Question

Using field and order axioms prove the following theorems:

(i) 0 is neither in P nor in - P

(ii) -(-A) = A (where A is a set, as defined in the axioms.

(iii) Suppose a and b are elements of R. Then a<=b if and only if a<b or a=b

(iv) Let x and y be elements of R. Then either x <= y or y <= x (or both).

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

Answer 1