# Homework Help: Finding equivalence class

1. Oct 17, 2011

### finalight

(x1, y1)Υ(x2, y2) ⇔ x1 × y2 = x2 × y1

for all x1, x2 ∈ Z and y1, y2 ∈ Z+ have been shown to be an equivalence relation in tutorial.

Specify the equivalence class [(2; 3)] as induced by Υ.

i don't understand what it means by 'Specify the equivalence class [(2; 3)] as induced by Υ.'

can anyhone guide me along?

2. Oct 17, 2011

### LCKurtz

The equivalence class generated by (2,3) is the collection of all the pairs under consideration that are related to (2,3) by Y. So you need to answer the question something like

[(2,3)] = {(a,b): some criteria having to do with (2,3) that (a,b) must satisfy to be in the equivalence class}. You have to replace the bold part with appropriate wording.

3. Oct 18, 2011

### finalight

so is the answer like [(3,2)]?

4. Oct 18, 2011

### LCKurtz

I'm not sure what you are getting at with that question.

[(2,3)] is the symbol for the equivalence class. Some pairs (a,b) are in the equivalence class [(2,3)] and some aren't. (3,2) isn't. So [(2,3)] and [(3,2)] are distinct equivalence classes.

But your problem is to describe what pairs (a,b) are in [(2,3)]. I'll get you started. Is (6,9) in [(2,3)]? Can you find others? What pairs are?

5. Oct 18, 2011

### Staff: Mentor

Moving thread to Calculus & Beyond subforum.

6. Oct 18, 2011

### finalight

the ordered pair i found so far is like this
{(3,2), (6,4), (9,6), (12,8)}

{(a, b)∈ Z x Z+: b/a = ±2/3}

haha, i think i got it correct
i'm not sure whether to include the '±', but since y1, y2 could be any ±Z, i have to take precaution

Last edited: Oct 18, 2011
7. Oct 18, 2011

### SammyS

Staff Emeritus
Is (0,0)Y(2,3) ?

8. Oct 18, 2011

### finalight

I don't think (0,0) is a valid ordered pair, since b must be positive integer

9. Oct 18, 2011

### Staff: Mentor

Why do you think that? I don't see any restriction like this anywhere in this thread.

10. Oct 18, 2011

### finalight

(x1, y1)Υ(x2, y2)
x1, x2 ∈ Z and y1, y2 ∈ Z+

so y1 ∈ Z+

as stated in the question

11. Oct 18, 2011

### LCKurtz

This problem usually arises when you are developing the theory of fractions given the integers. I don't know the context in which you present the problem, but if it is that context, it may be that the use of fractions to describe the equivalence classes wouldn't be appropriate. You will have to judge that for yourself, but it is easy enough to describe the set you are alluding to above without using fractions.

12. Oct 19, 2011

### finalight

i probably need hints on how to describe it without using the form a/b

13. Oct 19, 2011

### SammyS

Staff Emeritus
Solve your equation for a: b/a = ±2/3.

Are you sure that the " ± " belongs there?

14. Oct 19, 2011

### finalight

sory, just reaize the ± don't belong there

so it's a = 2/3b??? is tat correct?

15. Oct 19, 2011

### HallsofIvy

Or, if you don't want to use fractions, 3a= 2b. So every ordered pair in the same equivalence class with (2, 3) is of the form (2x, 3x) for some positive integer x.