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
Get Answers For Free
Most questions answered within 1 hours.