A graduate of MaEd Math, an LPT and academician. A published writer.
Something About Fields.
Definition. A field is a set F, containing at least two elements, on which two operations + and · (called addition and multiplication, respectively) are defined so that for each pair of elements x, y in F there are unique elements x + y and x · y (often written xy) in F for which the following conditions hold for all elements x, y, z in F
- There is an element 0 ∈ F , called zero, such that x + 0 = x. (existence of an additive identity)
- (iv) For each x, there is an element −x ∈ F such that x+(−x) = 0. (existence of additive inverses)
- (v) xy = yx (commutativity of multiplication)
- (vi) (x · y) · z = x · (y · z) (associativity of multiplication)
- (vii) (x+y)·z=x·z+y·z and x·(y+z)=x·y+x·z (distributivity)
- (viii) There is an element 1 ∈ F , such that x·1 = x. (existence of a multiplicative identity)
(ix) If x ̸= 0, then there is an element x^-1∈ F such that x·x^-1 = 1. (existence of
Proposition. Let F be a field.
(1) The additive identity in F is unique.
(2) The additive inverse of an element of F is unique.
(3) The multiplicative identity of F is unique.
(4) The multiplicative inverse of a nonzero element of F is unique.
Examples. The rational numbers Q, the real numbers R and the complex numbers C (discussed below) are examples of fields.
The set Z of integers is not a field. In Z, axioms (i)-(viii) all hold, but axiom (ix) does not: the multiplicative inverse of integers do not belong to the set of integers except 1 and -1.The only nonzero integers that have multiplicative inverses that are integers are 1 and −1. For example, 2 is a nonzero integer. If 2 had a multiplicative inverse in Z, there would be an integer n such that 2n = 1, which is impossible, since 1 is an odd integer, and not an even integer.
Example. Let F be a field. Using the axioms in the definition of field, prove that (−1) · x = −x for all x ∈ F. State which axioms are used in your proof.
Solution: We must show that (−1) · x is an additive inverse of x, that is, x + (−1) · x = 0.
x+(−1)·x=x+x·(−1) by (v) =x·1+x·(−1) by (viii)
=x·(1+(−1)) by (vii)
=x·0+0 by(iii) =x·0+(x·0+−(x·0)) by (iv) = (x·0+x·0)+−(x·0) by (ii) =x·(0+0)+−(x·0) by (vii) =x·0+−(x·0) by (iii)
Example: Show that the set Q (Rational Numbers) is a field:
2, 3, 5 are elements of Q
let a = 2 b = 3 c = 5
Check for additive inverse
- a = - 2 -b = - 3 - c = - 5
all additive inverses belongs to the set of Rational numbers
Check for multiplicative inverse:
1/a = 1/2 1/b = 1/3 c = 1/5 but 0 has no multiplicative inverse. Thus this break the rule for field.
all the multiplicative inverses belong to the set of Rational numbers but 1/0 is not defined.
check for associativity:
( 2 + 3) + 5 = 2 + ( 3 + 5 )
5 + 5 = 2 + 8
( 2 * 3) * 5 = 2 * ( 3 * 5 )
check for commutative property:
2 + 3 = 3 + 2
2 * 3 = 3 * 2
Check for distributive property of multiplication
( 2 + 3) * 5 = 2 * 5 + 3 * 5
5 * 5 = 10 + 15
check for additive identity:
2 + -2 = 0
3 + - 3 = 0
5 + -5 = 0
check for multiplicative identity:
2 (1/2) = 1
3 (1/3) = 1
5 (1/5) = 1
additive identity = 0 is an element of Rational Numbers
multiplicative identity = 1 is also an element of Rational Numbers
Since 0 has no multiplicative inverse the set of Rational Numbers on multiplication except 0 is a field.