November 25th, 2016, 12:50 AM  #1 
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Exercise with well defined operations.
Hi, I have some problems with an exercise where I don't understand from where some calculations come from. I post the theory part where I have some doubts, and then the correspondent exercise. Theory about well defined operations: Let $\displaystyle S = \{a,b,c, ...\}$ be a set on which a binary operation $\displaystyle \circ$ is defined, and let the relation $\displaystyle \mathcal{R}$ partition S into a set $\displaystyle E = \{[a], [b], [c], ... \}$ of equivalence classes. Let a binary operation $\displaystyle \oplus$ on $\displaystyle E$ be defined by $\displaystyle [a] \oplus [b] = [a \circ b] \mbox{ for every } [a], [b] \in E$ Now, it is not immediately clear that, for arbitrary $\displaystyle p,q \in [a] \mbox{ and } r,s \in [b]$, we have the following we can name (c) $\displaystyle [p \circ r] = [q \circ s] = [a \circ b]$ . We shall say that $\displaystyle \oplus$ is well defined on $\displaystyle E$, that is, $\displaystyle [p] \oplus [r] = [q] \oplus [s] = [a] \oplus [b]$ if and only if (c) holds. END I have some question here the theory, is there any misprint? On the above, if it is right to think that as verb, I think that we could have "partitions" instead of "partition", i.e. the relation $\displaystyle \mathcal{R}$ makes a partition of $\displaystyle S$. And also, if I understand the theory, it says that is possible to define a binary operation $\displaystyle \oplus$ on the partition E, so the set E is closed with respect to the operation $\displaystyle \oplus$, and the result is an element in S representative of an equivalence class in E, obtained if we make the operation $\displaystyle \circ$ between two element of S. Here is the exercise: The relation "has the same remainder when divided by 9 as" partitions $\displaystyle \mathbb{N}$ into nine equivalence classes $\displaystyle [1], [2], ... , [9]$. If $\displaystyle \circ$ is interpreted as addition on $\displaystyle \mathbb{N}$, we can show that $\displaystyle \oplus$ is well defined. For example, when $\displaystyle x \cdot y \in \mathbb{N}, \\ 9x + 2 \in [2] \mbox{ and } 9y + 5 \in [5] \mbox{; then } \\ [2] \oplus [5] = \\ [(9x+2)+(9y+5)] = \\ [9(x+y)+7] = \\ [7] = \\ [2+5] \mbox{ etc... }$ Honestly, I don't understand what happened here in the example; please, can you help me? Can you explain it better? Many thanks! Last edited by skipjack; November 25th, 2016 at 02:12 AM. 
November 25th, 2016, 01:39 AM  #2 
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 
In addition: If I understand, here $\displaystyle \mathcal{R}$ is the relation "has the same remainder when divided by 9 as", and it partitions $\displaystyle \mathbb{N}$ in some equivalence classes as: for example $\displaystyle 3 \mathcal{R} 21$, and $\displaystyle 21 \mathcal{R} 3$ because: 3:9 has remainder 3 21:9 has remainder 3 so $\displaystyle [3]_{\mathcal{R}} = [21]_{\mathcal{R}} = \{... 3, 21, ... \}$ and all the numbers that have the same remainder 3 when divided by 9 in the same way $\displaystyle 2 \mathcal{R} 20$ and $\displaystyle 20 \mathcal{R} 2$ because: 2:9 has remainder 2 20:9 has remainder 2 so $\displaystyle [2]_\mathcal{R} = [20]_\mathcal{R} = \{...2, 20, ... \}$ and all the numbers that have the same remainder 2 when divided by 9 one thing I don't understand is: Why $\displaystyle 9x+2$ must be an element in $\displaystyle [2]$ ? Last edited by beesee; November 25th, 2016 at 01:57 AM. 
November 25th, 2016, 02:23 AM  #3  
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619  The preceding use of the word "let" means that "partition" (subjunctive) is correct and "partitions" would be incorrect. Quote:
Note that if $x$ is an integer, dividing $9x + 2$ by 9 gives $x$ with remainder 2. Can you fully understand the example now?  
