Hello. I need help on determining this relation.
Determine the relation $\displaystyle \rho $ defined with $\displaystyle x\rho y \Leftrightarrow (\exists z)(y=zx)$ for $\displaystyle Z$ 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 $\displaystyle \mathbb{Z}$ when you wrote $\displaystyle Z$).
Here we go:
1. (r) $\displaystyle (\forall a\in A) a\rho a$
2. (s) $\displaystyle (\forall a,b \in A) a\rho b \Rightarrow b\rho a$
3. (t) $\displaystyle (\forall a,b,c \in A) a\rho b \wedge b\rho c \Rightarrow a\rho c$
4. (a)$\displaystyle (\forall a,b\in A)a\rho b \wedge b\rho a \Rightarrow a = b$
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
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)
$\displaystyle x\rho x \Leftrightarrow \exists k=1$ such that $\displaystyle x=kx \Rightarrow x=x $
it fills the reflexive condition
(2) (s) $\displaystyle x\rho y (\exists z)(y=zx)$, $\displaystyle \hspace{2cm}y\rho x (\exists u)(x=uy)$
if we multiply these we have $\displaystyle (\exists z)(\exists u) yx=z x u y$ For $\displaystyle z=u=1$ we have $\displaystyle yx=xy$ (Does this mean that we have a symmetry?
(3) (t)$\displaystyle \exists z(y=zx)\wedge \exists u(t=uy)$. I multiplied again
$\displaystyle (\exists u)(\exists z) t=(uz)x. $
sub. $\displaystyle uz=p \Rightarrow t=px.$
What do you think?
brilliant!
think of an example.(2) (s) $\displaystyle x\rho y (\exists z)(y=zx)$, $\displaystyle \hspace{2cm}y\rho x (\exists u)(x=uy)$
if we multiply these we have $\displaystyle (\exists z)(\exists u) yx=z x u y$ For $\displaystyle z=u=1$ we have $\displaystyle yx=xy$ (Does this mean that we have a symmetry?
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?
yup, that's nice. we do have transitivity(3) (t)$\displaystyle \exists z(y=zx)\wedge \exists u(t=uy)$. I multiplied again
$\displaystyle (\exists u)(\exists z) t=(uz)x. $
sub. $\displaystyle uz=p \Rightarrow t=px.$
What do you think?
no, anti-symmetric is defined by the fourth definition you have. after reading a few math books you should start to see how mathematicians define things. if anti-symmetric means "not symmetric" that is exactly how they would define it. it shouldn't be hard to come up with an example of a relation that is both not symmetric and not anti-symmetric. see here
you have to check if it is anti-symmetric. if it is, we have an order relation (since you already showed we have reflexivity and transitivity)