Field Axioms

Field Axioms of R

$\mathbb{R} \not = \emptyset$ with two binary operations $+$ and $\cdot$ satisfying the following properties

1. Closed under addition: $\forall a, b \in \mathbb{R} ; a + b \in \mathbb{R}$
2. Commutative: $\forall a, b \in \mathbb{R}; a + b = b + a$
3. Associative: $\forall a, b, c \in \mathbb{R}; (a + b) + c = a + (b + c)$
4. Additive identity: $\exists 0 \in \mathbb{R}\, \forall a \in \mathbb{R}; a + 0 = 0 + a = a$
5. Additive inverse: $\forall a \in \mathbb{R}\, \exists (-a); a + (-a) = (-a) + a = 0$
6. Closed under multiplication: $\forall a, b \in \mathbb{R} ; a \cdot b \in \mathbb{R}$
7. Commutative: $\forall a, b \in \mathbb{R}; a \cdot b = b \cdot a$
8. Associative: $\forall a, b, c \in \mathbb{R}; (a \cdot b) \cdot c = a \cdot (b \cdot c)$
9. Multiplicative identity: $\exists 1 \in \mathbb{R}\, \forall a \in \mathbb{R}; a \cdot 1 = 1 \cdot a = a$
10. Multiplicative inverse: $\forall a \in \mathbb{R}-\Set{0}\, \exists a^{-}; a \cdot a^{-} = a^{-} \cdot a = 1$
11. Multiplication is distributive over addition: $a \cdot (b + c) = a \cdot b + a \cdot c$

Required proofs

The below mentioned propositions can and should be proven using the above-mentioned axioms. $a, b, c \in \mathbb{R}$. $$

• $a\cdot0 = 0$ Hint: Start with $a(1+0)$
• $1\not = 0$
• Additive identity ($0$) is unique
• Multiplicative identity ($1$) is unique
• Additive inverse ($-a$) is unique for a given $a$
• Multiplicative inverse ($a^{-1}$) is unique for a given $a$
• $a + b = 0 \implies b = -a$
• $a + c = b + c \implies a = b$
• $-(a + b) = (-a) + (-b)$
• $-(-a) = a$
• $ac = bc \implies a = b$
• $ab = 0 \implies a = 0 \lor b = 0$
• $−(ab) = (−a)b = a(−b)$
• $(−a)(-b) = ab$
• $a \not = 0 \implies {(a^{-1})}^{-1} = a$
• $a, b \not = 0 \implies {ab}^{-1} = a^{-1}b^{-1}$

Field

Any set satisfying the above axioms with two binary operations (commonly $+$ and $\cdot$) is called a field. Written as:

$$(\mathbb{R}, +, \cdot) \;\text{is a Field} \;\; \text{but} \;\; (\mathbb{R}, \cdot, +)\;\text{is not a field}$$

Field or Not?

	Is field?	Reason (if not)
$(\mathbb{R},+,\cdot)$	True
$(\mathbb{R},\cdot,+)$	False	Axiom 11 is invalid
$(\mathbb{Z},+,\cdot)$	False	Multiplicative inverse doesn’t exist
$(\mathbb{Q},+,\cdot)$	True
$(\mathbb{Q}^c,+,\cdot)$	False	$\sqrt{2}\cdot\sqrt{2}\not\in\mathbb{Q}^c$
Boolean algebra	False	Additive inverse doesn’t exist
$(\set{0,1},+\;\text{mod}\;2,\cdot\;\text{mod}\;2)$	True
$(\set{0,1,2},+\;\text{mod}\;3,\cdot\;\text{mod}\;3)$	True
$(\set{0,1,2,3},+\;\text{mod}\;4,\cdot\;\text{mod}\;4)$	False	Multiplicative inverse doesn’t exist