Demonstrate the three axioms of a field that are not satisfied by the set {... -4, -2, 0, 2, 4, 6...} under the usual operations of multiplication and addition
no multiplicative identity
no multiplicative inverses
can't think of a third one ...
To demonstrate the three axioms of a field that are not satisfied by the set {... -4, -2, 0, 2, 4, 6...} under the usual operations of multiplication and addition, let's go through each axiom and see where this set fails to satisfy it.
1. Closure under Addition:
The closure property states that for any two elements a and b in the set, their sum a + b also belongs to the set. In the given set, if we take -4 and 4, their sum is 0 which is not present in the set. Therefore, this set fails to satisfy closure under addition.
2. Closure under Multiplication:
Similar to the closure property for addition, closure under multiplication means that for any two elements a and b in the set, their product ab also belongs to the set. Let's take -2 and 2 from the set. Their product -2 * 2 is -4, which is not in the given set. Hence, this set fails to satisfy closure under multiplication.
3. Existence of Additive and Multiplicative Inverses:
For every element a in the set (except the additive identity 0), there must exist an additive inverse -a in the set such that a + (-a) = 0. In the given set, there is no additive inverse for 6 since -6 is not present in the set. Hence, this set fails to satisfy the existence of additive inverses.
Additionally, for every element a in the set (except the multiplicative identity 1), there must exist a multiplicative inverse a^(-1) in the set such that a * a^(-1) = 1. In this set, there is no multiplicative inverse for 2 since it does not have an integer reciprocal in the set.
Therefore, the given set {... -4, -2, 0, 2, 4, 6...} under the usual operations of multiplication and addition does not satisfy the closure property for addition, closure property for multiplication, and the existence of additive and multiplicative inverses.
To demonstrate the three axioms of a field that are not satisfied by the set {..., -4, -2, 0, 2, 4, 6, ...} under the usual operations of multiplication and addition, we need to show that at least one of the axioms is violated.
1. Closure under addition: For any two elements a and b in the set, a + b must also be in the set.
Let's take -2 and 4 from the set. According to closure, -2 + 4 should be in the set. However, -2 + 4 = 2, which is not in the given set. Therefore, closure under addition is violated.
2. Existence of additive inverse: For every element a in the set, there must exist an element -a in the set such that a + (-a) = 0.
Consider the number 6 from the set. To satisfy this axiom, there should be an element in the set such that 6 + (-6) = 0. However, -6 is not in the set. Thus, the existence of an additive inverse is violated.
3. Associativity of multiplication: For any three elements a, b, and c in the set, (a * b) * c = a * (b * c).
Let's take -2, 4, and 6 from the set. If we calculate (-2 * 4) * 6, we get -8 * 6 = -48. However, if we calculate -2 * (4 * 6), we get -2 * 24 = -48 as well. In this case, associativity of multiplication is satisfied.
Therefore, the set {..., -4, -2, 0, 2, 4, 6, ...} violates the closure under addition and the existence of an additive inverse axioms of a field under the usual operations of multiplication and addition.