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April 2nd, 2018, 01:24 PM   #11
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Quote:
 Originally Posted by cjem Whoops! I misread the OP and interpreted it as follows: If the implication $ab = 1 \implies a, b = \pm 1$ holds, then it is required that we are in a commutative ring that is an integral domain with the specified order properties. To answer the actual OP: Let $d$ be a positive, squarefree integer and consider the (ordered) integral domain $R = \mathbb{Z}\left[\sqrt{d}\right] = \{a + b \sqrt{d} : a, b \in \mathbb{Z} \}$. By elementary number theory, there are infinitely many pairs of integers $(x,y)$ such that $x^2 - d y^2 = 1$. (If you're not familiar with this, look up Pell's equation.) So there is certainly such a pair $(x,y)$ not equal to (1,0) or (-1,0). Then $x+y \sqrt{d}, x-y \sqrt{d} \neq \pm 1$ but $(x + y \sqrt{d})(x - y \sqrt{d}) = x^2 - d y^2 = 1$.
1) You have misstated the OP.. The OP requires that you show the solutions to ab=1 are a,b =$\displaystyle \pm$1 using a commutative ring, or integral domain, or ordered integral domain. I thought about "divisors of 1" as a topic, but I wanted to minimize look-up jargon and maximize accessibility. A simple solution was found in my previous post by adding well-ordering requirement.

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
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Quote:
 Originally Posted by zylo The OP was crystal clear...
"was"?

April 2nd, 2018, 02:03 PM   #13
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Quote:
 Originally Posted by zylo 1) You have misstated the OP.. The OP requires that you show the solutions to ab=1 are a,b =$\displaystyle \pm$1 using a commutative ring, or integral domain, or ordered integral domain. I thought about "divisors of 1" as a topic, but I wanted to minimize look-up jargon and maximize accessibility. A simple solution was found in my previous post by adding well-ordering requirement.
Yes, I was just explaining my original misinterpretation of the OP.

Quote:
 Originally Posted by zylo 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.
I'm not sure what you're getting at here. $\mathbb{Z}[\sqrt{d}]$ is not a field! But if $d$ is a positive, squarefree integer, then it is an ordered integral domain in which $ab = 1$ has solutions $a, b \neq \pm 1$.

Quote:
 Originally Posted by zylo 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.
The results regarding Pell's equation show that for every positive, squarefree integer $d$, there are infinitely many integer solutions to this equation.

Quote:
 Originally Posted by zylo You answered OP by showing it couldn't be proved under given conditions. Actually very nice. Thanks
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
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Quote:
 Originally Posted by zylo 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)
Is this $(1,0)$ here a typo? It should just be $1$. But you're right, it would have been better to give one specific example rather than the general result! :P

Quote:
 Originally Posted by zylo I said ab=1, obviously meaning an integer. Not (1,0).
I'm not sure what you mean by this.

April 2nd, 2018, 02:41 PM   #16
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Quote:
 Originally Posted by cjem Is this $(1,0)$ here a typo? It should just be $1$. But you're right, it would have been better to give one specific example rather than the general result! :P I'm not sure what you mean by this.
(1,0) is the unity of the integral domain (unordered by the way) you proposed, not the 1 specified in OP. I wanted the answer for integer 1, a simple question really.

Ok, sour grapes to some extent. Jargon makes me cranky.

April 2nd, 2018, 03:02 PM   #17
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Quote:
 Originally Posted by zylo (1,0) is the unity of the integral domain (unordered by the way) you proposed, not the 1 specified in OP. I wanted the answer for integer 1, a simple question really. Ok, sour grapes to some extent. Jargon makes me cranky.
No, the integer $1$ is the unity of $\mathbb{Z}[\sqrt{d}] = \{a + b \sqrt{d} : a, b \in \mathbb{Z} \}$. How could you even think $(1,0)$ is an element of this integral domain?

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
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Quote:
 Originally Posted by cjem ... So there is certainly such a pair $(x,y)$ not equal to (1,0) or (-1,0)....
And $\displaystyle Z(\sqrt{d})$ is not ordered: what does (a,b)<(c,e) mean?

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:
 Originally Posted by zylo Found an answer in BM. ab=1 -> |ab|= |a||b|=1 |a|$\displaystyle \geq$1 and |b|$\displaystyle \geq1\rightarrow$|a|=|b|=1 But there is a catch: OP properties don't let you prove there is no element between 0 and 1. You either fudge it, "obviously there is no integer between 0 or 1," or you have to invoke the Well Ordering Principle for Integers: every subset of positive integers contains a least member, to show it.
To show there is no integer between 0 and 1, assume there is. Then by well ordering the set of integers less than 1 has a least member $\displaystyle 0<m<1$. Then $\displaystyle 0<m^{2}<m$ is a contradiction.

April 3rd, 2018, 07:12 AM   #19
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Quote:
 Originally Posted by zylo And $\displaystyle Z(\sqrt{d})$ is not ordered: what does (a,b)<(c,e) mean?
Reread what I wrote more carefully... (x,y), (1,0) and (-1,0) are not elements of $\mathbb{Z}[\sqrt{d}]$; they are solutions to an equation in two variables.

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}

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