1. ## Dedekind Cut

Let $\alpha = 0^* \cup \{p \in \mathbf{Q}: p \geq 0$ and $p^2 <2 \}$. Prove that $\alpha$ is a Dedekind cut and also that it has the property $\alpha \cdot \alpha = 2^*$; that is, the square of $\alpha$ is $2^*$. Note: This seems to be surprisingly tricky, as pointed out by Linda Hill and Robert J. Fisher at Idaho State University. Their solution is available from them or from the author.

2. Originally Posted by novice
Let $\alpha = 0^* \cup \{p \in \mathbf{Q}: p \geq 0$ and $p^2 <2 \}$. Prove that $\alpha$ is a Dedekind cut and also that it has the property $\alpha \cdot \alpha = 2^*$; that is, the square of $\alpha$ is $2^*$. Note: This seems to be surprisingly tricky, as pointed out by Linda Hill and Robert J. Fisher at Idaho State University. Their solution is available from them or from the author.
Since you have a solution in what way is this a question?

CB

3. Obviously a homework problem.

The only real difficulty is in showing that the set does NOT have a largest member.

Suppose x is the largest member of the set of all rational numbers whose square is less than 2. Since $1.4^2= 1.96< 2$ you can assume 1.4< x. Since $1.5^2= 2.25> 2$ you can assume x< 1.5. Let $d= 2- x^2$ (d is, of course, rational). Then $d< 2- 1.4^2= 2- 1.96= 0.04$. Now find an integer, n such that $(x+ d/n)^2= x^2+ 2 dx/n+ d^2/n^2< 2$ showing that x is NOT the largest number in that set.

4. Thank you, sir. You instruction is more than adequate. I knew the set has no maximum, but didn't know how to commucate my idea in mathematical term. The last line in particular is of great help.

5. Originally Posted by CaptainBlack
Since you have a solution in what way is this a question?

CB
Respectfully sir, I do not have the solution. It will take at least three weeks to get the book by the inter-library loan. I could eventually get my answer one way or the other, but I NOT am too proud to ask.

6. Originally Posted by novice
Respectfully sir, I do not have the solution. It will take at least three weeks to get the book by the inter-library loan. I could eventually get my answer one way or the other, but I am too proud to ask.
take a peek at wikipedia, eh?!

7. Originally Posted by HallsofIvy
Obviously a homework problem.

The only real difficulty is in showing that the set does NOT have a largest member.

Suppose x is the largest member of the set of all rational numbers whose square is less than 2. Since $1.4^2= 1.96< 2$ you can assume 1.4< x. Since $1.5^2= 2.25> 2$ you can assume x< 1.5. Let $d= 2- x^2$ (d is, of course, rational). Then $d< 2- 1.4^2= 2- 1.96= 0.04$. Now find an integer, n such that $(x+ d/n)^2= x^2+ 2 dx/n+ d^2/n^2< 2$ showing that x is NOT the largest number in that set.
Since the domain of $\sqrt{x}$ is $\mathbb{Z}^+ \cup {0}$, does it mean that the set $\{p\in \mathbb{Q}: p \geq 0$ and $p^2 <2 \}$ is not a Dedekin cut in itself, unless it is unioned with $0^*$?