A = {(x,y)| x,y belong R}

scalar multiplication and addition are defined as

a(x,y) = (ax,ay)

(x1 , y1) + (x2 , y2) = (x1 + x2 + 1, y1+y2+1)

1) can we define zero vector as (-1,-1)

2) can we define -(x,y) as (-x-1,-y-1)

3) is A a vector space

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- December 23rd 2012, 09:05 AMscnakandalaIs this a vector space
A = {(x,y)| x,y belong R}

scalar multiplication and addition are defined as

a(x,y) = (ax,ay)

(x1 , y1) + (x2 , y2) = (x1 + x2 + 1, y1+y2+1)

1) can we define zero vector as (-1,-1)

2) can we define -(x,y) as (-x-1,-y-1)

3) is A a vector space - December 23rd 2012, 10:45 AMemakarovRe: Is this a vector space
Good questions. Why don't you try answering them?

- December 23rd 2012, 10:53 AMHallsofIvyRe: Is this a vector space
Your scalar mutiplication must satisfy a(u+ v)= au+ av. Is that true?

- December 23rd 2012, 11:11 AMscnakandalaRe: Is this a vector space
yes

- December 23rd 2012, 11:14 AMscnakandalaRe: Is this a vector space
I don't see any reason why first and second statements are wrong.

But the theorems -1u = -u and 0u = 0 contradicts my answer. - December 23rd 2012, 11:31 AMemakarovRe: Is this a vector space
Please show us the proof.

Let's see. "Can we define -(x,y) as (-x-1,-y-1)?" If yes, then -(x, y) + (x, y) = (-x - 1, -y - 1) + (x, y) = (-x - 1 + x + 1, -y - 1 + y + 1) = (0, 0). Have you done this computation? How does it square with the answer to question 1?

The definition for unary minus can be adjusted so that -(x, y) + (x, y) = (-1, -1). However, post #3 points out a real problem.

Yes, -1(x, y) = (-x, -y) according by definition, but then -1(x, y) + 1(x, y) ≠ 0(x, y).