Exercise with proposed solution
So for this question, the functions I am getting for the additive inverse are the following:
-u and -6-u
following the process of the answer in the following question:
Finding the additive inverse in a vector space with unusual operations
Which will give me -3 and -3 for the vector.
For some reason it is still wrong.
I think I am misunderstanding something
Update: (u1,u2) element of V, (v1,v2) element of V: Addition is defined as (u1+v1-3, u2+v2+3) and not as simply (u1+v1, u2+v2). So I can't just say that the additive inverse is -(u2,u2). If I set u1+v1-3=3 and u2+v2+3=-3, I get v1=-u1 and v2 = -6-u2 which would represent the i functions as mentioned above. However, they don't seem to work.
$\endgroup$ 41 Answer
$\begingroup$The first several are correct but you seem to have misunderstood the last question. You are asked for the additive inverse of (x, y) which will depend on x and y, not constants. (x, y)+ (p, q) is defined as (x+ p- 3, y+ q+ 3). Since the additive identity is (3, -3) we must have x+ P- 3= 3 and y+ q+ 3= -3 so p= 6- x and q= -y- 6.
The additive inverse of (x, y) is (6- x, -y- 6).
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