Thread: 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 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

Why does the fact that is bounded above by imply that ?

The least upper bound property says the following: Suppose where is ordered, , and is bounded above. Then exists in .

Originally Posted by Sampras

Why does the fact that is bounded above by imply that ?

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 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

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 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