0+x = x

need a proof, that 0+x = x
while x ∈ V and (V, +, •) is a vector space over F

If by zero you mean the zero vector, then the above is the definiton of the zero (identity) element of V.  Do you mean a proof that "the zero scalar in F times something in V equals the zero vector in V"?

0 * x  = (0 + 0) * x = 0*x + 0*x
cancel 0*x from each side to get 0*x = 0_V

No, not scalar times vector
btw: Is this legal^^?
-> 0x = (1-1)x = x -x = 0

yes, Zero is the zero vector
is 0+x = x an axiom then? Oo

http://en.wikipedia.org/wiki/Vector_space
Axiom 3

>>4
And probably Axiom 2 as well, to get 0 + x = x

ok, thanks. but whats with -0 = 0 ?
this is not an axiom...

>>  0x = (1-1)x = x -x = 0
This is sort of legal, but it uses -1 * x = (-x), which is not an axiom.  By (-x) I mean the additive inverse of x.  The proof of -1 * x = (-x) is easy using 0*x = 0 though.

>>is 0+x = x an axiom then? Oo
Yep, because V is a (abelian) group with composition denoted by "+".

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let me add to my comment (>>7).  What you wrote:

>>  0x = (1-1)x = x -x = 0

is the idea of the proof that -1 * x = -x once you know 0 * x = 0.  The proof would be:

0 = 0*x = ( 1+ (-1) ) * x = 1*x + (-1) * x = x + (-1) * x

hmmm, is this wrong ->
-x =
0 - x =
x(0-1) =
x(-1) =
(-1)x

you used -x = -1 * x going from the 2nd line to the 3rd line; the distributivity axiom only states that multiplication distributes over +.

I feel stupid...

Using the wikipedia axioms:

>>6 -0=0
0 + (-0) = 0 (axiom 4, additive inverse)
-0 = 0 (axiom 3, additive identity)

>>10 -x = (-1)x
This one is a bit long, there is probably a shorter way.  First we need to show that 0x = 0.
Note that the two zeros are different though, one is a scalar, the other is a vector.
0 + 0 = 0 (field axiom additive identity, these are scalars)
(0 + 0)x = 0x (acting on x)
0x + 0x = 0x (axiom 6 distributivity)
(0x + 0x) + -(0x) = 0x + -(0x) (Add -(0x) to both sides, use axiom 4 for existence)
0x + (0x + -(0x)) = 0x + -(0x) (axiom 1 associativity)
0x + 0 = 0 (Axiom 4 additive inverse)
0x = 0 (Axiom 3 additive identity)

Now the rest follows easily
(1 + -1) x = 0 (field axiom, additive inverse)
1x + (-1)x = 0 (axiom 6 distributivity)
x + (-1)x = 0 (axiom 8 scalar multiplication identity)
Hence (-1)x is an additive inverse of x, so by axiom 4
(-1)x = -x

If by 0 you mean the zero vector, then simply,

0 + x
= (0, 0, 0,...) + (x_1, x_2, x_3, ...)
= (0 + x_1, 0 + x_2, 0 + x_3, ...)
= (x_1, x_2, x_3, ...)
= x.

ahh... thanx

ONLY IF x = 0

0+0=0

PROBLEM SOLVD

>need a proof, that 0+x = x

1.  Add nothing to everything and you get the same shit you started with.

Qed

42.  THE ANSWER IS 42.

>>14,17
fail

>>16,18
troll

well, x could be anything, if ur not adding anything to it, it wont change

>>20
No, x is an element of V, and you are adding something to it, namely the vector 0.

>>17
>>20

Fail. There is no intuition in math.

Hey guys, you all suck at math.

0 + x = x can't be proven for an arbitrary vector space, it's an axiom. You only prove that 0 + x = x when you're proving that something is a vector space.

>>23

Welcome to yesterday.


See:
>>4
>>5

INFINITE IS NOT A NUMBER!

typedef int infinity;

It is now!

x=0

Thread over.

>>19
thaknk you captain obvious

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Marijuana MUST be legalized.