Proving If a set is a Vector Free Space

[maiad]

Homework Statement

Determine whether set is a vector space. If not, give at least one axiom that is not satisfied.
the set of vectors <a1,a2>, addition and scalar multiplication defined by:
<a1,a2>+<b1,b2>=<a1+b1+1,a2+b2+1>
k<a1,a2>=<ka1+k-1,ka2+k-1>

The Attempt at a Solution

For Vector Addition:
1) well since a1 and a2 is not restricted, the vector spaces are all real entities in V

2) Rule:x+y=y+x;
<a1,a2>+<b1,b2>=<a1+b1+1,a2+b2+1>=<b1,b2>

3)Rule:(x+y)+z=(x+(y+z);
(<a1,a2>+<b1,b2>)+<0,0>= <a1,a2>+(<b1,b2>+<0,0>)

4)Not sure how to prove is this set has a unique vector O in V such that O+x=x+O

5)Rule: There exist a vector where x+(-x)=(-x)+x=O;
Since the set is not restricted, there exist a negative vector where a1+(-a1)=(-a1)+a1

For Scalar Multiplication:
6)Since the set is no restricted, any scalar would be in the space

7)Rule:k(x+y)=kx+ky;
Not sure how to prove this one...

8)Rule: (k1+k2)x=k1x+k2x
nor this one

9)Rule:k1(k2x)=(k1k2)x
or this one

10)Rule: 1x=x
This one obviously satisfies

Can someone give me hints on the ones i didn't get and also see if the ones i did is correct?

 

[jambaugh]

Do not assume the space satisfies all your axioms. Try the ones you are not sure about with some concrete numbers. 2<3,5> and such.

You left of an important part of 4). 0+x = x+0 = x

BTW, When you say something is "obvious" you should double check yourself. Most errors occur in the "obvious" cases because we stop considering them too soon. (I'm not saying you made an error. This is just good general advice.)