# Thread: proof of existence of z in x < z < y

1. ## proof of existence of z in x < z < y

If x and y are arbitrary real numbers with x < y, prove that there is at least one real z satisfying x < z < y.

At my disposal, the least-upper-bound axiom - every nonempty set S of reals bounded above has a supremum. And a handful of theorems including x > sup S - h where h > 0 and set S has a supremum.

Is my proof OK?

Let the set S be bounded above by y, so y = sup S. Let x $\displaystyle \in$ S and x + h = y = sup S. Then by the above theorem, there exists a z > sup S - h, therefore sup S = y > z > sup S - h = x. Q.E.D.

Thanks,
Scott

2. Why does the fact that $\displaystyle S$ is bounded above by $\displaystyle y$ imply that $\displaystyle y = \sup S$?

The least upper bound property says the following: Suppose $\displaystyle E \subset S$ where $\displaystyle S$ is ordered,$\displaystyle E \neq \emptyset$, and $\displaystyle E$ is bounded above. Then $\displaystyle \sup E$ exists in $\displaystyle S$.

3. Originally Posted by Sampras
Why does the fact that $\displaystyle S$ is bounded above by $\displaystyle y$ imply that $\displaystyle y = \sup S$?
A little loose with my wording... you're right, it doesn't.

Let y be the least upper bound of set S, so y = sup S. Let x $\displaystyle \in$ S and x + h = y = sup S. Then by the above theorem, there exists a z > sup S - h, therefore sup S = y > z > sup S - h = x. Q.E.D.

Scott

4. Originally Posted by ScottO
If x and y are arbitrary real numbers with x < y, prove that there is at least one real z satisfying x < z < y.

At my disposal, the least-upper-bound axiom - every nonempty set S of reals bounded above has a supremum. And a handful of theorems including x > sup S - h where h > 0 and set S has a supremum.

Is my proof OK?

Let the set S be bounded above by y, so y = sup S. Let x $\displaystyle \in$ S and x + h = y = sup S. Then by the above theorem, there exists a z > sup S - h, therefore sup S = y > z > sup S - h = x. Q.E.D.

Thanks,
Scott
you want to show that real Nos are dense. This you can show in the following way:

Let x<y, then by adding x to both sides you get :

2x<y+x............................................ .......................................1

by adding y to both sides you get:

x+y<2y............................................ ........................................2

from (1) AND (2) we have :

x<(x+y)/2 <y ,so there exists z,such that x<z<y and z=(x+y)/2