ORDERED FIELDS 15 Consult the list of properties A1 A5 M1

ORDERED FIELDS

#15. Consult the list of properties A1 - A5, M1 - M5, DL, O1 - O4 from my Ordered Field notes. Rather than considering those properties applied on the set R of real numbers, restrict the set as indicated below. In other words, check which properties still apply, when R is replaced by the set specified.

#15 (a) Which properties are not satisfied on the set N? (Just list the identifying labels.)

#15(b) Which properties are not satisfied on the set Z? (Just list the identifying labels.)

#15(c) Let S = {x in R such that 1 x 1} = [1 , 1]. Which properties are not satisfied on the set S? (Just list the identifying labels.)

#16. Prove that 0 < 1/2 < 1. Fill in the blanks of the proof. Refer to the field axioms and order axioms and the Theorem in my Ordered Field notes, pages 1-2.

Proof:

Note that 0 < 1 (by Theorem part ___) Adding1tobothsides,0+1<1+1 byorderaxiom___.

0+1=1byfieldaxiom_____, and1+1=successorof1,whichisdesignatedby2(Peanoaxiomin Induction notes).

So,wehave 1<2. Since0<1and1<2,wehave0<1<2. Then 0 < 1/2 < 1/1 by Theorem, part ___ Notethat 1/1=1(because11=1). Thus, we have 0 < 1/2 < 1, as desired.

Solution

15. a) All properties hold

b) Since Z is a prime no.

A1 (if a,b belongs to Z than a+b does not belong to Z)

M1 (if a,b belongs to Z than a*b does not belong to Z)

Therefore these properties does not hold.

c) A1 if a,b belongs to [-1,1] than a+b not necessarily belong to[-1,1]

Example 0.5+0.6=1.1 (out of bound)

16. Adding 1 to both sides, 0+1<1+1 by order axiom O3.

0+1 = 1 by field axiom A4, and 1+1 = successor of 1, which is designated by 2 (Peanoaxiomin Induction notes).

So,we have 1<2. Since 0<1 and1<2, we have 0<1<2. Then 0 < 1/2 < 1/1 by Theorem, Note that 1/1=1(because11=1) (Using field axiom M4). Thus, we have 0 < 1/2 < 1, as desired.