# Discrete set maths

• August 19th 2009, 07:21 PM
quah13579
Discrete set maths
On the set
$
\mathbb{P}\times{}\mathbb{P}
$

define a relation by (x,y)~(u,v) if and only if xv = yu.
(a) Show that ~ is an equivalence relation.
(b) Find the equivalence class of (2,3)

(Itwasntme)
thanks
• August 19th 2009, 07:45 PM
Gamma
not sure what your sets P are, but...
Im assuming we are in a commutative setting here.
Reflexive
$(a,b)\sim(a,b)$?
$ab=ba$
Symmetry
$(a,b)\sim (c,d)\Rightarrow ad=bc \Rightarrow cb=da \Rightarrow (c,d) \sim (a,b)$

Think you can do transitive?

(2,3), so the things equivalent are of the form $(a,b)$ and must satisfy $2b=3a\Rightarrow b=\frac{3}{2}a$, so the equivalence class is
$\{(a, \frac{3}{2}a)|a\in \mathbb{P})\}$
• August 24th 2009, 11:36 PM
yoonsi
I'm sorry but could you explain that again? I'm confused, ab = ba, how does that show that its reflexive? and I thought that (x,y) =/= (y,x)

Thank you =]
• August 25th 2009, 12:35 AM
quah13579
Quote:

Originally Posted by Gamma
Im assuming we are in a commutative setting here.
Reflexive
$(a,b)\sim(a,b)$?
$ab=ba$
Symmetry
$(a,b)\sim (c,d)\Rightarrow ad=bc \Rightarrow cb=da \Rightarrow (c,d) \sim (a,b)$

Think you can do transitive?

(2,3), so the things equivalent are of the form $(a,b)$ and must satisfy $2b=3a\Rightarrow b=\frac{3}{2}a$, so the equivalence class is
$\{(a, \frac{3}{2}a)|a\in \mathbb{P})\}$

I think p is positive integer.
• August 25th 2009, 02:09 AM
Defunkt
Quote:

Originally Posted by yoonsi
I'm sorry but could you explain that again? I'm confused, ab = ba, how does that show that its reflexive? and I thought that (x,y) =/= (y,x)

Thank you =]

$(x,y)\sim(u,v) \Leftrightarrow xv = yu$
~ is reflexive iff $\forall {a,b} \in {\mathbb{P}} : (a,b)\sim(a,b) \$
$(a,b) \sim (a,b) \Leftrightarrow ab = ba$
Since $\mathbb{P}$ is commutative, $ab=ba$ for any ${a,b} \in \mathbb{P}$
• August 26th 2009, 11:17 PM
yoonsi
oh well I understand the concept of reflexivity buuuut, how come P is commutative? I thought in ordered pairs, (2,3) does not equal (3,2)?

Sorry I must seem really silly...

Could someone also step me through showing how that is transitive? (Bow)
• August 27th 2009, 12:40 AM
HallsofIvy
Quote:

Originally Posted by Gamma
Im assuming we are in a commutative setting here.
Reflexive
$(a,b)\sim(a,b)$?
$ab=ba$
Symmetry
$(a,b)\sim (c,d)\Rightarrow ad=bc \Rightarrow cb=da \Rightarrow (c,d) \sim (a,b)$

Think you can do transitive?

(2,3), so the things equivalent are of the form $(a,b)$ and must satisfy $2b=3a\Rightarrow b=\frac{3}{2}a$, so the equivalence class is
$\{(a, \frac{3}{2}a)|a\in \mathbb{P})\}$

In other words (a, b) is in the same equivalence class as (2, 3) if and only if $\frac{b}{a}= \frac{3}{2}$. You could think of that class as all fractions that are equivalent to the fraction $\frac{3}{2}$ and so representing the rational number $\frac{3}{2}$.