1

(This is a follow-up to Need help with proof for Dedekind cuts on $\mathbb{Q}^+$ posted December 23.)

I am still working on the same proof about Dedekind cuts on the positive rational numbers. Now I am stuck on the final step of the proof on the following point and would appreciate any help.

Given $x\in \mathbb{Q}^+$ such that $x<2$, how can I prove the existence of $y\in \mathbb{Q}^+$ such that $y^2 < \frac{x^2}{2}$ and $y^2 < 2$. I suspect the last requirement may be redundant.

Community
  • 1
Dan Christensen
  • 15,710

1 Answers1

1

Probably the simplest way is to let $y=\frac{x}2$: then $y^2=\frac{x^2}4<\frac{x^2}2$.

Brian M. Scott
  • 631,399
  • Downvoting a correct answer that was posted at the explicit request of the OP? Tsk. – Brian M. Scott Jan 04 '17 at 23:01
  • 1
    It seems your suspicion that I was asking the wrong question was correct. The inequality should have been $\frac{x^2}{2} < y^2 < 2$ where $x<2$. Do you have any suggestions on how I might prove the existence of such a $y\in \mathbb{Q}^+$? – Dan Christensen Jan 05 '17 at 04:53
  • @Dan: (Sorry to take so long, but it’s been a busy few days.) I’m going to have to give it quite a bit more thought; I don’t immediately see how to do it. – Brian M. Scott Jan 05 '17 at 23:26
  • I've probably taken enough of your time, Brian. Someone at Quora posted an informal proof to related problem that looks promising. I haven't had a chance to look at the details yet. – Dan Christensen Jan 07 '17 at 19:03