April 2nd, 2018, 01:24 PM | #11 | |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 | Quote:
2) Pell's equation, Pell Equation -- from Wolfram MathWorld is ruled out by the occurrence of division in the proof. Again, fields are ruled out in this thread because they are totally irrelevant to OP. But that doesn't matter if you can show, by whatever means, there is a d st $x^2 - d y^2 = 1$ has a solution, which shows OP can't be proven in an ordered integral domain. You answered OP by showing it couldn't be proved under given conditions. Actually very nice. Thanks | |
April 2nd, 2018, 02:00 PM | #12 |
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,806 Thanks: 1045 Math Focus: Elementary mathematics and beyond | |
April 2nd, 2018, 02:03 PM | #13 | |||
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 186 Thanks: 55 Math Focus: Algebraic Number Theory, Arithmetic Geometry | Quote:
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You're welcome. | |||
April 2nd, 2018, 02:11 PM | #14 |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 |
Your dis-proof of OP as stands represents massive over-kill: A much simpler version, accessible to most, is: $\displaystyle Z(\sqrt{2})=a+b\sqrt{2}$ $\displaystyle (a+b\sqrt{2})(a-b\sqrt{2})= a^{2}-2b^{2} = (1,0)$ has the solution (3,2) Technically, Pell's equation doesn't hold for all positive square-free integers d and I said ab=1, obviously meaning an integer. Not (1,0). Last edited by zylo; April 2nd, 2018 at 02:18 PM. |
April 2nd, 2018, 02:23 PM | #15 | |
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 186 Thanks: 55 Math Focus: Algebraic Number Theory, Arithmetic Geometry | Quote:
I'm not sure what you mean by this. | |
April 2nd, 2018, 02:41 PM | #16 | |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 | Quote:
Ok, sour grapes to some extent. Jargon makes me cranky. | |
April 2nd, 2018, 03:02 PM | #17 | |
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 186 Thanks: 55 Math Focus: Algebraic Number Theory, Arithmetic Geometry | Quote:
Also, as $d$ is positive, $\mathbb{Z}[\sqrt{d}]$ is naturally an ordered set - it's a subset of the ordered set $\mathbb{R}$. | |
April 3rd, 2018, 06:57 AM | #18 | ||
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 | Quote:
Just to put things in perspective, the OP asks, what is required to show the only solutions to ab=1 are a,b=$\displaystyle \pm$1? Commutative Ring, Integral Domain, or Ordered Domain? cjem showed (above and post 14) that in the integal domain $\displaystyle Z(\sqrt{d})$ you could have (a,b) $\displaystyle \neq \pm(1,0)\equiv1$, so more than Integral Domain is required. Note this doesn't rule out ordered domain satisfying OP because $\displaystyle Z(\sqrt{d})$ is not ordered (no <>). It turns out you need the concept of well-ordering of the integers. So OP answer requires Ordered Domain plus well-ordering. Quote:
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April 3rd, 2018, 07:12 AM | #19 | |
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 186 Thanks: 55 Math Focus: Algebraic Number Theory, Arithmetic Geometry | Quote:
Again: the elements of $\mathbb{Z}[\sqrt{d}]$ are numbers of the form $a + b \sqrt{d}$ where $a$ and $b$ are integers. If $d$ is positive, then $\sqrt{d}$ is a real number, so all elements of $\mathbb{Z}[\sqrt{d}]$ are real numbers. Thus $\mathbb{Z}[\sqrt{d}]$, being a subset of the reals, is ordered. Last edited by cjem; April 3rd, 2018 at 07:26 AM. | |
April 3rd, 2018, 07:47 AM | #20 |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 |
a+b$\displaystyle \sqrt{d}\equiv$ (a,b) Why do you suddenly bring up the reals? I have said again and again fields are ruled out of the discussion. It makes ab=1 a totally different situation. And it still isn't ordered, just as the complex numbers are a field but are not ordered. For a complex number, what does $\displaystyle z_{1}<z_{2}$ mean? |