Equivalence Relations

[rooski]

1. The problem statement, all variables and given/known data

For each of the relations on the set R x R - (0,0) (ie. no origin) :

- prove it is an equivalence
- give the # of equivalence cases
- give a geometric interpretation of the equivalence cases assuming an element of R x R is a point on a plane

a) {((a,b),(c,d)) | a/c > 0 and b/d > 0 }
b) {((a,b),(c,d)) | b/a = d/c }


3. The attempt at a solution

I don't even know how to start this question. :yuck:

I know the definition of equivalence means reflexivity, symmetry and transitivity. So how would i go about proving each of these on the set?

[tiny-tim]

hi rooski! :smile:

start with the geometric interpretation …

in other words, draw (on an x,y coordinate plane) the equivalence classes in each of the two cases …

how to prove the rest of the question will then be fairly obvious :wink:

what do you think the equivalence classes look like? :smile:

[rooski]

Well question a would have an empty diagonal (since 1/1 = 0, 2/2 = 0, etc.), but everything else would be valid. Am i supposed to just shade in all the valid areas (since we're using real numbers)?

Question b would have a proper diagonal (since 1/1 = 1/1, etc.) but there's also things like 1/2 = 2/4, 1/2 = 4/8...

I think i'm confused by the fact that i'm being given 4 variables for a relation. Let's say a,b,c,d = 1,2,4,8... Where would i put the dot on my graph?

[tiny-tim]

hi rooski! :smile:
Well question a would have an empty diagonal (since 1/1 = 0, 2/2 = 0, etc.), but everything else would be valid. Am i supposed to just shade in all the valid areas (since we're using real numbers)?

1/1 isn't 0, it's 1, 2/2 likewise … what are you talking about? :confused:

start again …

take the point (1,2) … is it in the same class as (1,3), (-1,3), (-1,-3), or (1,-3) for example?
Question b would have a proper diagonal (since 1/1 = 1/1, etc.) but there's also things like 1/2 = 2/4, 1/2 = 4/8...

ok, so draw all of them (well, not all obviously, but a good selection :wink:)
I think i'm confused by the fact that i'm being given 4 variables for a relation. Let's say a,b,c,d = 1,2,4,8... Where would i put the dot on my graph?

1,2,4,8 means the two points (1,2) and (4,8) …

you put them on the graph in the usual position, the only question is whether you join them up in some way (put them in the same shaded area for example), or leave them separate

[rooski]

Oh sorry, i haven't been thinking straight.

Basically any values in question a that are both in the top left and bottom right quadrant are going to be shaded (since neg/neg = positive and pos/neg or neg/pos = negative).

So now for question a i must show it is reflexive. How would i prove this without showing examples for every possible x and y value? Can i just say "since for all xRx there exists a value in R, we conclude it is reflexive"? or is there a more rigorous proof i need?

[tiny-tim]

hi rooski! :smile:
Basically any values in question a that are both in the top left and bottom right quadrant are going to be shaded (since neg/neg = positive and pos/neg or neg/pos = negative).

that doesn't look right to me :redface:

i'm not sure exactly what you mean, but anyway what equivalence class do you say, for example, (-1,3) would belong to?

[rooski]

hi rooski! :smile:


that doesn't look right to me :redface:

i'm not sure exactly what you mean, but anyway what equivalence class do you say, for example, (-1,3) would belong to?

well -1/3 < 0 so it would not be graphed. i still can't figure out the concept behind equivalence classes.

and i thought my solution for a was correct. In order for a/c and b/d to be greater than 0, all values of a,b,c,d must be either positive or negative. Thus there cannot be any coordinates in the top left or bottom right quadrants, right?

[tiny-tim]

well -1/3 < 0 so it would not be graphed. i still can't figure out the concept behind equivalence classes.

no, that would be a/b, not a/c

a/c is the ratio of the two x-coordinates of two different points in RxR

try again :smile:

[rooski]

Well if a needs to match c and b needs to match d (in terms of positivity or negativity).

So if a is negative then c is negative. If b is negative then d is negative. And vice versa.

So we could have something like a,b,c,d = -1,2,-1,2 and it would be valid.

[tiny-tim]

Well if a needs to match c and b needs to match d (in terms of positivity or negativity).

yes, so then what is the whole equivalence class of (-1,2) ?

ie all the points (x,y) such that (-1,2) ~ (x,y)

[rooski]

Well x would be the set of all negative real numbers and y would be the set of all positive numbers.

[tiny-tim]

hi rooski! :smile:

(just got up :zzz: …)
Well x would be the set of all negative real numbers and y would be the set of all positive numbers.

hmm … that's sort-of right, but it doesn't really make any sense :redface:

you're asked for the set of points (x,y), ie for the set of pairs of numbers, not for sets of individual numbers

if you mean all (x,y) such that x < 0 and y > 0, in other words the interior of the second quadrant, then yes, that whole interior would be in the same equivalence class as (-1,2) :smile:

and obviously each of the other three quadrants is an equivalence class also …

but what about the axes (apart from the origin) … is (1,0) or (0,1) related to anything? :wink: