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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

$$-(-x) = x.$$

amWhy
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Alexander John
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    Looks good to me. – B. Mehta Aug 09 '18 at 22:41
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    This is correct. The only thing I'd suggest is, when using associativity, add the parentheses directly into your work. For example, when adding $-x$ and $-(-x)$, write $x + \left[(-x) + (-(-x))\right] = x + 0 = x$. You might also want to note that this final step uses the additive identity axiom. But, this proof is certainly correct. –  Aug 09 '18 at 22:56

2 Answers2

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Yes it seems to be correct indeed by axioms we have

$$-x-(-x) = 0 $$

then

$$x=x\iff x+[-x-(-x)] = x+0 \iff -(-x) = x $$

user
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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:

$$-(-x)=x$$

Michael McGovern
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