Vector Space: Valid Addition Defined

[Shackleford]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110606_200153.jpg

The book says x + y = 0 is not satisfied. That is, for each x in V, there is a y that satisfies the equation, which is an additive inverse. How vector addition is defined, for each x, I could simply set b₁ = -a₁ and b₂ = 0 to satisfy the additive inverse.

 

[micromass]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110606_200153.jpg

The book says x + y = 0 is not satisfied. That is, for each x in V, there is a y that satisfies the equation, which is an additive inverse. How vector addition is defined, for each x, I could simply set b₁ = -a₁ and b₂ = 0 to satisfy the additive identity.

Hi Shackleford!

The problem is that the additive identity is not (0,0) here. Indeed, the additive identity (a,b) must satisfy

$$(x,y)+(a,b)=(x,y)$$

and thus

$$(x+a,yb)=(x,y)$$

but with (a,b)=(0,0), we have

$$(x,y)+(0,0)=(x+0,y0)=(x,0)$$

which is not what we wanted. So (0,0) is not the additive identity. Can you figure out what is?

 

[Dick]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110606_200153.jpg

The book says x + y = 0 is not satisfied. That is, for each x in V, there is a y that satisfies the equation, which is an additive inverse. How vector addition is defined, for each x, I could simply set b₁ = -a₁ and b₂ = 0 to satisfy the additive identity.

But the identity isn't (0,0) is it? What is the identity?

 

[Shackleford]

Wait second. I meant additive inverse, not identity.

 

[micromass]

Wait second. I meant additive inverse, not identity.

I know you did. But you assumed that (0,0) was the additive identity, which is not true!

 

[Shackleford]

I know you did. But you assumed that (0,0) was the additive identity, which is not true!

No. I didn't. I did not state what the additive identity is. The additive identity would have to be (0,1).

 

[Shackleford]

Oh, I see now.

I completely overlooked that part.

 

[Shackleford]

Even still, I could define b₂ = (a₂)^-1.

Oops. What if it's zero. Then, it doesn't work.