## dedekind cut

How do I show that A={r in Q: r^3<2} is a Dedekind cut.

Here is the definition I am working with.

A subset A of Q is a Dedekind cut if and only if A satisfies the
following 3 properties:

(i) A is a proper nonempty subset of Q.
(ii) If r is in A, s in Q, and s<r, then s is in A.
(iii) A contains no greatest rational.

I showed that A satisfies (i) and (ii). I noticed that 5/4 is in A
and I tried to find a rational greater than 5/4 whose cube is less
than 2. I looked at the sequence (5n+1)/4n, and I found that n=26
works. That is, (4/5)<(131/104) and (131/104)^3 < 2.

So I think that if a/b is any rational with b>0 and (a/b)^3 < 2, then
there should be some positive integer n such that [(an+1)/bn]^3 < 2. But I don't know
how to show that this n exists.

 Recognitions: Gold Member Science Advisor Staff Emeritus I would start with some specific numbers. Since the cube root of 2 is approximately 1.26 (from a calculator) note that 1.253= 1.953125< 2 so 1.25 is in the set. On the other hand 1.263= 2.000376> 2 so 1.26 is not in the set. The point of that is that any r in the set that is less than 1.25 has 1.25 larger so if there were a largest member it would have to be between 1.25 and 1.26. Let $\delta= 2- x^3$. Since 1.253= 1.953125, $\delta< 2- 1.95325= 0.045875$. Now, try to find some number, n (a positive integer just for simplicity) so that $(x+ \delta/n)^3= x^3+ 3(\delta/n)x^2+ 3(\delta/n)^2 x+ (\delta/n)^3= x^3+ \delta(3x^2/n+ 3\delta x/n^2+ \delta^2/n^3)< 2$ Remembering the bounds on the sized for x and $\delta$ that should be easy. Proof by contradiction: Suppose there were some number, x, which were the largest member of {x| x3< 2}. That, as above, [itex]1.25< x< 1.27