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I inserted the following in the calculator:

$$--2$$

and the calculator gave me a result of $2$.

Is $--2$ by itself a valid Math expression, or did the calculator added $0$ or something before it (e.g.. the calculator might have turned $--2$ into $0--2$)?

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  • $\begingroup$ You can think of $"-"$ as an operator you can apply to any number to obtain the additive inverse. So you can apply it to $-2$. This is formalised in group theory, for example, where one of the group axioms is that every element has an inverse [technical note: other axioms exist, but they are equivalent]. $\endgroup$
    – Mark Bennet
    Commented Sep 28, 2020 at 17:41

3 Answers 3

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There is nothing wrong with this. It may be interpreted as $-(-2)$, which is $2$. If the calculator somehow inserted a $0$, it is still right since $0 - (-2)= 0+2=2$.

If you're okay with negatives, think of $-(-2)$ as $(-1)(-1 \cdot 2)$ and note that $(-1)\cdot(-1)=1$, so you just get $2$ at the end.

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It is the additive inverse of the additive inverse of two. That is, we have $$(--2)+(-2)=0$$ by definition, so that $$(--2)-2=0\implies --2=2.$$

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$--2$ is equivalent to $-(-2)$ or $-1 \cdot (-1 \cdot 2)$.
$-1 \cdot -1$ is just $1$, so we get $1 \cdot 2$, which is equal to $2$.

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