1. ## equivalence class

Wazzup,

I hope that everybody is A+ up in here.

Check it:

I have have a subset $Q$ of $\mathbb{Z} \times \mathbb{Z}$ :

$Q=\left\lbrace \left( a,b \right)\in \mathbb{Z}\times \mathbb{Z}\mid b \neq 0\right\rbrace$

I have a relation $\sim$ defined on $Q$:

$\left( a,b \right) \sim \left( c,d \right)\Leftrightarrow ad=bc$

I've shown that the relation is an equivalence-relation.

Now I wanna "find" $\left[ (2,3)\right]$ and in general $\left[ (a,b)\right]$.

Is the following enough:

The equivalence class $\left[ (2,3)\right]$ consists of all pairs of numbers $(c,d)$ such that $2d=3c$, meaning $\left[ (2,3) \right]= \left\lbrace (c,d) \mid 2d=3c \right\rbrace$.

The equivalence class $\left[ (a,b)\right]$ consists of all pairs of numbers $(c,d)$ such that $ad=bc$, meaning $\left[ (a,b) \right]= \left\lbrace (c,d) \mid ad=bc \right\rbrace$.

Get back 2 me
Peace out

2. ## Re: equivalence class

Originally Posted by Meelas
Wazzup,

I hope that everybody is A+ up in here.

Check it:

I have have a subset $Q$ of $\mathbb{Z} \times \mathbb{Z}$ :

$Q=\left\lbrace \left( a,b \right)\in \mathbb{Z}\times \mathbb{Z}\mid b \neq 0\right\rbrace$

I have a relation $\sim$ defined on $Q$:

$\left( a,b \right) \sim \left( c,d \right)\Leftrightarrow ad=bc$

I've shown that the relation is an equivalence-relation.

Now I wanna "find" $\left[ (2,3)\right]$ and in general $\left[ (a,b)\right]$.

Is the following enough:

The equivalence class $\left[ (2,3)\right]$ consists of all pairs of numbers $(c,d)$ such that $2d=3c$, meaning $\left[ (2,3) \right]= \left\lbrace (c,d) \mid 2d=3c \right\rbrace$.

The equivalence class $\left[ (a,b)\right]$ consists of all pairs of numbers $(c,d)$ such that $ad=bc$, meaning $\left[ (a,b) \right]= \left\lbrace (c,d) \mid ad=bc \right\rbrace$.

Get back 2 me
Peace out
i'd take it a bit further

$d = \dfrac 3 2 c \Rightarrow \left(c, \dfrac 3 2 c\right) = c\left(1,\dfrac 3 2\right),~c \in \mathbb{Z}$

Now since the elements of $[(2,3)]$ must be integers it must be that $c$ is even in the formula above so

$[(2,3)] = k\left(2,3 \right),~k \in \mathbb{Z}$

similarly

$[(a,b)] = k\left(a,b\right),~k \in \mathbb{Z}$

this reveals the true nature of the equivalence class

3. ## Re: equivalence class

Originally Posted by romsek
i'd take it a bit further
$d = \dfrac 3 2 c \Rightarrow \left(c, \dfrac 3 2 c\right) = c\left(1,\dfrac 3 2\right),~c \in \mathbb{Z}$

this reveals the true nature of the equivalence class Does it?
What would you do the the equivalence class, $[(0,1)]~?$

4. ## Re: equivalence class

Originally Posted by Plato
What would you do the the equivalence class, $[(0,1)]~?$
you're right

$[(0,b)]=(a,0),~a\in\mathbb{Z}$

5. ## Re: equivalence class

Originally Posted by romsek
you're right

$[(0,b)]=(a,0),~a\in\mathbb{Z}$ WRONG!
In the definition, $\{(a,b)|b\not= 0$.

6. ## Re: equivalence class

Originally Posted by Plato
In the definition, $\{(a,b)|b\not= 0$.
yep, i got the order mixed up

$[(0,b)]=(0,a), a \in \mathbb{Z}-\{0\}$

7. ## Re: equivalence class

W'zup yall,

Great feedback but Romsek, you said:

$\left( c,d \right) = \left( c,\frac{3}{2}c \right) =c\left(1, \frac{3}{2} \right) \hspace{0,3cm} , \hspace{0,3cm} c \in \mathbb{Z}$

So you divide by a, but a could take on the value 0, as could c and d. The only element which (by definition) cannot take on the value 0 is b, so wouldn't it be more correct to divide by b?

Wazzup with it?

8. ## Re: equivalence class

Originally Posted by Meelas
W'zup yall,

Great feedback but Romsek, you said:

$\left( c,d \right) = \left( c,\frac{3}{2}c \right) =c\left(1, \frac{3}{2} \right) \hspace{0,3cm} , \hspace{0,3cm} c \in \mathbb{Z}$

So you divide by a, but a could take on the value 0, as could c and d. The only element which (by definition) cannot take on the value 0 is b, so wouldn't it be more correct to divide by b?

Wazzup with it?
I guess I can agree with that.

you'd end up with $c=\dfrac{a}{b} d$

$\left(\dfrac{a}{b}d,d\right) = d\left(\dfrac{a}{b},1\right)=d b(a, b) = k(a,b),~ k \in \mathbb{Z} - \{0\}$

and you get the same answer.

The correct answer for $(0,b)$ is in post #6