# Is numerical equivalence in the complex numbers actually an equivalence relation?

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