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 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 , , ... , $. If $\displaystyle \circ$ is interpreted as addition on $\displaystyle \mathbb{N}$, we can show that $\displaystyle \oplus$ is well defined. For example, when $\displaystyle x \cdot y \in \mathbb{N}, \\ 9x + 2 \in  \mbox{ and } 9y + 5 \in  \mbox{; then } \\  \oplus  = \\ [(9x+2)+(9y+5)] = \\ [9(x+y)+7] = \\  = \\ [2+5] \mbox{ etc... }$ 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 $\displaystyle _{\mathcal{R}} = _{\mathcal{R}} = \{... 3, 21, ... \}$ 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 $\displaystyle _\mathcal{R} = _\mathcal{R} = \{...2, 20, ... \}$ 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 $ ? Last edited by beesee; November 25th, 2016 at 01:57 AM. November 25th, 2016, 02:23 AM   #3
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Quote:
 Originally Posted by beesee I think that we could have "partitions" instead of "partition"
The preceding use of the word "let" means that "partition" (subjunctive) is correct and "partitions" would be incorrect.

Quote:
 Originally Posted by beesee . . . the result is an element in S representative of an equivalence class in E
No, the result is an equivalence class in E. The equation [$a$] $\oplus$ [$b$] = [$a \circ b$] uses arbitrary elements $a$ and $b$ of S to identify the members of E that are referred to. For consistency, the result needs to be the same if $a$ is replaced by some other member of [$a$] and/or $b$ is replaced by some other member of [$b$]. If (and only if) the definition is always consistent, $\oplus$ is termed well-defined (on E).

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
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Quote:
 Originally Posted by skipjack Note that if $x$ is an integer, dividing $9x + 2$ by 9 gives $x$ with remainder 2.
Many thanks!

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,
$\mbox{for x = 1) } 9(1) + 2 = 9 + 2 = 11 \\ \mbox{for x = 2) } 9(2) + 2 = 18 + 2 = 20 \\  = \{... , 11, 20, ... \}$

the same for remainder 5:
$\mbox{for x = 1) } 9(1) + 5 = 9 + 5 = 14 \\ \mbox{for x = 2) } 9(2) + 5 = 18 + 5 = 23 \\  = \{... , 14, 23, ... \}$

Therefore if $\circ = +$:
if $a = 2, b = 3$ we obtain for example:
$ \oplus  = [2 + 3] = $
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
$... = [9(x+y)+7] =  = [2 + 5]$
can you help me?

Last edited by skipjack; November 26th, 2016 at 02:56 AM. November 26th, 2016, 12:31 AM   #5
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Quote:
 Originally Posted by skipjack The equation [$a$] $\oplus$ [$b$] = [$a \circ b$] uses arbitrary elements $a$ and $b$ of S to identify the members of E that are referred to.
if I understand, $[a], [b]$ are elements in $E$, and in $[a \circ b], a \circ b$ is a operation between two numbers in $S$ and the result is in $S$, the class of equivalence of that result $[a \circ b]$ is in $E$

so in $ \oplus  = [(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 , (9y+5) \in $. 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
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Quote:
 Originally Posted by beesee . . . it is stated that $(9x+2) \in , (9y+5) \in $. Why we have that?
As  is the equivalence class that includes all members of $\mathbb{N}$ that have a remainder of 2 when divided by 9, $9x + 2$ qualifies - it gives a quotient of $x$ and a remainder of 2 when divided by 9. Similarly, $9y + 5 \in$ . These explanations assume that $x,y \in \mathbb{N}$ (see below).

Quote:
 Originally Posted by beesee . . . why is it imposed that $x \cdot y \in \mathbb{N}$, and not $x,y \in \mathbb{N}$?
I think the second form was intended. November 26th, 2016, 04:55 AM   #7
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Quote:
 Originally Posted by skipjack As  is the equivalence class that includes all members of $\mathbb{N}$ that have a remainder of 2 when divided by 9, $9x + 2$ qualifies - it gives a quotient of $x$ and a remainder of 2 when divided by 9. Similarly, $9y + 5 \in$ . These explanations assume that $x,y \in \mathbb{N}$ (see below).
ok!

Quote:
 Originally Posted by skipjack These explanations assume that $x,y \in \mathbb{N}$ ... I think the second form was intended.
ok! So that is a misprint!

maybe, now I'll tell you why I continue to do wrong:
looking at the rule we have seen before:
Quote:
 $[a] \oplus [b] = [a \circ b]$ for every $[a], [b] \in E$
Once we have chosen the numbers $a,b$, that I don't know from what set they come from, I think from $\mathbb{N}$, so $a,b \in \mathbb{N}$, and let $a=2$ and $b=5$, as in the above sample exercise, I am automatically inducted to proceed in this way:

$ \oplus  = [2 + 5]$

where in the left side $, \in E$, 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 $$ is $\in E$

instead what I see from the exercise is that the rule probably says instead
$ \oplus  = [ \mbox{ one element from the equivalence class  } + \mbox{ one element from the equivalence class  } ]$
but, as written here, $\mbox{ one element from the equivalence class  }$ 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
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Quote:
 Originally Posted by beesee and also I don't understand how to obtain the next passages $... = [9(x+y)+7] =  = [2 + 5]$ can you help me?
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 $$? maybe because is it imposed $x, y = 0$? but, if I go to impose $x=1, y=0$ I obtain $[9+7]$ and not $[2+5]$ as in the exercise. November 26th, 2016, 10:56 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2218 In the example, $(9x+2)$ is added to $(9y+5)$ to give $9x + 9y + 2 + 5$, i.e. $9(x + y) + 7$. November 26th, 2016, 02:29 PM   #10
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Quote:
 Originally Posted by skipjack In the example, $(9x+2)$ is added to $(9y+5)$ to give $9x + 9y + 2 + 5$, i.e. $9(x + y) + 7$.
yes, it's clear! But, why from $[9(x+y)+7]$ I obtain $$? Are we assigning any particular values to $x,y$? Tags defined, exercise, operations Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post DanRayson Number Theory 5 September 18th, 2014 10:12 PM hbkmad Real Analysis 3 June 10th, 2014 06:43 AM zaff9 Abstract Algebra 6 January 22nd, 2013 08:21 PM durky Abstract Algebra 6 March 28th, 2012 01:29 PM cos5000 Abstract Algebra 2 May 19th, 2008 06:07 AM

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