Difference between revisions of "09-240/Classnotes for Tuesday September 15"

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(Examples: Addition and multiplication tables.)
(Examples: Expand, fix table format.)
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# <math>\,\!F_2 = \{ 0, 1 \}</math>
 
# <math>\,\!F_2 = \{ 0, 1 \}</math>
 
# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math>
 
# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math>
# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field (counterexample)
+
# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field because not every element has a multiplicative inverse.
 +
#: Let <math>a = 2.</math>
 +
#: Then <math>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 ''F''<sub>6</sub> s.t. ''a · b'' = 1
  
 
{|
 
{|
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! + !! 0 !! 1
 
! + !! 0 !! 1
 
|-
 
|-
! 0 || 0 || 1
+
! 0
 +
| 0 || 1
 
|-
 
|-
! 1 || 1 || 0
+
! 1
 +
| 1 || 0
 
|-
 
|-
 
|}
 
|}
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! × !! 0 !! 1
 
! × !! 0 !! 1
 
|-
 
|-
! 0 || 0 || 0
+
! 0
 +
| 0 || 0
 
|-
 
|-
! 1 || 0 || 1
+
! 1
 +
| 0 || 1
 
|}
 
|}
  
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! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
 
! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
 
|-
 
|-
! 0 || 0 || 1 || 2 || 3 || 4 || 5 || 6
+
! 0
 +
| 0 || 1 || 2 || 3 || 4 || 5 || 6
 
|-
 
|-
! 1 || 1 || 2 || 3 || 4 || 5 || 6 || 0
+
! 1
 +
| 1 || 2 || 3 || 4 || 5 || 6 || 0
 
|-
 
|-
! 2 || 2 || 3 || 4 || 5 || 6 || 0 || 1
+
! 2
 +
| 2 || 3 || 4 || 5 || 6 || 0 || 1
 
|-
 
|-
! 3 || 3 || 4 || 5 || 6 || 0 || 1 || 2
+
! 3
 +
| 3 || 4 || 5 || 6 || 0 || 1 || 2
 
|-
 
|-
! 4 || 4 || 5 || 6 || 0 || 1 || 2 || 3
+
! 4
 +
| 4 || 5 || 6 || 0 || 1 || 2 || 3
 
|-
 
|-
! 5 || 5 || 6 || 0 || 1 || 2 || 3 || 4
+
! 5
 +
| 5 || 6 || 0 || 1 || 2 || 3 || 4
 
|-
 
|-
! 6 || 6 || 0 || 1 || 2 || 3 || 4 || 5
+
! 6
 +
| 6 || 0 || 1 || 2 || 3 || 4 || 5
 
|}
 
|}
  
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! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
 
! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
 
|-
 
|-
! 0 || 0 || 0 || 0 || 0 || 0 || 0 || 0
+
! 0
 +
| 0 || 0 || 0 || 0 || 0 || 0 || 0
 
|-
 
|-
! 1 || 0 || 1 || 2 || 3 || 4 || 5 || 0
+
! 1
 +
| 0 || 1 || 2 || 3 || 4 || 5 || 0
 
|-
 
|-
! 2 || 0 || 2 || 4 || 6 || 1 || 3 || 1
+
! 2
 +
| 0 || 2 || 4 || 6 || 1 || 3 || 1
 
|-
 
|-
! 3 || 0 || 3 || 6 || 2 || 5 || 1 || 2
+
! 3
 +
| 0 || 3 || 6 || 2 || 5 || 1 || 2
 
|-
 
|-
! 4 || 0 || 4 || 1 || 5 || 2 || 6 || 3
+
! 4
 +
| 0 || 4 || 1 || 5 || 2 || 6 || 3
 
|-
 
|-
! 5 || 0 || 5 || 3 || 1 || 6 || 4 || 4
+
! 5
 +
| 0 || 5 || 3 || 1 || 6 || 4 || 4
 
|-
 
|-
! 6 || 0 || 6 || 5 || 4 || 3 || 2 || 5
+
! 6
 +
| 0 || 6 || 5 || 4 || 3 || 2 || 5
 
|}
 
|}
 
|}
 
|}

Revision as of 23:19, 15 September 2009

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×FF, \times\,\!: F×FF 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 because not every element has a multiplicative inverse.
    Let a = 2.
    Then a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4
    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: \,\!F_P for p>1 is a field iff (if and only if) 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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