April 2nd, 2018, 09:00 AM | #1 |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 | 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]. |
April 2nd, 2018, 10:02 AM | #2 |
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry |
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. |
April 2nd, 2018, 10:44 AM | #3 | |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 | Quote:
Prove "blahblah" Let S be the set of elements that satisfy "blahblah." | |
April 2nd, 2018, 10:51 AM | #4 | ||
Math Team Joined: May 2013 From: The Astral plane Posts: 1,922 Thanks: 774 Math Focus: Wibbly wobbly timey-wimey stuff. | Quote:
Quote:
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 | ||
April 2nd, 2018, 11:18 AM | #5 | |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 | Quote:
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. | |
April 2nd, 2018, 11:52 AM | #6 | |
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry | Quote:
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 $(x + y \sqrt{d})(x - y \sqrt{d}) = x^2 - d y^2 = 1$. | |
April 2nd, 2018, 12:10 PM | #7 |
Senior Member Joined: Dec 2015 From: Earth Posts: 276 Thanks: 32 |
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 12:14 PM. |
April 2nd, 2018, 12:32 PM | #8 |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 |
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. |
April 2nd, 2018, 01:03 PM | #9 |
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry | |
April 2nd, 2018, 01:38 PM | #10 |
Math Team Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 Math Focus: Mainly analysis and algebra | |