My Math Forum Exercise with well defined operations.

 Abstract Algebra Abstract Algebra Math Forum

 November 25th, 2016, 01:50 AM #1 Senior Member   Joined: Jan 2013 From: Italy Posts: 137 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 03:12 AM.
 November 25th, 2016, 02:39 AM #2 Senior Member   Joined: Jan 2013 From: Italy Posts: 137 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 02:57 AM.
November 25th, 2016, 03:23 AM   #3
Global Moderator

Joined: Dec 2006

Posts: 16,225
Thanks: 1150

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, 10:23 AM   #4
Senior Member

Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

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

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 03:56 AM.

November 26th, 2016, 01:31 AM   #5
Senior Member

Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

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 $[2] \oplus [5] = [(9x+2)+(9y+5)]$, 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, 04:19 AM   #6
Global Moderator

Joined: Dec 2006

Posts: 16,225
Thanks: 1150

Quote:
 Originally Posted by beesee . . . it is stated that $(9x+2) \in [2], (9y+5) \in [5]$. Why we have that?
As [2] 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, $9y + 5 \in$ [5]. 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, 05:55 AM   #7
Senior Member

Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

Quote:
 Originally Posted by skipjack As [2] 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, $9y + 5 \in$ [5]. 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:

$[2] \oplus [5] = [2 + 5]$

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, 09:47 AM   #8
Senior Member

Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

Quote:
 Originally Posted by beesee and also I don't understand how to obtain the next passages $... = [9(x+y)+7] = [7] = [2 + 5]$ 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 $[7]$? 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, 11:56 AM #9 Global Moderator   Joined: Dec 2006 Posts: 16,225 Thanks: 1150 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, 03:29 PM   #10
Senior Member

Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

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 $[7]$? Are we assigning any particular values to $x,y$?

 Tags defined, exercise, operations

 Thread Tools Display Modes Linear 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 09:21 PM durky Abstract Algebra 6 March 28th, 2012 01:29 PM cos5000 Abstract Algebra 2 May 19th, 2008 06:07 AM

 Contact - Home - Forums - Top