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It seems that either numerical equivalence is not an equivalence relation or complex numbers do not have an inverse.

We know that $-1 = i \cdot i$

We also know that $-1 = -i \cdot -i$

If we assume "=" to be an equivalence relation we can get $i \cdot i = -i \cdot -i$

And $i \cdot i = -(i \cdot i)$

If we assume $i^{-1}$ exists, then so does $(i \cdot i)^{-1}$

Thus $ (i \cdot i) (i \cdot i)^{-1} = -(i \cdot i)(i \cdot i)^{-1}$

And $1 = -1$, a contradiction.

But this can't be right, can it?

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up vote 9 down vote accepted

Your step $i\cdot i = -i \cdot (-i)$ implies (?) $i\cdot i = -(i \cdot i)$ is false: $2\cdot 2 = -2\cdot (-2)$, but $2\cdot 2 = 4 \neq -4 = -(2\cdot 2)$.

Or, if you want with the original complex numbers: $i\cdot i = -1 \neq 1 = -(-1) =-(i \cdot i)$.

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The kink is in the step where you "factor out" the (-1) from both "i" s.

$$-(i\cdot i)=(-1)(i\cdot i)=(-1)(-1)=1$$

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