Hi so we've started learning about relations and I'm completely lost.. I don't get how to represent the equivalence classes and I'm not even sure the way I've proved the equivalence relation is correct..
Question 1
a.
Note that if for every $a,b ∈ R$ if $a = b = 0 \ $ ➜ $a*b$ $\not>$ $0$, but in this case it's already part of the relation.
Also note that in order for this inequality to work, $a,b$ must be both negative, or both positive.
Reflexive:
Since $a$ obviously has the same sign as itself, $a * a > 0$.
Therefore for every $a ∈ R/(0) $ $ \ a R a \ $
Symmetric:
By associativity, if $a * b > 0 ➜ b * a > 0$.
Therefore for every $a,b ∈ R/(0) $ if $aRb ➜ bRa$.
Transitive:
Since $a,b,c$ must share the same sign, if $a * b > 0$ and $b * c > 0 ➜ a * c > 0$.
Therefore for every $a,b,c ∈ R/(0) $ if $aRb ∧ bRc ➜ aRc$.
Equivalence classes:
$\left\{R^{+},R^{-},\left\{0\right\}\right\}$
b.
Note that if $8 ∉ XΔY$ then either $8 ∈ X ∧ 8 ∈ Y \ $ or $ \ 8 ∉ X ∧ 8 ∉ X$.
Reflexive:
Since $XΔX$ is the $∅$, then it is obvious $8 ∉ XΔX$.
Therefore for every $X ∈ P(NxN)$ $XRX$.
Symmetric:
Δ is associative, therefore if $8 ∉ XΔY ➜ 8 ∉ YΔX $
Therefore for every $X,Y ∈ P(NxN) \ XRY➜YRX$.
Transitive:
Since $X,Y,Z$ must all contain or all not contain $8$ for the relation to work,
then if $8 ∉ XΔY ∧ 8 ∉ YΔZ ➜ 8 ∉ XΔZ$.
Therefore for every $X,Y,Z ∈ P(NxN) \ $ if $XRY ∧ YRZ ➜ XRZ$.
Equivalence class:
$\left\{\left\{X\ ∈\ \ P(N)\ :\ 8\ ∉\ \ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 8\ ∈\ X\ \right\}\right\}$
c.
Note that we're proven in class equality is both reflexive, symmetric and transitive.
Reflexive: Obviously $X ∩ {1,2} = X ∩ {1,2}$.
Symmetric:
It is also obvious $X ∩ {1,2} = Y ∩ {1,2} ➜ Y ∩ {1,2} = X ∩ {1,2}$.
Transitive:
Again, obviously if $X ∩ {1,2} = Y ∩ {1,2} ∧ Y ∩ {1,2} = Z ∩ {1,2} \ ➜ \ X ∩ {1,2} = Z ∩ {1,2}$.
Equivalence classes:
$\left\{\left\{X\ ∈\ \ P(N)\ :\ 1,2\ ∉\ \ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 1\ ∉\ \ X\ \ ∧\ 2\ ∈\ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 2\ ∉\ \ X\ \ ∧\ 1\ ∈\ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 1,2\ ∈\ \ X\ \right\}\right\}$
Instead of 'describing the equivalence classes' it might be more appropriate to rather ask for a set $B$ and a surjective function $f:A\to B$ for each example such that the given equivalence relation on $A$ is just its 'kernel' $\{(a,a'):f(a)=f(a')\}$.
This set $B$ represents the best the equivalence classes.
For 1., the positive numbers are in relation with all each other, and so are the negative numbers, and there's also the $0$, so the function we're looking for is the signum function ${\rm sign}:\Bbb R\to\{-1,0,+1\}$.
For 2. and 3., observe that condition in 2. holds for sets $X,Y$ if and only if either both of them contains $8$ or neither of them, so that this can be translated to the same condition as for 3.: $$X\sim Y\iff X\cap\{8\}=Y\cap\{8\}\,.$$ Generalizing this, for any set $U$ and any subset $A\subseteq U$, the relation $X\sim Y\iff X\cap A=Y\cap A$ is an equivalence relation on $P(U)$, and it is the kernel of the 'restriction' function $$f:P(U)\to P(A)\quad X\mapsto X\cap A\,.$$
So, the equivalence classes of 2. are represented by elements of $P(\{8\})=\big\{\emptyset,\,\{8\}\big\}$ and similarly, those of 3. by $P(\{1,2\})=\big\{\emptyset,\,\{1\},\, \{2\},\, \{1,2\}\big\}$.