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Transcript
```Fields
Let (F; +; ) be a …eld.
For the sake of brevity, we will omit reference to the use of commutative
axioms. The dot ‘’in multiplication is sometimes omitted. x2 is de…ned to be
x x:
Proposition 1.16(a). For any x 2 F we have 0 x = 0: (And hence 0 = 0 x:)
Proof. By (D) and (A4), we have 0 x + 0 x = (0 + 0) x = 0 x: Since
0 x + 0 x = 0 + 0 x;
Proposition 1.14(a) [i.e., x + y = x + z ) y = z] implies 0 x = 0.
The number 1 is de…ned to be the additive inverse of 1.
2
Lemma. ( 1) = ( 1) ( 1) = 1:
Proof. By de…nition, 0 = 1 + 1: By Proposition 1.16(a), Axioms (A5), (D),
and (M4), we have
0 = 0 ( 1) = ( 1 + 1) ( 1) = ( 1) ( 1) + 1 ( 1) = ( 1) ( 1) + ( 1) :
Now, since
1 + ( 1) = 0 = ( 1) ( 1) + ( 1) ;
we conclude from Proposition 1.14(a) that 1 = ( 1) ( 1).
Lemma. For any x 2 F we have x = ( 1) x:
Proof. We have
0 = 0 x = ( 1 + 1) x = ( 1) x + 1 x = ( 1) x + x:
Since
x + x = 0 = ( 1) x + x;
we conclude from Proposition 1.14(a) that x = ( 1) x.
Lemma.1 For any x; y 2 F we have ( x) ( y) = xy:
Proof. We compute
2
( x) ( y) = ( 1) x ( 1) y = ( 1) xy = xy
2
since ( 1) = 1:
The real …eld
Theorem 1.20(a). If x; y 2 R and x > 0, then there is a positive integer n
such that
nx > y:
Proof. See p. 9 of the book.
Proposition. For any real number x there exists a greatest integer n0 with
n0 x.
1I
used this statement without proof in class.
1
Z.
Proof. De…ne the set S = fn 2 Z : n
xg, which we consider as a subset of
Claim. S is nonempty and bounded from above.
Proof of claim. (1) By Theorem 1.20(a), since 1 > 0 and x 2 R, there
exists a positive integer n such that n = n 1 > x. That is, n < x; which
says that n 2 S. So S is nonempty.
(2). By Theorem 1.20(a) again, since 1 > 0 and x 2 R, there exists a positive
integer m such that m = m 1 > x. Then, for each p 2 S we have p x < m:
Thus m is an upper bound for S: This completes the proof of the claim.
Now recall from Math 109 the following fact which is equivalent to the Principle of Mathematical Induction.
Least Natural Number Principle. Each set of positive integers contains
a least element.
From this it can easily be proved that:
‘Greatest Negative Integer Principle.’ Each set of negative integers
contains a greatest element.
We can extend this to:
Statement. Each set of integers which is bounded above contains a greatest
element.
The statement now implies that the set S has a greatest element, which we
call n0 : By the de…nition of S, this means that n0 is the greatest integer with
n0 x.
Constructing the reals and Dedekind cuts
Given r 2 Q, de…ne R = fp 2 Q : p < rg = ( 1; r).
We say that a subset S of Q is a Dedekind cut if
(1) If x 2 S and y 2 Q satisfy y < x, then y 2 S.
(2) If x 2 S, then there exists z 2 S with z > x.
Exercise: Determine where the hidden universal quanti…es are in the above
conditions.
Clearly if r 2 Q, then ( 1; r) is a Dedekind
p cut.
Here the Dedekind cut corresponding to 2. De…ne S Q the union of all
nonpositive rationals and the positive rationals r with r2 < 2. What we showed
in class on Monday Jan 5 is that S is a Dedekind cut. We also show that there
is no rational number r0 with r02 = 2.
Now, for each set of the form ( 1; r), where r 2 Q, the number r is the
least upper bound for ( 1; r) :
Lemma. S does not have a least upper bound in Q. Hence, S cannot be of
the form ( 1; r), where r 2 Q.
Proof. Suppose s 2 Q is an upper bound for S. Then s 2
= S since S has
no greatest element. This implies that s2 > 2. But then we can …nd q 2 Q
with q < s and q 2 > 2 (see p. 2 of the book). Thus s cannot be the least upper
bound.
2
```