Question

Using field and order axioms prove the following theorems:

(i) Let x, y, and z be elements of R, the

a. If 0 < x, and y < z, then xy < xz

b. If x < 0 and y < z, then xz < xy

(ii) If x, y are elements of R and 0 < x < y, then 0 < y ^ -1 < x ^ -1

(iii) If x,y are elements of R and x < y, there exists a number x in R such that x < z < y

The order axioms given are :

-A = (x is an element of R such that -x is an element of A)

-P intersection P = null set (where -P is negative numbers and P is positive numbers)

-P union {0} union P = R

If a and b are elements of P, then a + b is an element of P and ab is an element of P

If b - a is an element of P, then b > a

Answer 1