# Thread: Show there is at least one n in Z

1. ## Show there is at least one n in Z

(a) If x-y>1, show that there is at least one n $\in$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.

2. Originally Posted by tn11631
(a) If x-y>1, show that there is at least one n $\in$Z (integers) such that y<n<x.
(b) if y<x, show there is a rational number z such that y<z<x.
For the first one, note that $y+1.

The floor function has the property that $\left\lfloor y \right\rfloor \leqslant y < \left\lfloor y \right\rfloor + 1 \leqslant y + 1$

For the second one you know that $\left( {\exists n\in \mathbb{Z}^+ \right)\left( {n > \frac{1}{{x - y}}} \right)$. WHY is that?

Notice that means $nx-ny>1$. Apply the first part to that.

3. 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