Question

Let r be any rational number and define L = { x in Q: x < r }, the set of rational numbers less than r. Show that L is a Dedekind cut by proving the following properties:

A. There exists a rational number x in L and there exists a rational number y not in L. ( This proves L is nonempty and L is not equal to Q)

B. If x in L, then there exists z in L, such thatt x<z. ( Thus L does not contain a largest element)

C. If x in L, and if y is any rational number less than x, then y in L. ( Thus if L contains x, then L contains all the rational numbers less than x)

Answer 1

A. since r is a rational number both 2r and r/2 are rational numbers and r/2 < r (if r >0) and 2r < r (if r<0)

hence, take x = r/2 (if r>0) and take x=2r (if r<0) hence x belongs to L

now, r/2 > r (if r <0) and 2r > r (if r>0)

hence, take y = r/2 (if r<0) and take y = 2r (if r>0) hence y is a rational but does not belongs to L

B. if x is in L , then, x < r. since both x and r are rational so is (x+r) and so is (x+r)/2 and we know, x < (x+r)/2 < r

hence, take, z = (x+r)/2

C. x is in L , i.e. x<r and y<x

then, y<x and x<r and by order property and transitivity of "<" we have, y < r

hence y belongs to L