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Sahithyan's S1
Sahithyan's S1 — Mathematics

Field Axioms

Field Axioms of R

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

  1. Closed under addition: a,bR;a+bR\forall a, b \in \mathbb{R} ; a + b \in \mathbb{R}
  2. Commutative: a,bR;a+b=b+a\forall a, b \in \mathbb{R}; a + b = b + a
  3. Associative: a,b,cR;(a+b)+c=a+(b+c)\forall a, b, c \in \mathbb{R}; (a + b) + c = a + (b + c)
  4. Additive identity: 0RaR;a+0=0+a=a\exists 0 \in \mathbb{R}\, \forall a \in \mathbb{R}; a + 0 = 0 + a = a
  5. Additive inverse: aR(a);a+(a)=(a)+a=0\forall a \in \mathbb{R}\, \exists (-a); a + (-a) = (-a) + a = 0
  6. Closed under multiplication: a,bR;abR\forall a, b \in \mathbb{R} ; a \cdot b \in \mathbb{R}
  7. Commutative: a,bR;ab=ba\forall a, b \in \mathbb{R}; a \cdot b = b \cdot a
  8. Associative: a,b,cR;(ab)c=a(bc)\forall a, b, c \in \mathbb{R}; (a \cdot b) \cdot c = a \cdot (b \cdot c)
  9. Multiplicative identity: 1RaR;a1=1a=a\exists 1 \in \mathbb{R}\, \forall a \in \mathbb{R}; a \cdot 1 = 1 \cdot a = a
  10. Multiplicative inverse: aR{0}a;aa=aa=1\forall a \in \mathbb{R}-\Set{0}\, \exists a^{-}; a \cdot a^{-} = a^{-} \cdot a = 1
  11. Multiplication is distributive over addition: a(b+c)=ab+aca \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,cRa, b, c \in \mathbb{R}.

  • a0=0a\cdot0 = 0 Hint: Start with a(1+0)a(1+0)
  • 101\not = 0
  • Additive identity (00) is unique
  • Multiplicative identity (11) is unique
  • Additive inverse (a-a) is unique for a given aa
  • Multiplicative inverse (a1a^{-1}) is unique for a given aa
  • a+b=0    b=aa + b = 0 \implies b = -a
  • a+c=b+c    a=ba + c = b + c \implies a = b
  • (a+b)=(a)+(b)-(a + b) = (-a) + (-b)
  • (a)=a-(-a) = a
  • ac=bc    a=bac = bc \implies a = b
  • ab=0    a=0b=0ab = 0 \implies a = 0 \lor b = 0
  • (ab)=(a)b=a(b)−(ab) = (−a)b = a(−b)
  • (a)(b)=ab(−a)(-b) = ab
  • a0    (a1)1=aa \not = 0 \implies {(a^{-1})}^{-1} = a
  • a,b0    ab1=a1b1a, 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:

(R,+,)  is a Field    but    (R,,+)  is not a field(\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)
(R,+,)(\mathbb{R},+,\cdot)True
(R,,+)(\mathbb{R},\cdot,+)FalseAxiom 11 is invalid
(Z,+,)(\mathbb{Z},+,\cdot)FalseMultiplicative inverse doesn’t exist
(Q,+,)(\mathbb{Q},+,\cdot)True
(Qc,+,)(\mathbb{Q}^c,+,\cdot)False22∉Qc\sqrt{2}\cdot\sqrt{2}\not\in\mathbb{Q}^c
Boolean algebraFalseAdditive inverse doesn’t exist
({0,1},+  mod  2,  mod  2)(\set{0,1},+\;\text{mod}\;2,\cdot\;\text{mod}\;2)True
({0,1,2},+  mod  3,  mod  3)(\set{0,1,2},+\;\text{mod}\;3,\cdot\;\text{mod}\;3)True
({0,1,2,3},+  mod  4,  mod  4)(\set{0,1,2,3},+\;\text{mod}\;4,\cdot\;\text{mod}\;4)FalseMultiplicative inverse doesn’t exist