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I have problem with proofs in vector space. First is

$\vec x+(-(\vec y+\vec z))=(\vec x+(-\vec y))+(-\vec z)$

and the second

$a\cdot \vec x+b \cdot \vec y=b \cdot \vec x+a\cdot \vec y \Leftrightarrow a=b \vee \vec x=\vec y $

Could anyone help me with this? I'm sorry for my bad english.

In the second task I have:

$a\cdot \vec x+b \cdot \vec y $ I have sentece that $ \vec x = \vec y$ or $a=b$

So i'm changing $ \vec y $ on $ \vec x$

$a\cdot \vec x+b \cdot \vec x = (a+b)\vec x = (a+b) \vec y =...$

And i don't know what can i do next.

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I'm not sure what the variables are, can you specify them? –  xiamx Dec 1 '12 at 22:41
I have specified what variables are vectors and which ones aren't. Could you please check that it is correct? –  Arthur Dec 1 '12 at 22:46
What is $\cdot$ ? –  Belgi Dec 1 '12 at 22:46
@Belgi I assume it's multiplication (by scalar) –  Arthur Dec 1 '12 at 22:47
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Julian Kuelshammer Dec 1 '12 at 22:54

1 Answer 1


For the first one show that, by definition, $$-(v_{1}+v_{2})=-v_{1}-v_{2}$$

where $v_{i}$ are vectors.

For the second use $$\alpha v-\beta v=(\alpha-\beta)v$$

where $\alpha,\beta$ are scalars, $v$ is a vector.

Also use $a-b=-(b-a)$

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Hmm ok I'm trying do first. I'm starting convert left side but I'm ending with $(x+(z-y)$ ;( –  Emil Dec 1 '12 at 23:15
@Emil - How come ? you should have $x+(-y-z)=x-y-z$ –  Belgi Dec 1 '12 at 23:18
$x+(-y+z)=x-(y-z)=x+(z-y)=x-y+z$ (BTW, should I writte it here or in new post?) –  Emil Dec 1 '12 at 23:23
@Emil - thats not he left side on the post...correct me if I'm wrong –  Belgi Dec 1 '12 at 23:25
Oh damn it, you're right, ok now I can do that and start with second ;) –  Emil Dec 1 '12 at 23:30

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