Equivalence Relations Questions

[RJLiberator]

Homework Statement

For the set ℤ, define ~ as a ~ b whenever a-b is divisible by 12. You may assume that ~ is an equivalence relation and may also assume that addition and multiplication of equivalence classes is well defined where e define [a]+[ b ] = [a+b] and [a]*[ b ] = [ab] for all [a],[ b ].

Find a positive integer d such that
[d]+[5]=[0]

find a positive integer t such that
[t]+[8] =[3]

Homework Equations

3. The Attempt at a Solution [/B]

These problems seem like a lot of fun. However, I'm not quite getting it.

I feel like once I understand one of these, i'll be able to understand all of the easy ones like this.

We define a ~b whenever a-b is divisible by 12.
So we are saying in the first problem d-5 has to be divisible by 12?

If d = 17 then we have 17-5 which is 12 and that is divisible by 12.
But how would [17]+[5]=[0]
In fact, how would any positive integer satisfy that? Since we have well defined addition as [a]+ = [a+b]
this would mean [d]+[5] = [d+5]
and this means [d+5] = [0], but since s must be a positive integer this could not happen...I feel like there must be something clear here that I'm missing and once I get it it will be an easily solvable problem.

 

[Dick]

Homework Statement

For the set ℤ, define ~ as a ~ b whenever a-b is divisible by 12. We define [a]+ = [a+b] and [a]* = [ab] as well defined for all [a],.

Find a positive integer d such that
[d]+[5]=[0]

find a positive integer t such that
[t]+[8] =[3]

Homework Equations

3. The Attempt at a Solution [/B]

These problems seem like a lot of fun. However, I'm not quite getting it.

I feel like once I understand one of these, i'll be able to understand all of the easy ones like this.

We define a ~b whenever a-b is divisible by 12.
So we are saying in the first problem d-5 has to be divisible by 12?

If d = 17 then we have 17-5 which is 12 and that is divisible by 12.
But how would [17]+[5]=[0]
In fact, how would any positive integer satisfy that? Since we have well defined addition as [a]+ = [a+b]
this would mean [d]+[5] = [d+5]
and this means [d+5] = [0], but since s must be a positive integer this could not happen...I feel like there must be something clear here that I'm missing and once I get it it will be an easily solvable problem.

I don't see how you can define [a]+ as [a+b]. What's b? The definition you want is [ a ] + [ b ] = [ a+b ]. So [d]+[5]=[0] becomes [d+5]=[0]. Try taking it from there. And you don't really define something as 'well defined'. You have to SHOW your definition is well defined.

 

[RJLiberator]

I'm sorry, there was some formatting errors in my initial post since I used [ s ] initially and made it strike through everything :p. I do indeed want [a]+ [ b ]= [a+b] and [a]*[ b ] = [ab].

Also, the problem stated that they could be assumed to be well defined in more eloquent way. Perhaps I should have posted the entirety of the problem.

I will try to edit my initial post now.

 

[RJLiberator]

As far as:

[d]+[5]=[0] becomes [d+5]=[0]

We are saying some positive integer d such that [d+5] = [0]

Does this mean that
d+5 - 0 = must be divisible by 12?

I assume that the meaning of equivalence classes suggests that for equivalence to be uphold it must be divisible by 12 in this scenario.
So (d+5) - 0 = must be divisible by 12.
So if I choose d = 7 then we have (7+5)- 0 = 12 which is divisible by 12.

Is this the correct way to proceed with these problems? :D

 

[Dick]

As far as:We are saying some positive integer d such that [d+5] = [0]

Does this mean that
d+5 - 0 = must be divisible by 12?

I assume that the meaning of equivalence classes suggests that for equivalence to be uphold it must be divisible by 12 in this scenario.
So (d+5) - 0 = must be divisible by 12.
So if I choose d = 7 then we have (7+5)- 0 = 12 which is divisible by 12.

Is this the correct way to proceed with these problems? :D

That is exactly the correct way. Yeah, and I was having similar formatting problems myself.

 

[RJLiberator]

Perfect, now I have confirmation on my understanding of equivalence relationships.
This thread is solved.