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

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• May 23rd 2009, 05:11 PM
ScottO
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 $\in$ 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
• May 23rd 2009, 05:39 PM
Sampras
Why does the fact that $S$ is bounded above by $y$ imply that $y = \sup S$?

The least upper bound property says the following: Suppose $E \subset S$ where $S$ is ordered, $E \neq \emptyset$, and $E$ is bounded above. Then $\sup E$ exists in $S$.
• May 23rd 2009, 06:44 PM
ScottO
Quote:

Originally Posted by Sampras
Why does the fact that $S$ is bounded above by $y$ imply that $y = \sup S$?

A little loose with my wording... you're right, it doesn't.

How's this, instead?

Let y be the least upper bound of set S, so y = sup S. Let x $\in$ 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
• May 24th 2009, 10:54 AM
xalk
Quote:

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 $\in$ 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