09-240/Classnotes for Tuesday September 15

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The real numbers A set [math]\displaystyle{ \mathbb R }[/math] with two binary operators and two special elements [math]\displaystyle{ 0, 1 \in \mathbb R }[/math] s.t.

[math]\displaystyle{ F1.\quad \forall a, b \in \mathbb R, a + b = b + a \mbox{ and } a \cdot b = b \cdot a }[/math]

[math]\displaystyle{ F2.\quad \forall a, b, c, (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c) }[/math]

[math]\displaystyle{ \mbox{(So for any real numbers } a_1, a_2, ..., a_n, \mbox{ one can sum them in any order and achieve the same result.} }[/math]

[math]\displaystyle{ F3.\quad \forall a, a + 0 = a \mbox{ and } a \cdot 0 = 0 \mbox{ and } a \cdot 1 = a }[/math]

[math]\displaystyle{ F4.\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1 }[/math]

[math]\displaystyle{ \mbox{So } a + (-a) = 0 \mbox{ and } a \cdot a^{-1} = 1 }[/math]

[math]\displaystyle{ \mbox{(So } (a + b) \cdot (a - b) = a^2 - b^2) }[/math]

[math]\displaystyle{ \forall a, \exists x, x \cdot x = a \mbox{ or } a + x \cdot x = 0 }[/math]

Note: or means inclusive or in math.

[math]\displaystyle{ F5.\quad (a + b) \cdot c = a \cdot c + b \cdot c }[/math]

Definition: A field is a set F with two binary operators [math]\displaystyle{ \,\!+ }[/math]: F×F → F, [math]\displaystyle{ \times\,\! }[/math]: F×F → F and two elements [math]\displaystyle{ 0, 1 \in \mathbb R }[/math] s.t.

[math]\displaystyle{ F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F }[/math]

[math]\displaystyle{ F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c) }[/math]

[math]\displaystyle{ F3\quad a + 0 = a, a \cdot 1 = a }[/math]

[math]\displaystyle{ F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1 }[/math]

[math]\displaystyle{ F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c }[/math]

Examples

1. [math]\displaystyle{ F = \mathbb R }[/math]
2. [math]\displaystyle{ F = \mathbb Q }[/math]
3. [math]\displaystyle{ \mathbb C = \{ a + bi : a, b \in \mathbb R \} }[/math]

   [math]\displaystyle{ i = \sqrt{-1} }[/math]

   [math]\displaystyle{ \,\!(a + bi) + (c + di) = (a + c) + (b + d)i }[/math]

   [math]\displaystyle{ \,\!0 = 0 + 0i, 1 = 1 + 0i }[/math]

4. [math]\displaystyle{ \,\!F_2 = \{ 0, 1 \} }[/math]
5. [math]\displaystyle{ \,\!F_7 = \{ 0, 1,2,3,4,5,6 \} }[/math]
6. [math]\displaystyle{ \,\!F_6 = \{ 0, 1,2,3,4,5 \} }[/math] is not a field because not every element has a multiplicative inverse.

   Let [math]\displaystyle{ a = 2. }[/math]

   Then [math]\displaystyle{ a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4 }[/math]

   Therefore F4 fails; there is no number b in F₆ s.t. a · b = 1

Ex. 4 + 0 1 0 0 1 1 1 0	Ex. 4 × 0 1 0 0 0 1 0 1
Ex. 5 + 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5	Ex. 5 × 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 0 2 0 2 4 6 1 3 1 3 0 3 6 2 5 1 2 4 0 4 1 5 2 6 3 5 0 5 3 1 6 4 4 6 0 6 5 4 3 2 5

Theorem: [math]\displaystyle{ \,\!F_P }[/math] for [math]\displaystyle{ p\gt 1 }[/math] is a field iff (if and only if) [math]\displaystyle{ p }[/math] is a prime number

Tedious Theorem

1. [math]\displaystyle{ a + b = c + d \Rightarrow a = c }[/math] "cancellation property"

   Proof:

   By F4, [math]\displaystyle{ \exists d \mbox{ s.t. } b + d = 0 }[/math]

   [math]\displaystyle{ \,\! (a + b) + d = (c + b) + d }[/math]

   [math]\displaystyle{ \Rightarrow a + (b + d) = c + (b + d) }[/math] by F2

   [math]\displaystyle{ \Rightarrow a + 0 = c + 0 }[/math] by choice of d

   [math]\displaystyle{ \Rightarrow a = c }[/math] by F3

2. [math]\displaystyle{ a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c }[/math]
3. [math]\displaystyle{ a + O' = a \Rightarrow O' = 0 }[/math]

   Proof:

   [math]\displaystyle{ \,\! a + O' = a }[/math]

   [math]\displaystyle{ \Rightarrow a + O' = a + 0 }[/math] by F3

   [math]\displaystyle{ \Rightarrow O' = 0 }[/math] by adding the additive inverse of a to both sides

4. [math]\displaystyle{ a \cdot l' = a, a \ne 0 \Rightarrow l' = 1 }[/math]
5. [math]\displaystyle{ a + b = 0 = a + b' \Rightarrow b = b' }[/math]
6. [math]\displaystyle{ a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1} }[/math]

   [math]\displaystyle{ \,\! \mbox{Aside: } a - b = a + (-b) }[/math]

   [math]\displaystyle{ \frac ab = a \cdot b^{-1} }[/math]

7. [math]\displaystyle{ \,\! -(-a) = a, (a^{-1})^{-1} }[/math]
8. [math]\displaystyle{ a \cdot 0 = 0 }[/math]

   Proof:

   [math]\displaystyle{ a \cdot 0 = a(0 + 0) }[/math] by F3

   [math]\displaystyle{ = a \cdot 0 + a \cdot 0 }[/math] by F5

   [math]\displaystyle{ = 0 = a \cdot 0 }[/math]

9. [math]\displaystyle{ \forall b, 0 \cdot b \ne 1 }[/math]

   So there is no 0⁻¹

10. [math]\displaystyle{ (-a) \cdot b = a \cdot (-b) = -(a \cdot b) }[/math]
11. [math]\displaystyle{ (-a) \cdot (-b) = a \cdot b }[/math]
12. (Bonus) [math]\displaystyle{ \,\! (a + b)(a - b) = a^2 - b^2 }[/math]