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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.

a6i09per5f
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 17 '22 at 22:54
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    If $(u,v)=-(x,y)$ is the additive inverse of $(x,y)$ then we must have $(x,y)+(u,v)=0_V$ I'm getting $$-(x,y)=(-x+6,-y-6)$$ – Matthew H. Apr 17 '22 at 22:57
  • (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). So your solution didn't work because of that. – a6i09per5f Apr 17 '22 at 23:13
  • 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. – a6i09per5f Apr 17 '22 at 23:20

1 Answers1

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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).

George Ivey
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  • I dont thinkI actually misunderstood it. I just made an algebra mistake tbh. However, your solution does not work for some reason. It gives me (3, -3) for the additive inverse which is set as wrong on webwork. – a6i09per5f Apr 18 '22 at 21:51
  • Nevermind. I did misunderstand something. The whole thing that confused me was that I thought I have to put in an absolute number. Your solution worked. Thank you sir. – a6i09per5f Apr 18 '22 at 21:53