Does V = {RXR} Satisfy the Vector Space Condition?

[Catchfire]

Homework Statement

Let V = {RXR} with addition defined as (a₁,a₂) + (b₁,b₂) = (a₁ + b₁, a₂b₂)

Show the vector space condition, for each x in V there exists a y in V such that x + y = 0, fails for the V defined above.

Homework Equations

The Attempt at a Solution

a + d = (a₁,a₂) + (d₁,d₂) = (a₁ + d₁, a₂d₂), so let d₁ = -a₁ and d₂ = 0, therefore (a₁ - a₁, 0*a₂) = (0,0) = 0

Seems to me that the condition holds. What have I missed?

 

[jbunniii]

Homework Statement

Let V = {RXR} with addition defined as (a₁,a₂) + (b₁,b₂) = (a₁ + b₁, a₂b₂)

Show the vector space condition, for each x in V there exists a y in V such that x + y = 0, fails for the V defined above.

Homework Equations

The Attempt at a Solution

a + d = (a₁,a₂) + (d₁,d₂) = (a₁ + d₁, a₂d₂), so let d₁ = -a₁ and d₂ = 0, therefore (a₁ - a₁, 0*a₂) = (0,0) = 0

Seems to me that the condition holds. What have I missed?

I agree that it does hold. However, there is another vector space axiom that is not satisfied. Can you see which one?

 

[HallsofIvy]

When you write "x+ y= 0", exacty what do you mean by "0"? You have, apparently, assumed that it meant (0, 0) but there "vector space condition" requiring that, for each x, there exist y such that x+y= (0, 0).

There is a vector space condition that there exist a "zero vector", 0, such that x+ 0= 0 and a condition that, given any x, there exist a y such that x+ y= 0. (The existence of an "additive identity" and the existence of "additive inverses".)

Since we are defining (a₁, a₁)+ (b₁, b₂)= (a₁+ b₂, a₂b₂), (0, 0) is NOT the "additive identiy": (a₁, a₂)+ (0 , 0)= (a₁, 0), NOT (a₁, a₂).

What is the additive identity for this space?

 

[Catchfire]

Thanks for the replies.

(a1, a2)+ (0, 1)= (a1+ 0, a2*1) = (a1, a2) is the additive identity so 0 = (0,1)

(a1,a2) + (b1,b2) = (a1+ b1, a2b2) = (0, 1) means b1 = -a1 and b2 = 1/a2... still missing something.

 

[HallsofIvy]

So what would be the additive inverse of (a, 0)?

 

[Catchfire]

(-a,1/0)!

Haha thanks, appreciate the help.