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I just read in a book the following dialogue:

"'Two negatives make a positive, am I correct?' Andret smiled. 'You are indeed. At least for operations in which the identity element is one.'"

Anyone care to elaborate as to why this is true?

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    Could "operations in which..." just be an elaborate way to say multiplication? Taken literally it's not true; under the operation $x\oplus y=x+y-1$, $\mathbb R$ is a group with identity $1$, yet $(-1)\oplus(-1)=-3$. – stewbasic Jul 12 '16 at 03:51
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    @stewbasic It's an amateur math nerd's joke, so can't fault them too much. I agree it's meant to indicate multiplication. – Matt Samuel Jul 12 '16 at 03:57
  • That is a misleading statement for any reader other than one who already knows the exact meaning. It is misleading because it seems we're talking about one operation (at a time). If there were only one operation, then the negation is inversion with respect to that operation. In that case, either it is not true: $a^{-1} * b^{-1} \ne a * b$ (in general); or it is true in a way that is clearly not the intended one: $(a^{-1})^{-1} = a$ [this is not the intended meaning, for this works even when the identity is not "one": $-(-a) = a$]. – M. Vinay Jul 12 '16 at 03:58
  • The intended meaning is that there are two operations: $+$ and $$, where negation is defined with respect to $+$ and the identity for $$ is $1$. Then, $(-a)(-b) = ab$, but only with some more assumptions on $+$ and $*$. – M. Vinay Jul 12 '16 at 04:02
  • Thank you all for clearing that up for me. –  Jul 12 '16 at 04:03
  • Oh, and even the terms "negative" and "positive" have to be interpreted in a particular way. For example: $(-i) \times (-i) = -1$ [though of course, it's still true that $(-i) \times (-i) = +i^2$]. The problem here is that the complex numbers do not form an ordered field. – M. Vinay Jul 12 '16 at 04:09

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