Is the Set of Pairs of Real Numbers with Non-Negative First Term a Vector Space?

[trixitium]

Homework Statement

Determine if the following set is a vector space under the given operations. List all the axioms that fail to hold.

The set of all pairs of real numbers of the form (x,y), where x >= 0, with the standard operations on R^2

Homework Equations

The Attempt at a Solution

By the axioms of a vector space the set fail on hold this axiom:

for each u in V, there is an object -u in V, called negative of u, such that u + (-u) = (-u) + u = 0.

If the x term in the pair (x,y) is positive (or zero) then -u = (-x, -y) can not exists. Thus, the negative of u does not exist, and V is not a vector space.

Is this correct?

 

[Mark44]

Homework Statement

Determine if the following set is a vector space under the given operations. List all the axioms that fail to hold.

The set of all pairs of real numbers of the form (x,y), where x >= 0, with the standard operations on R^2

Homework Equations

The Attempt at a Solution

By the axioms of a vector space the set fail on hold this axiom:

for each u in V, there is an object -u in V, called negative of u, such that u + (-u) = (-u) + u = 0.

If the x term in the pair (x,y) is positive (or zero) then -u = (-x, -y) can not exists. Thus, the negative of u does not exist, and V is not a vector space.

Is this correct?

It's OK as far as you went, but you have some more work to do. You need to check all the axioms. There is at least one more that isn't satisfied.

 

[trixitium]

It also fails in:

K is any scalar, u is in V, ku is in V.

u = (x,y)

If I choose k < 0, then ku = k(x,y) = (kx,ky) and kx < 0 and ku is not in V.

 

[HallsofIvy]

Note that if the problem had asked only if this was a vector space, you could have stopped after showing one axiom did not hold. But this problem specifically asks you to list all axioms that do not hold.