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?
Is using the - symbol with the Associative Law of multiplication invalid?
1
$\begingroup$
abstract-algebra
proof-writing
axioms
-
0If you're not assuming commutativity I feel like it should be $-(x+y) = -y -x$. – 2012-07-10
-
0@Cocopuffs Commutativity is assumed in the reals. – 2012-07-10
-
0@PeterTamaroff Oh, I didn't see that. Then why not use it? Just $(x+y) + -x - y = (y + x) + -x - y = 0$ by associativity. – 2012-07-10
-
2There'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 $-(x+y)=(-1)\cdot(x+y)=(-1)\cdot x+(-1)\cdot y$ – 2012-07-10
2 Answers
1
$$(x+y)+(-x-y)=(y+x)+(-x-y)=y+(x-x)+(-y)=y+0+(-y)=0$$ So $(x+y)$ is the additive inverse of $(-x-y)$. Hence $-(x+y)=-x-y$
2
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.