Answer:
Properties 1,2, 5(a) and 5(c) are satisfied, the rest of the properties arent valid.
Step-by-step explanation:
Note that both sum and scalar multiplication involves in exchanging the order from that main coefficient with the independent term before doing the standard sum/scalar multiplication.
Property 1 and 2 apply because by exchanging the order of 2 coefficients of a polynomial we obtain a polynomial of degree at most 2, and then we can conclude both properties are valid becuase standard sum of 2 polynomials of degree 2 or less or standard scalar multiplication of a polynomial with a real number will give as a result a polynomial of degree 2 or less.
Property 3 does not apply: Suppose that Property 3 is valid, lets call v = ax² +bx +c the neuter of V. Since v is the neuter, then 0 should be fixed by the neuted, therefore 0 = 0+v = (0x² + 0x + 0) + (ax² +bx +c) = cx²+b²+a.
0 is fixed by v only if c = b = a = 0. Thus, v = 0. If 0 is the neuter, then it should fix x², however 0 + x² = (0x²+0x+0) + (x²+0x+0) = 1. This is a contradiction because x² is not 1. We conclude that V doesnt have a neuter vector. This also means that property 4 doesn't apply either. A set without zero cant have additive inverse
Let v = v2x² + v1x + v0, w = w2x² + w1x + w0. We have that
- v + w = (v0+w0) * x² + (v1*w1) * x + (v2*w2)
- w + v = (w0+v0) * x² + (w1*v1) * x + (w2*v2)
Since the sum of real numbers is commutative, we conclude that v+w = w+v. Therefore, property 5(a) is valid.
Property 5(b) is not valid: we will introduce a counter example. lets use v = x², w = x²+1, z = 1, then
- (v+w)+z = (x²+2)+1 = 3x² + 1
- v + (w+z) = x² + (2x²+1) = x²+3
Since 3x²+1 ≠ x²+3, then the associativity rule doesnt hold.
Property 5(c) does apply. If v = v2x²+v1x+v0 and w = w2x²+w1x+w0, then we have that, for a real number c
- c*(v+w) = c*( (v0+w0)x² + (v1+w1)x + (v2+w2) ) = c*(v2+w2) x² + c*(v1+w1) x + c(v0+w0)
- c*v + c*w = (cv0x²+cv1x+cv2)*(cw0x²+cw1x+cw2) = (cv2+cw2)x²+(cv1+cw1)x+(cv0+cw0)
Note that both expressions are equal due to the distributive rule of real numbers. Also, you can notice that his property holds because in both cases we <em>'swap variables twice'. </em>For this same argument neither of properties 5d and 5e apply, because on one term we swap variables just once and on the other term we swap variables twice. I will give an example with the vector x² + x and the scalars 1 and 2.
- (1+2)*(x²+x) = 3*(x² + x) = 3x + 3
- 1*(x²+x)+2*(x²+x) = (x+1)+(2x+2) = 3x²+x (≠ 3x + 3)
- (1*2)*(x²+x) = 2*(x²+x) = 2x+2
- 1*(2*(x²+x)) = 1*(2x+2) = 2x²+2x (≠ 2x+2)
Property f doesnt apply due to the swap of variables. for example, if v = x², 1* v = 1*x² = 1 ≠ v.
