(a) If x-y>1, show that there is at least one n Z (integers) such that y<n<x.
For this should I try showing that there are no n's in Z? Essentially a proof by contradiction.
(b) if y<x, show there is a rational number z such that y<z<x.
At first i was thinking the transitive property (i think thats the name of the one i''m thinking of) but I don't seem to confident in it and i'm only recalling about relations like (a,b) (c,d) (a,d) etc.
Fundamentally, you are using the "Archimedean principle"- if z is any real number, there exist an integer, N, such that N> z- and the fact that the set of positive integers is "well ordered"- any non-empty set of positive integers contains a smallest integer.
Certainly if x- y> 1, x- y is positive. Let z= x- y. Then there exist a positive integer N> z= x- y so the set of "all integers N, greater than x- y" is non-empty and so has a smallest integer, n. Since n is the smallest integer in the set, n- 1 is NOT in the set. Consider the two cases:
1) n- 1= x- y
2) n-1< x- y