# Math Help - Relation

1. ## Relation

Hello. I need help on determining this relation.
Determine the relation $\rho$ defined with $x\rho y \Leftrightarrow (\exists z)(y=zx)$ for $Z$ numbers

2. Originally Posted by javax
Hello. I need help on determining this relation.
Determine the relation $\rho$ defined with $x\rho y \Leftrightarrow (\exists z)(y=zx)$ for $Z$ numbers
what do you mean "determine the relation"? isn't it already determined. it is given.

note that, geometrically speaking, the relation describes the set of all straight lines that pass through the origin with integer slope (i suppose you mean $\mathbb{Z}$ when you wrote $Z$).

3. Originally Posted by Jhevon
what do you mean "determine the relation"? isn't it already determined. it is given.

note that, geometrically speaking, the relation describes the set of all straight lines that pass through the origin with integer slope (i suppose you mean $\mathbb{Z}$ when you wrote $Z$).
Sorry for my english, I haven't had the chance to see these terms in english, I guess. Yes I mean $\mathbb{Z}$
I mean determine if it is equivalence relation or order relation. That's what it is required!

Thanks

4. Originally Posted by javax
Sorry for my english, I haven't had the chance to see these terms in english, I guess. Yes I mean $\mathbb{Z}$
I mean determine if it is equivalence relation or order relation. That's what it is required!

Thanks
it is not an equivalence relation. it is an (partial) order relation.

can you see why?

5. Originally Posted by Jhevon
it is not an equivalence relation. it is an (partial) order relation.

can you see why?
Nope, I need that prove. C'mon help me

6. Originally Posted by javax
Nope, I need that prove. C'mon help me

here's how to start, what ae the definitions of "equivalence relation" and "(partial) order relations"? how do we know a relation is one of these or not?

7. Originally Posted by Jhevon

here's how to start, what ae the definitions of "equivalence relation" and "(partial) order relations"? how do we know a relation is one of these or not?
Here we go:
1. (r) $(\forall a\in A) a\rho a$
2. (s) $(\forall a,b \in A) a\rho b \Rightarrow b\rho a$
3. (t) $(\forall a,b,c \in A) a\rho b \wedge b\rho c \Rightarrow a\rho c$
4. (a) $(\forall a,b\in A)a\rho b \wedge b\rho a \Rightarrow a = b$

The relation that fulfills conditions 1, 2 & 3 is called equivalence relation. Relation that fulfills 1, 3 & 4 is called a partial order relation. That's what I know

8. Originally Posted by javax
Here we go:
1. (r) $(\forall a\in A) a\rho a$
2. (s) $(\forall a,b \in A) a\rho b \Rightarrow b\rho a$
3. (t) $(\forall a,b,c \in A) a\rho b \wedge b\rho c \Rightarrow a\rho c$
4. (a) $(\forall a,b\in A)a\rho b \wedge b\rho a \Rightarrow a = b$

The relation that fulfills conditions 1, 2 & 3 is called equivalence relation. Relation that fulfills 1, 2 & 4 is called a partial order relation. That's what I know
actually, 1, 3 & 4 makes a partial order relation.

ok, so check if it is an equivalence relation:

(1) does x relate to itself? can you find and integer such that x = zx?
(2) do we have symmetry. if we can find and integer z so that y = zx, will we always be able to find an integer n so that x = ny?
(3) do we have transitivity? if we have integers n and m so that a = mb and b = nc, can we find an integer k so that a = kc?

9. Originally Posted by Jhevon
actually, 1, 3 & 4 makes a partial order relation.

ok, so check if it is an equivalence relation:

(1) does x relate to itself? can you find and integer such that x = zx?
(2) do we have symmetry. if we can find and integer z so that y = zx, will we always be able to find an integer n so that x = ny?
(3) do we have transitivity? if we have integers n and m so that a = mb and b = nc, can we find an integer k so that a = kc?
After exploring my notebook, here's my attempt:

(1) (r)
$x\rho x \Leftrightarrow \exists k=1$ such that $x=kx \Rightarrow x=x$
it fills the reflexive condition

(2) (s) $x\rho y (\exists z)(y=zx)$, $\hspace{2cm}y\rho x (\exists u)(x=uy)$
if we multiply these we have $(\exists z)(\exists u) yx=z x u y$ For $z=u=1$ we have $yx=xy$ (Does this mean that we have a symmetry?

(3) (t) $\exists z(y=zx)\wedge \exists u(t=uy)$. I multiplied again
$(\exists u)(\exists z) t=(uz)x.$
sub. $uz=p \Rightarrow t=px.$

What do you think?

10. Originally Posted by javax
After exploring my notebook, here's my attempt:

(1) (r)
$x\rho x \Leftrightarrow \exists k=1$ such that $x=kx \Rightarrow x=x$
it fills the reflexive condition
brilliant!

(2) (s) $x\rho y (\exists z)(y=zx)$, $\hspace{2cm}y\rho x (\exists u)(x=uy)$
if we multiply these we have $(\exists z)(\exists u) yx=z x u y$ For $z=u=1$ we have $yx=xy$ (Does this mean that we have a symmetry?
think of an example.

take y = 2 and x = 1. clearly y relates to x, since y = zx (where z = 2).

but does x relate to y? can we find an integer k so that x = ky, that is, 1 = k*2?

(3) (t) $\exists z(y=zx)\wedge \exists u(t=uy)$. I multiplied again
$(\exists u)(\exists z) t=(uz)x.$
sub. $uz=p \Rightarrow t=px.$

What do you think?
yup, that's nice. we do have transitivity

11. Originally Posted by Jhevon

think of an example.

take y = 2 and x = 1. clearly y relates to x, since y = zx (where z = 2).

but does x relate to y? can we find an integer k so that x = ky
y=zx, x=ky
Ok like you said if we have y=x=z=k=1, we have a symmetry don't we?

12. Originally Posted by javax
y=zx, x=ky
Ok like you said if we have y=x=z=k=1, we have a symmetry don't we?
i gave you an example where it didn't work, didn't i? look back at your definitions, it must work for ALL x and y, not just specific ones

13. Originally Posted by Jhevon
i gave you an example where it didn't work, didn't i? look back at your definitions, it must work for ALL x and y, not just specific ones
ohhhh cool
ok so if it is not symmetric does that mean that it is antisymmetric?