Download Math 3333: Fields, Ordering, Completeness and the Real Numbers

Math 3333: Fields, Ordering, Completeness and the Real Numbers
A field is a set F of elements, called numbers, together with two binary operations, addition
and multiplication. (A binary operation on F acts on any two elements from F and returns
an element from F .) Addition and multiplication must satisfy the following enumerated
conditions.
For any two elements x and y in F addition, denoted by x + y , is required to satisfy
A1: x + y ∈ F and if r = x and s = y then r + s = x + y .
A2: x + y = y + x . That is, addition is commutative.
A3: x + (y + z) = (x + y) + z . That is, addition is associative.
A4: There is a unique number 0 ∈ F such that x + 0 = x for every x ∈ F . That is, the
field contains a unique additive identity.
A5: For every x ∈ F there is a unique number −x ∈ F such that x + (−x) = 0. That is,
each number has a unique additive inverse.
For any two elements x and y in F multiplication, denoted by x · y , is required to satisfy
M1: x · y ∈ F and if r = x and s = y then r · s = x · y .
M2: x · y = y · x . That is, multiplication is commutative.
M3: x · (y · z) = (x · y) · z . That is, multiplication is associative.
M4: There is a unique number 1 ∈ F such that x · 1 = x for every x ∈ F . That is, the
field contains a unique multiplicative identity.
M5: For every x ∈ F other than 0 there is a unique 1/x ∈ F such that x · (1/x) = 1.
That is, each number except for the additive identity has a unique multiplicative inverse.
Finally, addition and multiplication must jointly satisfy the distributive law
DL: For every x , y and z in F we must have x · (y + z) = x · y + x · z .
Here’s an example of a field which has only three numbers. Say F = {a, b, c}, and let
addition and multiplication be defined by
a+a=a a+b=b a+c=c
b+a=b b+b=c b+c=a
c + a = c c + b = a c + c = b,
a·a=a
b·a=a
c·a=a
a·b=a
b·b=b
c·b=c
a·c=a
b·c=c
c · c = b.
In this example the additive identity (0) is a . The multiplicative identity (1) is b . The
additive inverse for a is a ≡ −a , for b is c ≡ −b and for c is b ≡ −c . The multiplicative
inverse for b is b ≡ 1/b and for c is c ≡ 1/c .
Here are a couple of exercises.
1. Let F be a field, and so the eleven enumerated properties listed above need be satisfied.
(a) Let 0 denote F ’s additive identity. Prove that 0 · x = 0 for any x ∈ F , (b) Let −1
1
denote the additive inverse of F ’s multiplicative identity 1. For any x ∈ F prove that
y ≡ (−1) · x is the additive inverse of x . (c) Verify these two facts using the example field
given above.
2. Suppose a set F consists of two letters F = {a, b}, and addition and multiplication
between letters is defined as follows.
a+a=a a+b=b
a·a=a a·b=a
b + a = b b + b = a,
b · a = a b · b = b.
(a) Determine the additive and multiplicative identities. (b) What is the additive inverse
for the letter a and for the letter b ? (c) Conclude this system defines a field.
As we all know, the set of natural numbers is given by N = {1, 2, 3, . . .}, and the set
of integers is given by Z = {0, ±1, ±2, . . .}. However, using the usual notion of integer
addition and multiplication, it is easy to conclude that neither of these sets define a field.
The two finite field discussed earlier are not particularly useful. On the other hand, the
next field we discuss is.
The set of rational numbers is given by
p
: with p ∈ Z, q ∈ N coprime .
Q=
q
In particular, the number 3/4 is in Q since 3 ∈ Z and 4 ∈ N are coprime. Strictly speaking
however, the number 6/8 is not. It should be identified with 3/4. Let x1 ≡ p1 /q1 ∈ Q
and x2 ≡ p2 /q2 ∈ Q . We say x1 = x2 if and only if p1 = p2 and q1 = q2 . To define
addition of x1 to x2 first use integer arithmetic to evaluate
p1
p2
p1 q2 + p2 q1
+
=
,
q1
q2
q1 q2
and then determine the greatest common divisor of the top ( p1 q2 + p2 q1 ) and the bottom
( q1 q2 ), call this GCD z ∈ N , to write
p3
x1 + x2 ≡ x3 ≡
where p3 z = p1 q2 + p2 q1 and q3 z = q1 q2 .
q3
Since p3 and q3 are coprime this gives a sum x3 ∈ Q . Multiplication of two rationals is
defined similarly.
An ordered field is a field which also has a binary relation, say < , which satisfies the
following enumerated conditions.
For any two elements x and y in F , x < y is either true or false and is required to satisfy
O1: Exactly one from x = y or x < y or y < x is true.
O2: If x < y and y < z then x < z .
O3: If x < y then x + z < y + z for any z .
O4: If x < y then x · z < y · z for any z satisfying 0 < z .
