Now, let S = {cx^3 | c is any real number}.
1. Let ax^3 and bx^3 be in S.
ax^3 + bx^3 = (a+b)x^3.
Since a and b are real numbers, a+b is a real number, so (a+b)x^3 is in S.
6. Let a be a scalar and let bx^3 be in S.
a * bx^3 = (ab)x^3.
Since a and b are real number, ab is a real number. Therefore, (ab)x^3 is in S.
In this example, in order to prove 5, you must prove 8 first.
8. Let a and b be scalars and let cx^3 be in S.
(a + b) * cx^3 = [(a+b)c]x^3 = (ac+bc)x^3 = acx^3 + bcx^3
because of how real numbers work. Therefore, the distribution of vectors holds.
5. Let ax^3 and bx^3 be in S.
ax^3 + bx^3
= (a+b)x^3
= (b+a)x^3 (since real numbers are commutative under addition)
= bx^3 + ax^3 (by 8)
So commutativity of addition holds.
2. Let ax^3, bx^3, and cx^3 be in S.
(ax^3 + bx^3) + cx^3
= (a+b)x^3 + cx^3
= [(a+b)+c]x^3
= [a+(b+c)]x^3 (since a,b,c are real numbers)
= ax^3 + (b+c)x^3 (by 8)
= ax^3 + (bx^3 + cx^3) (by 8).
Therefore associativity holds.
3. Claim: 0x^3 is the "0" vector.
Proof:
Let ax^3 be in S.
0x^3 + ax^3
= (0+a)x^3
= ax^3 (by the identity property of addition).
By 5, this is commutative.
Therefore, 0x^3 is the additive identity.
4. Let ax^3 be in S. Claim: then -ax^3 is the additive inverse of ax^3.
-ax^3 + ax^3
= (-a+a)x^3
= 0x^3 (since -a and a are real numbers).
By 5, this is commutative.
Since 0x^3 is the identity element, there is an inverse element for all elements in S. Namely, -ax^3 is the inverse of ax^3.
7. Let a be a scalar and let bx^3 and cx^3 be in S.
a(bx^3 + cx^3)
= a*(b+c)x^3
= (ab+ac)x^3 (since a, b, and c are real numbers)
= abx^3 + acx^3.
Therefore elements in S have scalar distribution.
9. Let a and b be scalars and let cx^3 be in S.
a(bcx^3)
= abcx^3
= (ab)cx^3 since a,b,c are real numbers.
Therefore, associativity of scalar multiples holds.
10. Let ax^3 be in S.
1 * ax^3 = ax^3 (since a is a real number).
Therefore, axiom 10 holds.
Since all 10 axioms hold, the set S = {ax^3 | a is a real number} is in the vector space P3. Q.E.D.