My Math Forum  

Go Back   My Math Forum > College Math Forum > Abstract Algebra

Abstract Algebra Abstract Algebra Math Forum


Thanks Tree4Thanks
Reply
 
LinkBack Thread Tools Display Modes
November 25th, 2016, 01:50 AM   #1
Senior Member
 
Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

Question 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.
beesee is offline  
 
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.
beesee is offline  
November 25th, 2016, 03:23 AM   #3
Global Moderator
 
Joined: Dec 2006

Posts: 16,225
Thanks: 1150

Quote:
Originally Posted by beesee View Post
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 View Post
. . . 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?
Thanks from beesee
skipjack is offline  
November 25th, 2016, 10:23 AM   #4
Senior Member
 
Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

Quote:
Originally Posted by skipjack View Post
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.
beesee is offline  
November 26th, 2016, 01:31 AM   #5
Senior Member
 
Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

Quote:
Originally Posted by skipjack View Post
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!
beesee is offline  
November 26th, 2016, 04:19 AM   #6
Global Moderator
 
Joined: Dec 2006

Posts: 16,225
Thanks: 1150

Quote:
Originally Posted by beesee View Post
. . . 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 View Post
. . . 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.
Thanks from beesee
skipjack is offline  
November 26th, 2016, 05:55 AM   #7
Senior Member
 
Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

Quote:
Originally Posted by skipjack View Post
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 View Post
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!
beesee is offline  
November 26th, 2016, 09:47 AM   #8
Senior Member
 
Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

Quote:
Originally Posted by beesee View Post
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.
beesee is offline  
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. $9(x + y) + 7$.
skipjack is offline  
November 26th, 2016, 03:29 PM   #10
Senior Member
 
Joined: Jan 2013
From: Italy

Posts: 137
Thanks: 7

Quote:
Originally Posted by skipjack View Post
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 $x,y$?
beesee is offline  
Reply

  My Math Forum > College Math Forum > Abstract Algebra

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 09:21 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





Copyright © 2017 My Math Forum. All rights reserved.