Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

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

share|improve this question
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. – Peter 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
2  
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 $-(x+y)=(-1)\cdot(x+y)=(-1)\cdot x+(-1)\cdot y$ – Peter Tamaroff Jul 10 '12 at 17:48

2 Answers

up vote 1 down vote accepted

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

share|improve this answer
up vote 2 down vote

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.