What is the Induced Equivalence Class for (2, 3) in Relation Υ?

[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?

 

[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.

 

[finalight]

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

 

[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.

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

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?

 

[Mark44]

Moving thread to Calculus & Beyond subforum.

 

[finalight]

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

so the answer would be

{(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

 

[SammyS]

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

 

[finalight]

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

 

[Mark44]

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

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

 

[finalight]

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

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

so y1 ∈ Z+

as stated in the question

 

[LCKurtz]

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

so the answer would be

{(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

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.

 

[finalight]

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.

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

 

[SammyS]

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

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

Are you sure that the " ± " belongs there?

 

[finalight]

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

Are you sure that the " ± " belongs there?

sory, just reaize the ± don't belong there

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

 

[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.