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I'm terrible with these kinds of proofs and this might be a duplicate but I didn't find anything.

I need to prove that $\forall x,y \in \mathbb{R}: -(-x)=x$ and $-(x+y) = -x - y$

I know that this is basically trivial but I just don't see how these follow from the basic field axioms of $\mathbb{R}$. Can anybody help me out please?

Axioms for a field $\mathbb{K}$ are with respect to the additive operation $+ : \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$:

1) $\forall x,y,z\in \mathbb{K}: (x+y)+z = x+(y+z)$

2) $\forall x,y \in \mathbb{K}: x+y = y+x$

3) $\exists 0 \in \mathbb{K} $such that$ : x+0=x$

4) $\forall x \in \mathbb{K} \exists y \in \mathbb{K}: x+y=0 \rightarrow y = -x$

for the multiplicative operation:

1) $\forall x,y,z \in \mathbb{K}: (xy)z = x(yz)$

2) $\forall x,y \in \mathbb{K}: xy=yx$

3) $\exists 1 \in K : \forall x \in \mathbb{K}: x1=x$

4) $\forall x\neq 0 \in \mathbb{K} \exists y \in \mathbb{K}: xy=1 \rightarrow y = x^{-1}= \frac{1}{x}$

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If you list your axioms, you'll get quicker and better answers. – Git Gud Apr 27 '14 at 15:07
Start with $0-(-x)=-(-x)$ and also $0=x+(-x)$ – EgoKilla Apr 27 '14 at 15:13
Is the third product axiom really written like that? No quantification on $y$ on the fourth? The third additive axiom is also wrong. And there's no mention whatsoever to the existence of an additive entity. – Git Gud Apr 27 '14 at 15:24
I edited my post. @EgoKilla thanks for the response, I figured it out. Could you give me a hint for how to start with proofs like these? How did you come up with these two equations? – eager2learn Apr 27 '14 at 15:28
Honestly it's just about looking at what axioms seem relevant and playing around. Also sometimes starting with your answer and working backwards can be helpful. But when I did these (last semester) I really just played around with the axioms till I figured them out. – EgoKilla Apr 27 '14 at 15:32
up vote 3 down vote accepted

You need to use uniqueness in some axioms, for example:

$$-x+x\stackrel{\text{dist.}}=(-1+1)x=0\cdot x=0$$

so by the uniqueness of additive inverse, $\;x\;$ is the additive inverse of $\;-x\;$, whose additive inverse, by definition, $\;-(-x)\;$ , and thus $\;-(-x)=x\;$ .

Try to argue something very, very similar for $\;-(x+y)\;$ ...

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By definition $-(-x)+(-x)=0$ and $x+(-x)=0$ and the inverse of each element is unique, leading to $-(-x)=x$. Likewise you can solve the other. Both sides are the inverse of $x+y$.

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