November 25th, 2016, 09:23 AM  #4  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Quote:
Maybe I understand. With that, we go to calculate the set of numbers that has been divided by 9 with a quotient x and remainder 2. For example, $\mbox{for x = 1) } 9(1) + 2 = 9 + 2 = 11 \\ \mbox{for x = 2) } 9(2) + 2 = 18 + 2 = 20 \\ [2] = \{... , 11, 20, ... \}$ the same for remainder 5: $\mbox{for x = 1) } 9(1) + 5 = 9 + 5 = 14 \\ \mbox{for x = 2) } 9(2) + 5 = 18 + 5 = 23 \\ [5] = \{... , 14, 23, ... \}$ Therefore if $\circ = +$: if $a = 2, b = 3$ we obtain for example: $[2] \oplus [3] = [2 + 3] = [5]$ i.e. if I pick a number that divided by 9 has a quotient of x and a remainder of 2 and, if I pick a number that divided by 9 has a quotient x and a remainder of 3, and I go to sum them, I obtain a number that divided by 9 has a quotient x and a remainder 5, therefore a number of another equivalence class still in E. or not? what do you think? and also I don't understand how to obtain the next passages $... = [9(x+y)+7] = [7] = [2 + 5]$ can you help me? Last edited by skipjack; November 26th, 2016 at 02:56 AM.  
November 26th, 2016, 12:31 AM  #5  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Quote:
so in $[2] \oplus [5] = [(9x+2)+(9y+5)]$, and having $\circ = +$ if $a = (9x+2) \mbox{ and } b = (9y+5)$, $a,b$ would be elements in $S$, and $a+b$ a result in $S$, while $[a+b]$ an element in $E$ (or $[\mbox{ result in S }]$ an element in E), but, in the exercise it is stated that $(9x+2) \in [2], (9y+5) \in [5]$. Why we have that? and also, why is it imposed that $x \cdot y \in \mathbb{N}$, and not $x,y \in \mathbb{N}$? So frustrating..... I have the sensation that I am losing within a glass of water!  
November 26th, 2016, 03:19 AM  #6  
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619  Quote:
I think the second form was intended.  
November 26th, 2016, 04:55 AM  #7  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Quote:
Quote:
maybe, now I'll tell you why I continue to do wrong: looking at the rule we have seen before: Quote:
$[2] \oplus [5] = [2 + 5]$ where in the left side $[2],[5] \in E$, and in the right side $2, 5 \in \mathbb{N}$ are the operands of the + operator and the result $7$ is still in $\mathbb{N}$, but $[7]$ is $\in E$ instead what I see from the exercise is that the rule probably says instead $[2] \oplus [5] = [ \mbox{ one element from the equivalence class [2] } + \mbox{ one element from the equivalence class [5] } ]$ but, as written here, $\mbox{ one element from the equivalence class [2] }$ is $\in E$, and, considering what I have said before, it would be in $\mathbb{N}$. It is here that still I am unsure! Excuse me, I'm very very sorry, I am putting to the test your patience!  
November 26th, 2016, 08:47 AM  #8 
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  I understand that $x + y$ is the sum of quotients of two numbers divided by $9$ and with remainder $7$, but, why after the first passage there is $[7]$? maybe because is it imposed $x, y = 0$? but, if I go to impose $x=1, y=0$ I obtain $[9+7]$ and not $[2+5]$ as in the exercise.

November 26th, 2016, 10:56 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
In the example, $(9x+2)$ is added to $(9y+5)$ to give $9x + 9y + 2 + 5$, i.e. $9(x + y) + 7$.

November 26th, 2016, 02:29 PM  #10 
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  

Tags 
defined, exercise, operations 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Fundamental operations and recursive operations  DanRayson  Number Theory  5  September 18th, 2014 10:12 PM 
A function defined in R^2  hbkmad  Real Analysis  3  June 10th, 2014 06:43 AM 
Well Defined/Injective  zaff9  Abstract Algebra  6  January 22nd, 2013 08:21 PM 
Set defined by { X e P(A)  X is odd}  durky  Abstract Algebra  6  March 28th, 2012 01:29 PM 
When is a function (not) well defined?  cos5000  Abstract Algebra  2  May 19th, 2008 06:07 AM 