Proofs field axioms 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) $\forall x\neq 0 \in \mathbb{K} \exists y \in \mathbb{K}: xy=1 \rightarrow y = x^{-1}= \frac{1}{x}$
 A: 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 $\;-(x+y)\;$ ...
A: 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 $x+y$.
A: We are so used to the operators in arithmetic that we don't realise when they are being used in a way that has yet to be defined.
In the 4th axiom for the additive operator there is an identification of y with -x, and thereafter the minus sign is blithely tossed around and assigned elsewhere as we know it can be from the arithmetic we have been doing for years. But there is nothing in the axioms that says that you can divorce the minus sign from the real number in this way: the -x is an indivisible whole just as the y is.
However, we can divorce the sign from its number and toss it about as we know we can if we can prove that -1.x = -x; and, in general, that -x.y = x.-y (note that we rely on x.0 = 0 via x.(y + 0) = x.y + x.0):
     (x + -x).y = x.y + -x.y = 0

hence
     -x.y = -(x.y) ;

also
     x.(y + -y) = x.y + x.-y = 0

hence
     x.-y = -(x.y) ;

thus
     -x.y = x.-y

In particular
     -1.x = -x.1 = -x

