Is the following a complete proof? It is from the book, $Prealgebra$ by R. Rusczyk, D. Patrick, and R. Boppana
Consider the sum
$$x + (-x) + (-(-x)).$$
That is, we are adding x, its negation -x, and the negation of -x. By associative property of addition, we can add these three in any order. If we start by adding the first two, we have $x + (-x) = 0$, so
$$x + (-x) + (-(-x)) = 0 + (-(-x)) = -(-x).$$
However, suppose we start by adding $(-x) + (-(-x))$ first. Since $(-(-x))$ is the negation of $-x$, we have $(-x) + (-(-x)) = 0.$ So, we find
$$x + (-x) + (-(-x)) = x + 0 = x.$$
We just showed that $x + (-x) + (-(-x))$ equals both $-(-x)$ and $x$, so we must have
$$-(-x) = x.$$
$\endgroup$ 22 Answers
$\begingroup$Yes it seems to be correct indeed by axioms we have
$$-x-(-x) = 0 $$
then
$$x=x\iff x+[-x-(-x)] = x+0 \iff -(-x) = x $$
$\endgroup$ $\begingroup$If we explicitly acknowledge that that existence of additive inverses means that there exists an element $-(-x)$ such that $-x+-(-x)=0$, we have,by the associativity of addition and the reflexivity of equality:
$$(x+-x)+-(-x)=x+(-x+-(-x))$$
By the definition of additive inverse $x+-x=0$ and $-x+-(-x)=0$ so:
$$0+-(-x)=x+0$$
By the definition of additive identity $-(-x)+0=-(-x)$ and $x+0=x$ therefore:
$$-(-x)=x$$
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