Download MAT 3749 1.1 Handout Definition A Field F is a set of objects for

MAT 3749 1.1 Handout
Definition
A Field F is a set of objects for which an addition “+” and multiplication “  ” is defined
satisfying the following rules:
1. Commutative Laws
a  b  b  a and ab  ba
2. Associative Laws
 a  b  c  a  b  c  and  ab  c  a bc 
3. Distributive Law
a  b  c   ab  ac
4. Zero and One
There are two distinct elements in F, denoted by 0 and 1, such that
a 1  a and a  0  a
5. Negatives and Inverses
For every a  F , there is an element a  F such that a   a   0 .
For every a  F , a  0 , there is an element a 1  F (inverse of a ) such that a  a 1  1 .
Real Numbers R
The set
of Real Numbers is a field.
Example 1
(a)
is a field.
(b)
is not a field.
Ordered Field
A field is an ordered field if there is an relation  defined on it such that
1. Transitive
If a  b , and b  c , then a  c .
2. For every two elements a , b , exactly one of the following holds.
ab, a b, a b
3. For any c , if a  b , then a  c  b  c .
4. For any c  0 , if a  b , then ac  bc .
Real Numbers R
The set
of Real Numbers is an ordered field.
Example 2
is a field but not an ordered field.
Last Edit: 5/4/2017 2:32 AM
Upper (Lower) Bounds
A set E  is bounded above if M  , called the upper bound of E such that
x  M , x  E .
If the upper bound M belongs to E, then M is called the maximum element of E, denoted
by max E .
Example 3
1. Determine which of the following sets are bounded above.
2. Determine which of the following sets have a maximum element.
(a) E1  0, 2
Analysis
1
(b) E2   n 
n
Analysis
Solution




Solution
2
(c) E3 
Analysis
Solution
Archimedean Property
If a , then n   such that a  n .
Density Property
If a, b  and a  b , then x 
and y  such that
a  x  b and a  y  b .
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Least Upper Bounds
A real number M is the least upper bound, or Supremum, of a set E 
1. M is an upper bound of E;
2. M is less than or equal to every upper bound of E.
if
Equivalent Statement
2. If N  M , then N is not an upper bound of E.
Example 4
Show that sup 0, 2  2 .
Analysis
Proof
4
Example 5
 1
Determine the supremum of E  1  2 n 
 n
Analysis


.

Solution
The Completeness Axiom
Let E be a non-empty subset of .
1. If E is bounded above, then E has a least upper bound.
2. If E is bounded below, then E has a greatest lower bound.
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