Question

Testing the 8 Axioms for A={[x, y] where x<0 and y is greater than or equal to 0} showing which fails

Answer 1

We presume that the given set A={[x, y] where x < 0 and y ≥ 0} is to be tested for being a vector space.

The conditions that a vector space V must satisfy are as under:

1. For all X, Y ∈ V, X+Y ∈ V( closure under vector addition).

2. For all X, Y , X+Y = Y+X ( commutativity of vector addition).     

3. For all X, Y, Z , (X+Y)+Z=X+(Y+Z) (Associativity of vector addition).

4. For all x, 0+X = X+0 = X ( Existence of Additive identity)

5. For any X, there exists a -X such that X+(-X)= 0 (Existence of additive inverse)

6. For any scalar k and the vector v ∈ V, the vector kv ∈ V( closure under scalar multiplication).

7. For all scalars r and vectors X,Y, r(X+Y)=rX+rY (Distributivity of vector addition).

8. For all scalars r,s and vectors X , (r+s)X=rX+sX (Distributivity of scalar addition).

9. For all scalars r,s and vectors X, r(sX)=(rs)X( Associativity of scalar multiplication).

10. For all vectors X, 1X=X ( Existence of Scalar multiplication identity).

(i). For, (x,y) and (p,q) ∈ A, (x+p, y+q) ∈ A as x+p >< 0 and y+q ≥ 0 so that the 1^st axiom also does not fail.

(ii). Since vector addition is commutative, the 2^nd axiom also does not fail.

(iii). Associativity of vector addition, the 3^rd axiom also does not fail.

(iv). (0,0) is the additive identity so the 4^th axiom also does not fail.

(v). (-x,-y) is the additive inverse of (x,y) so the 5^th axiom also does not fail.

(vi). If k is negaive or 0, then k(x,y) = (kx,ky) and kx = 0 or kx < 0. Also, ky ≤ 0. Hence the 6^th axiom fails.

(vii). Because of distributivity of vector addition, the 7^th axiom also does not fail.

(viii). Because of distributivity of scalar addition, the 8^th axiom also does not fail.

(ix). Because of associativity of scalar addition, the 9^th axiom also does not fail.

(x). 1(x,y) = (x,y) so that the Scalar multiplication identity exists and the 10^th axiom also does not fail.

Thus, only the 6^th axiom fails for A.