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I was trying to prove that $-(x + y) = -x - y$ and as you can see in the image below, I took the liberty of using the $-$ symbol as a number and applying the associative law with it. Is it kosher in all rigorousness given the axioms professional mathematicians use? enter image description here

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If you're not assuming commutativity I feel like it should be $-(x+y) = -y -x$. –  Cocopuffs Jul 10 '12 at 17:43
    
@Cocopuffs Commutativity is assumed in the reals. –  Pedro Tamaroff Jul 10 '12 at 17:44
    
@PeterTamaroff Oh, I didn't see that. Then why not use it? Just $(x+y) + -x - y = (y + x) + -x - y = 0$ by associativity. –  Cocopuffs Jul 10 '12 at 17:45
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There's a problem in there. You're "pulling out" the $-$ under the "associative law". I guess you know that, given $x$, $(-1) \cdot x =-x$ is its additive inverse. So what you want to do is to use the distributive law as $-(x+y)=(-1)\cdot(x+y)=(-1)\cdot x+(-1)\cdot y$ –  Pedro Tamaroff Jul 10 '12 at 17:48

2 Answers 2

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$$(x+y)+(-x-y)=(y+x)+(-x-y)=y+(x-x)+(-y)=y+0+(-y)=0$$ So $(x+y)$ is the additive inverse of $(-x-y)$. Hence $-(x+y)=-x-y$

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Well the LHS of your equation is just saying "the additive inverse of $x+y$".

So all you have to show is that the additive inverse of $x+y$ really is the RHS of the equation, i.e. $-x-y$, then by uniqueness of inverses in a group the two must be equal.

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