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

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One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.
 
One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.
  
 
+
You may also think of it as 27-n=23 23*23 + 23*n = 27*23.
 
Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 2<sup>3</sup> = 8, but 3<sup>2</sup> = 9.
 
Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 2<sup>3</sup> = 8, but 3<sup>2</sup> = 9.
  

Revision as of 17:46, 17 September 2009

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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 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 6
2 0 2 4 6 1 3 5
3 0 3 6 2 5 1 4
4 0 4 1 5 2 6 3
5 0 5 3 1 6 4 2
6 0 6 5 4 3 2 1

Theorem: F2 is a field.

In order to prove that the associative property holds, make a table (similar to a truth table) for a, b and c.

a b c  
0 0 0  
0 0 1  
0 1 0  
0 1 1 (0 + 1) + 1 =? 0 + (1 + 1)
1 + 1 =? 0 + 0
0 = 0
1 0 0  
1 0 1  
1 1 0  
1 1 1  


Theorem: \,\! F_p for p > 1 is a field iff (if and only if) p is a prime number

Proof:

Given a finite set with m elements in \mathbb Z, an element a will have a multiplicative inverse iff gcd(a,m) = 1

This can be shown using Bézout's identity:

\exists x, y \mbox{ s.t. } ax + my = 1
\left(ax + my\right) \pmod{m} = 1\pmod{m}
ax = 1
x = a^{-1}

We have shown that a has a multiplicative inverse if a and m are relatively prime. It is therefore a natural conclusion that if m is prime all elements in the set will satisfy gcd(a, m) = 1


Multiplication is repeated addition.

23 \times 27 = \begin{matrix} 27 \\ \overbrace{23 + 23 + 23 + \cdots + 23} \end{matrix} = 621

27 \times 23 = \begin{matrix} 23 \\ \overbrace{27 + 27 + 27 + \cdots + 27} \end{matrix} = 621

One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.

You may also think of it as 27-n=23 23*23 + 23*n = 27*23. Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 23 = 8, but 32 = 9.

Tedious Theorem

  1. a + b = c + d \Rightarrow a = c "cancellation property"
    Proof:
    By F4, \exists d \mbox{ s.t. } b + d = 0
    \,\! (a + b) + d = (c + b) + d
    \Rightarrow a + (b + d) = c + (b + d) by F2
    \Rightarrow a + 0 = c + 0 by choice of d
    \Rightarrow a = c by F3
  2.  a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c
  3. a + O' = a \Rightarrow O' = 0
    Proof:
    \,\! a + O' = a
    \Rightarrow a + O' = a + 0 by F3
    \Rightarrow O' = 0 by adding the additive inverse of a to both sides
  4. a \cdot l' = a, a \ne 0 \Rightarrow l' = 1
  5. a + b = 0 = a + b' \Rightarrow b = b'
  6. a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}
    \,\! \mbox{Aside: } a - b = a + (-b)
    \frac ab = a \cdot b^{-1}
  7. \,\! -(-a) = a, (a^{-1})^{-1}
  8. a \cdot 0 = 0
    Proof:
    a \cdot 0 = a(0 + 0) by F3
    = a \cdot 0 + a \cdot 0 by F5
    = 0 = a \cdot 0
  9. \forall b, 0 \cdot b \ne 1
    So there is no 0−1
  10. (-a) \cdot b = a \cdot (-b) = -(a \cdot b)
  11. (-a) \cdot (-b) = a \cdot b
  12. (Bonus) \,\! (a + b)(a - b) = a^2 - b^2

Quotation of the Day

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