09-240/Classnotes for Tuesday September 15

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The real numbers A set $\mathbb R$ with two binary operators and two special elements $0, 1 \in \mathbb R$ s.t.

$F1.\quad \forall a, b \in \mathbb R, a + b = b + a \mbox{ and } a \cdot b = b \cdot a$

$F2.\quad \forall a, b, c, (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)$

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

$F3.\quad \forall a, a + 0 = a \mbox{ and } a \cdot 0 = 0 \mbox{ and } a \cdot 1 = a$

$F4.\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1$

$\mbox{So } a + (-a) = 0 \mbox{ and } a \cdot a^{-1} = 1$

$\mbox{(So } (a + b) \cdot (a - b) = a^2 - b^2)$

$\forall a, \exists x, x \cdot x = a \mbox{ or } a + x \cdot x = 0$

Note: or means inclusive or in math.

$F5.\quad (a + b) \cdot c = a \cdot c + b \cdot c$

Definition: A field is a set F with two binary operators $\,\!+$: F×F → F, $\times\,\!$: F×F → F and two elements $0, 1 \in \mathbb R$ s.t.

$F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F$

$F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)$

$F3\quad a + 0 = a, a \cdot 1 = a$

$F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1$

$F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c$

Examples

1. $F = \mathbb R$
2. $F = \mathbb Q$
3. $\mathbb C = \{ a + bi : a, b \in \mathbb R \}$

   $i = \sqrt{-1}$

   $\,\!(a + bi) + (c + di) = (a + c) + (b + d)i$

   $\,\!0 = 0 + 0i, 1 = 1 + 0i$

4. $\,\!F_2 = \{ 0, 1 \}$
5. $\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}$
6. $\,\!F_6 = \{ 0, 1,2,3,4,5 \}$ is not a field (counterexample)

Theorem: $\,\!F_P$ for $p>1$ is a field IFF $p$ is a prime number

Tedious Theorem

1. $a + b = c + d \Rightarrow a = c$ "cancellation property"
2. $a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c$

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