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April 2nd, 2018, 08:00 AM   #1
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ab=1

ab=1
Show only solutions are a,b=$\displaystyle \pm$1

What is required?
i] commutative ring:
add,mult,assoc,comm,dist,0,1,and additive inverse.

ii] Integral domain:
i] and cancellation

iii] Order
Positive elements st for any b, b=0 or b pos, or -b pos. And definition of <,>.

For example, ab=0 $\displaystyle \rightarrow$ a or b =0 by ii].
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April 2nd, 2018, 09:02 AM   #2
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Let $S$ be any set containing $1$ and $-1$. Define a binary operation $S \times S \to S, (a,b) \mapsto ab$ on $S$ by

\[
ab =
\begin{cases}
\hfill 1 \hfill & \text{if $a$ and $b$ are both in $\{1, -1\}$} \\
\hfill -1 \hfill & \text{otherwise} \\
\end{cases}
\]

Then this satisfies $ab = 1 \implies a,b = \pm 1$ but none of the "required" properties hold.

_________

Even if you're only interested in integral domains, then you still don't need all of the properties you've listed.

For example, take the ring of integers $\mathcal{O}_K$ of any quadratic imaginary field $K$ (except for $K = \mathbb{Q}(i)$ or $K = \mathbb{Q}\left(\sqrt{-3}\right)$). Then $\mathcal{O}_K$ is an integral domain in which the implication $ab = 1 \implies a, b = \pm 1$ holds, but it's impossible to give $\mathcal{O}_K$ a coherent notion of positive/negative numbers.
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April 2nd, 2018, 09:44 AM   #3
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Quote:
Originally Posted by cjem View Post
Let $S$ be any set containing $1$ and $-1$. Define a binary operation $S \times S \to S, (a,b) \mapsto ab$ on $S$ by

\[
ab =
\begin{cases}
\hfill 1 \hfill & \text{if $a$ and $b$ are both in $\{1, -1\}$} \\
\hfill -1 \hfill & \text{otherwise} \\
\end{cases}
\]

Then this satisfies $ab = 1 \implies a,b = \pm 1$ but none of the "required" properties hold.

_________

Even if you're only interested in integral domains, then you still don't need all of the properties you've listed.
You are defining a set whose elements satisfy what you are trying to prove. I can prove anything that way.

Prove "blahblah"
Let S be the set of elements that satisfy "blahblah."
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April 2nd, 2018, 09:51 AM   #4
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Quote:
Originally Posted by zylo View Post
You are defining a set whose elements satisfy what you are trying to prove. I can prove anything that way.

Prove "blahblah"
Let S be the set of elements that satisfy "blahblah."
What's your point? You yourself started the thread with
Quote:
ab = 1
Show the only solutions are $\displaystyle a,b = \pm 1$
Given your information the solution is all b such that $\displaystyle b = a^{-1}$.

Clearly you have something in mind beyond cjem's comment. Don't get picky about the answers you get if you don't provide the necessary information!

-Dan
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April 2nd, 2018, 10:18 AM   #5
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Quote:
Originally Posted by topsquark View Post
What's your point?

Given your information the solution is all b such that $\displaystyle b = a^{-1}$.

Clearly you have something in mind beyond cjem's comment. Don't get picky about the answers you get if you don't provide the necessary information!

-Dan
My point was that cjem's reply was meaningless, and I explained why. If you disagree, please explain why, instead of responding in deprecatory generalities.

The OP was crystal clear:
Find sols of ab=1 using properties of commutative ring and, if necessary, integral domain and properties of order. I specifically did not allow a field, which is an integral domain with an inverse, for example the rationals.

In a field, obviously and trivially, ab=1 has a solution for any a.
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April 2nd, 2018, 10:52 AM   #6
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Quote:
Originally Posted by zylo View Post
My point was that cjem's reply was meaningless, and I explained why. If you disagree, please explain why, instead of responding in deprecatory generalities.

The OP was crystal clear:
Find sols of ab=1 using properties of commutative ring and, if necessary, integral domain and properties of order. I specifically did not allow a field, which is an integral domain with an inverse, for example the rationals.

In a field, obviously and trivially, ab=1 has a solution for any a.
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 $(x + y \sqrt{d})(x - y \sqrt{d}) = x^2 - d y^2 = 1$.
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April 2nd, 2018, 11:10 AM   #7
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For positive integers
$\displaystyle a\in N \Rightarrow b^{-1} \in N$
$\displaystyle 1 | b \Rightarrow $ $\displaystyle b\leq 1$
$\displaystyle (a,b)=(1,1)$

Last edited by idontknow; April 2nd, 2018 at 11:14 AM.
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April 2nd, 2018, 11:32 AM   #8
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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.

So, unless someone has a solution in the strict context of an ordered domain, the OP can't be proved. My apologies.

Again, a field (integral domain plus inverse}, is irrelevant.
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April 2nd, 2018, 12:03 PM   #9
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Quote:
Originally Posted by zylo View Post
x
My second post (post #6) contains a counterexample to the statement:
It's an ordered integral domain (that isn't a field) in which $ab = 1$ has infinitely many solutions with $a, b \neq \pm 1$.
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April 2nd, 2018, 12:38 PM   #10
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Quote:
Originally Posted by zylo View Post
The OP was crystal clear
With all due respect, it is not for you to say that. The fact that people reading it interpreted it differently clearly indicates that it wasn't.
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