Consequences of the Field and Order Axioms

 

The order axioms.

0.     There exists a relation “<” obeying the following axioms. (If  x<y, we may also choose to write y>x.)

1.     For any x, y, exactly one of the following is true:

x=y or x<y or y<x

2.     If x<y and y<z then x<z

3.     If x<y then for any a,  x+a < y+a

4.     If x<y and a>0 then x∙a<y∙a

 

Facts (theorems) and proofs (of some).  All steps use either order axioms or field axioms or previous results.

 

I.       If x+y=x+z then y=z. Proof: Exercise. (Use additive inverses.)

II.    If x<y and a<b then x+a<y+b Proof: x+a<y+a (Axiom 3). a+y<b+y (Axiom 3 again). Thus x+a<y+b (Axiom 2 and field axioms)

III.   x∙0=0

Proof: 0 + x∙0 = x∙0 = x∙ (0+0) = x∙0 + x∙0 . Therefore by I, x∙0 = 0

IV.                         (-1)∙x=-x Proof: x+(-1)∙x=x∙1+(-1)∙x=x∙(1+(-1))=x∙0=0 (by III) Therefore -1∙x is the additive inverse of x.  I.e. (-1)∙x=-x

V.   x∙y=0 ↔ x=0 or y=0

Proof: The “←” part is immediate from III. The converse is equivalent to: If x∙y=0 and x≠ 0 then y=0.  Since x≠0, there exists 1/x.  Then we have: 0=(1/x)∙0=(1/x)(x∙y)=1∙y=y

VI.                        x<y iff –y<-x.  Proof: Suppose x<y. Then x+(-x+(-y))<y+(-x+(-y)) and –y<-x quickly follows.  Conversely if –y<-x , add y+x to both sides.

VII.                     (-x) ∙ y= -(x∙y) Proof: (-x) ∙ y = (-1∙x)∙y = (-1)∙(x∙y)=-(x∙y) (Using IV and field axioms.)

VIII.                  If x<y and z<0, then x∙z>y∙z Proof: If z<0 then adding –z to both sides implies that –z>0.  Therefore (-z) ∙ x < (-z) ∙y  Thus,

using VII: -x∙z<-y∙z and then VI. yields x∙z>y∙z

IX.                        –(-x)=x  Proof: -x+x=0.  Therefore x is the additive inverse   of -x.  I.e. x=-(-x)

 

X.   If x≠0 then x-1 ≠0 and (x-1 ) -1  = x Proof: If x-1 = 0 then:

1 = x∙x-1 = 0∙x = 0, contradiction. (x-1 ) -1∙(x-1)=1 But x∙(x-1)=1 Therefore by the uniqueness of inverses, x=(x-1 ) -1 

 

XI.                        x∙y=(-x) ∙ (-y)

Proof: (-x) ∙ (-y)

= -(x∙ (-y)  ) (by VII.)

= -(-( x∙y )) (VII with field axioms)

= x∙y (IX)

XII.        If x≠0, then x2 > 0.  Proof:

If x>0, then x2 > 0∙ x = 0 (Using axiom 4). So x2 > 0

If x<0, then –x>0. (Add -x to both sides, using axiom 3.)

Then x2 = (-x)2 > 0∙ (-x) = 0 and again x2 > 0

 

XIII.    0<1

Proof one:  We use axiom 1.  If 0>1 then -1>0 (add -1 to both sides). And then: (-1)2 >0 but  (-1)2 = 1 (XI and field axiom).  So 0<1.  Contradiction. So  0>1 leads to contradiction. Also 0=1 contradicts field axiom.  So 0<1 is the only choice.

Proof two: 1 = 12 > 0 (Using XII)