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

From Drorbn
Jump to navigationJump to search
(→‎Examples: Expand, fix table format.)
Line 132: Line 132:
== Tedious Theorem ==
== Tedious Theorem ==
# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property"
# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property"
#: Proof:
#: By F4, <math>\exists d \mbox{ s.t. } b + d = 0</math>
#: <math>\,\! (a + b) + d = (c + b) + d</math>
#: <math>\Rightarrow a + (b + d) = c + (b + d)</math> by F2
#: <math>\Rightarrow a + 0 = c + 0</math> by choice of ''d''
#: <math>\Rightarrow a = c</math> by F3
# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math>
# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math>
# <math>a + O' = a \Rightarrow O' = 0</math>

#: <math>a + O' = a</math>
...
#: <math>\Rightarrow a + O' = a + 0</math> by F3
#: <math>\Rightarrow O' = 0</math> by adding the additive inverse of ''a'' to both sides
# <math>a \cdot l' = a, a \ne 0 \Rightarrow l' = 1</math>
# <math>a + b = 0 = a + b' \Rightarrow b = b'</math>
# <math>a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}</math>
#: <math>\,\! \mbox{Aside: } a - b = a + (-b)</math>
#: <math>\frac ab = a \cdot b^{-1}</math>
# <math>\,\! -(-a) = a, (a^{-1})^{-1}</math>
# <math>a \cdot 0 = 0</math>
#: Proof:
#: <math>a \cdot 0 = a(0 + 0)</math> by F3
#: <math>= a \cdot 0 + a \cdot 0</math> by F5
#: <math>= 0 = a \cdot 0</math>
# <math>\forall b, 0 \cdot b \ne 1</math>
#: So there is no 0<sup>&minus;1</sup>
# <math>(-a) \cdot b = a \cdot (-b) = -(a \cdot b)</math>
# <math>(-a) \cdot (-b) = a \cdot b</math>
# (Bonus) <math>\,\! (a + b)(a - b) = a^2 - b^2</math>

Revision as of 22:35, 15 September 2009

The real numbers A set with two binary operators and two special elements s.t.

Note: or means inclusive or in math.

Definition: A field is a set F with two binary operators : F×FF, : F×FF and two elements s.t.

Examples

  1. is not a field because not every element has a multiplicative inverse.
    Let
    Then
    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: for is a field iff (if and only if) is a prime number

Tedious Theorem

  1. "cancellation property"
    Proof:
    By F4,
    by F2
    by choice of d
    by F3
  2. by F3
    by adding the additive inverse of a to both sides
  3. Proof:
    by F3
    by F5
  4. So there is no 0−1
  5. (Bonus)