Comments

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TopNewestSee Proof that -1 = 1

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This question was asked to me by one of my friends and I couldn't answer him and he told me that it was ajoke. But because of you I have come to know the solution. Thanks a lot.

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but can we use the idea of log of a negative no. in the proof when it is not defined....???

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hey,revan check out ur mail (revandon007@gmail.com),a surprise is waiting for u...........

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Wow .. :D Amazing

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just add more problems like this...

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