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

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Homework Help Overview

The discussion revolves around the vector space defined as V = {RXR} with a specific addition operation. Participants are tasked with examining whether this structure satisfies the vector space conditions, particularly focusing on the existence of additive identities and inverses.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of the addition operation and its implications for the existence of a zero vector and additive inverses. There are attempts to identify the additive identity and to clarify the meaning of the zero vector in this context.

Discussion Status

Some participants express confusion regarding the additive identity and its role in the vector space conditions. There is ongoing exploration of the implications of the defined addition operation, with some suggesting that the additive identity may not be (0, 0) as initially assumed. The discussion is active, with participants questioning each other's reasoning and definitions.

Contextual Notes

Participants are navigating the definitions and properties of vector spaces, particularly in relation to the unconventional addition operation defined for this space. There is a focus on clarifying the conditions that must be satisfied for a set to qualify as a vector space.

Catchfire
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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?
 
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Catchfire said:

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?
 
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?
 
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.
 
So what would be the additive inverse of (a, 0)?
 
(-a,1/0)!

Haha thanks, appreciate the help.
 

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