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On an ordered field the relation < is called less than. There is an associated relation >
called greater than which is defined by x > y ≡ y < x .
The notation x ≤ y is used to signify either x < y or x = y . The notation x ≥ y is used
to signify either x > y or x = y .
The rationals numbers is an ordered field with the relation < defined as follows.
p1
p2
For x1 =
∈ Q and x2 =
∈ Q then x1 < x2 ⇐⇒ p2 q1 − p1 q2 ∈ N.
q1
q2
3. Suppose F is an ordered field. Prove the following. (a) x < y if and only if −y < −x .
(Recall −y denotes the additive inverse for y and −x for x .) (b) x > 0 if and only if
−x < 0. (c) Suppose x < y and z < 0 then x·z > y ·z . (d) Prove that 1 > 0. (e) Suppose
x > y > 0 then 1/y > 1/x > 0. (Recall 1/y denotes the multiplicative inverse for y and
1/x for x .)
4. It’s not hard to show that the rationals as defined above form an ordered field. Assume
this and then prove the following. (a) Given any two rationals x1 and x2 with x1 < x2
there is at least one rational x3 satisfying x1 < x3 < x2 . (b) Conclude there are in fact
an infinite number of distinct rationals between x1 and x2 .
5. Suppose F is an arbitrary ordered field. (a) Prove that x ∈ F implies x · x + 1 > 0.
(b) See if you can use this fact to deduce that there can be no order relation on the field
of complex numbers.
Suppose F is an ordered field.
A subset S ⊆ F is said to be bounded below if there exists a number a ∈ F such that
x ≥ a for every x ∈ S . A subset S ⊆ F is said to be bounded above if there exists a
number b ∈ F such that x ≤ b for every x ∈ S . We say S ⊆ F is bounded if it is bounded
above and below.
The infimum of a subset S ⊆ F is denoted by inf S and its value is given as follows.
(1) The number a is a lower bound for S.
a = inf S ⇒
(2) Any c ∈ F with c > a is not a lower bound for S.
Try to understand from above why the infimum is often called the greatest lower bound.
The supremum of a subset S ⊆ F is denoted by sup S and its value is given as follows.
(1) The number b is an upper bound for S.
b = sup S ⇒
(2) Any c ∈ F with c < b is not an upper bound for S.
Again, try to see why the supremum is often called the least upper bound.
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Carefully consider this next example. The subset of rationals S = {x ∈ Q : x2 ≤ 2} ⊆ Q
is clearly bounded since e.g. −3/2 ≤ x ≤ 3/2 for every x ∈ S . Moreover, it can be shown
that if r = sup S then this number must satisfy r2 = 2; see exercise 6 below. In class
however we showed there is no rational number whose square is two. This exemplifies the
fundamental flaw of the ordered field Q . In general bounded subsets of rationals do not
have rational supremums and/or infimums! It is often loosely said that the set of rational
numbers has lots and lots of holes in it.
This is where the so-called real numbers come in. The real numbers, denoted by R , is an
ordered field which contains the rational numbers as a subfield and also has the additional
important property that:
Given any nonempty S ⊆ R which is bounded below then inf S ∈ R exists.
Given any nonempty S ⊆ R which is bounded above then sup S ∈ R exists.
Surprisingly the rigorous development of the real number system was not discovered until
sometime around the mid nineteenth century. There are now several common procedures
to logically construct the reals. See for example [1]
6. Let S = {x ∈ Q : x2 ≤ 2 & x ≥ 0}, and suppose there is a rational number r = sup S .
(a) Show it is impossible for r2 < 2. (b) Show it is impossible for r2 > 2. Therefore we
are left with r2 = 2 which we also know is impossible.
7. Suppose S is a bounded set of real numbers, and let k be a fixed real number. From
these define two new sets S + k ≡ {x + k : x ∈ S} and kS ≡ {kx : x ∈ S}. Prove the
following.
(a) inf(S + k) = (inf S) + k
(c) If k ≥ 0 inf(kS) = k(inf S)
(b) sup(S + k) = (sup S) + k
(d) If k ≥ 0 sup(kS) = k(sup S)
Let T be some other bounded subset of reals and define S + T ≡ {x + y : x ∈ S, y ∈ T }.
(e) Is it true in general that sup(S + T ) = sup S + sup T ? (If not give a counter example.)
8. For each given subset of R , determine its sup and inf .
(a) {1/n : n ∈ N}
(d) ∩∞
n=1 {x ∈ R : 1 − 1/n < x < 1 + 1/n}
(b) {(−1)n + 1/n : n ∈ N}
(e) {x ∈ Q : |x| < 1}
(c) {n/(n + 1) : n ∈ N}
(f) {x 6∈ Q : |x| < 1}
[1] http://en.wikipedia.org/wiki/Construction of the real numbers
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