Equivalence Relations on Z: Proving m~n and Describing the Partition

[nitenglo]

Prove that the following is an equivalence relation on the indicated set. Then describe the partition associated with the equivalence relation.

1. In Z, let m~n iff m-n is a multiple of 10.2. The attempt at a solution

Reflexive: m-n = 0

0 ∈ Z, and 0 is a multiple of every number, therefore, m~n

Symmetric:
1. m~n
2. m-n = 0 ∈ Z, and 0 is a multiple of every number
3. -(m-n) = 0 ∈ Z, and 0 is a multiple of every number

4. -(m-n) = n-m = 0 ∈ Z, and 0 is a multiple of every number
5. Therefore, n~m

Transitive:
1. m~n and n~t
2. m-n = 0 ∈ Z, and 0 is a multiple of every number. and n-t = 0 ∈ Z, and 0 is a multiple of every number
3. (m-n)+(n-t) = m-t = 0 ∈ Z, and 0 is a multiple of every number
4. Therefore, m~t

 

[Stephen Tashi]

You have have asserted that various numbers are "in the indicated set", but you haven't given any proof they are.

 

[nitenglo]

I edited the problem.

 

[fourier jr]

For part 2 I think what you mean to say is n - n = 0 (which is clearly divisible by 10). In this problem you're not looking for the difference to be 0, but rather a multiple of 10 (say 10x for some integer x), or that 10 divides their difference.

 

[nitenglo]

Thank you fourier...can you take a look at the work I did for proving m~n and see if it is correct.

 

[Stephen Tashi]

2. The attempt at a solution

Reflexive: m-n = 0

To prove the relation is reflexive, what you must prove is that m \sim m.

Symmetric:
1. m~n
2. m-n = 0 ∈ Z, and 0 is a multiple of every number

It isn't true that m-n must be zero. You probably mean that m-n = 0 (mod \ 10).

Transitive:
1. m~n and n~t
2. m-n = 0 ∈ Z, and 0 is a multiple of every number

.

It is not true that m-n must be zero.

 

[SammyS]

Prove that the following is an equivalence relation on the indicated set. Then describe the partition associated with the equivalence relation.

1. In Z, let m~n iff m-n is a multiple of 10.2. The attempt at a solution

Reflexive: m-n = 0

0 ∈ Z, and 0 is a multiple of every number, therefore, m~n

Symmetric:
1. m~n
2. m-n = 0 ∈ Z, and 0 is a multiple of every number
3. -(m-n) = 0 ∈ Z, and 0 is a multiple of every number

4. -(m-n) = n-m = 0 ∈ Z, and 0 is a multiple of every number
5. Therefore, n~m

Transitive:
1. m~n and n~t
2. m-n = 0 ∈ Z, and 0 is a multiple of every number. and n-t = 0 ∈ Z, and 0 is a multiple of every number
3. (m-n)+(n-t) = m-t = 0 ∈ Z, and 0 is a multiple of every number
4. Therefore, m~t

Please, only edit the Original Post to correct typos and/or errors in the statement of the problem. Additionally, edit the OP to supply any essential information you may have initially left out. It may help clarify matters, if you indicate what is changed in the edit.

If you are making corrections to your solution of a problem, it's best to add another post to the thread. Otherwise it's nearly impossible to follow the sequence of events in the thread.

 

[fourier jr]

Thank you fourier...can you take a look at the work I did for proving m~n and see if it is correct.

for part 2 you start with $m \sim n$ which in this case means m - n = 10x for some x ∈ ℤ & you have to show $n \sim m$. Part 3 is a bit more complicated because you have $m \sim n$ which means m - n = 10x for x ∈ ℤ & $n \sim t$ which means n - t = 10y for y ∈ ℤ.

 

[HallsofIvy]

Prove that the following is an equivalence relation on the indicated set. Then describe the partition associated with the equivalence relation.

1. In Z, let m~n iff m-n is a multiple of 10.2. The attempt at a solution

Reflexive: m-n = 0

You mean m~m. Then m- m= 0.

0 ∈ Z, and

0 is a multiple of every number, therefore, m~n

Symmetric:
1. m~n
2. m-n = 0 ∈ Z, and 0 is a multiple of every number

No, if m~n then m- n is a multiple of 10, not necessarily 0.

3. -(m-n) = 0 ∈ Z, and 0 is a multiple of every number

4. -(m-n) = n-m = 0 ∈ Z, and 0 is a multiple of every number
5. Therefore, n~m

Transitive:
1. m~n and n~t
2. m-n = 0 ∈ Z, and 0 is a multiple of every number

[/quote]

. and n-t = 0 ∈ Z, and 0 is a multiple of every number

Again, no! m-n is a multiple of 10, not necessarily 0. m-n= 10k, for some integer k and n- t= 10j for some integer j. So m- t is equal to what?

3. (m-n)+(n-t) = m-t = 0 ∈ Z, and 0 is a multiple of every number
4. Therefore, m~t

 

[nitenglo]

Sorry I forgot to post this earlier...this is what I have came up with so far...

Reflexive: m-n = m=n mod 10. ∈ ℤ and is a divisible by 10 (same as being a multiple) Therefore, m~n

Symmetric: m~n = m-n = n mod 10. ∈ ℤ and is a divisible by 10. Then -(m-n)= n-m = n=m mod 10. Therefore, n~m

Transitive: ?? I am kind of stuck here... I know that if m~n and n~t then show m~t.

I think that it goes like this...

n= t= m mod 10. and m=n= t mod 10 therefore m~t.

But I am not sure.

 

[fourier jr]

You're trying to prove $m \sim n$ is an equivalence relation, where $m \sim n$ if $m \equiv n\, (mod 10)$, so what does it mean when $m \equiv n\, (mod 10)$? For symmetry for example the problem is to show if $m \equiv n\, (mod 10)$ then $n \equiv m\, (mod 10)$, how would you do that?

 

[nitenglo]

So I redid it again...here it goes.

Reflexive: let n∈ℤ, then for any n∈ℤ n-n=0 is a multiple of 10. ∴ n~n

Symmetric: Let m, n∈ℤ and m~n. Then m-n is a multiple of 10. That is m-n ≡ 10k, k∈ℤ and n-m ≡ 10(-k), -k∈ℤ. ∴ n-m is a multiple of 10 and n~m.

Transitive: Let m, n, p∈ℤ and m~n, n~p. Then m-n, n-p are multiples of 10. That is m-n ≡ 10k1 and n-p≡10k2, k1, k2∈ℤ. Now (m-n)+(n-p) ≡ 10k1 + 10k2, k1, k2∈ℤ ≡ 10 (k1 + k2), k1, k2 ∈ℤ ≡ m-p ∴ m-p is a multiple of 10 and m~n, n~p shows that m~p.

Partition:

Fix n∈ℤ. Then m~n and m-n is a multiple of 10.
m-n = 10k, k∈ℤ
m= n+10k, k∈ℤ
∴ n= {m: m= n+10k, k∈ℤ} =n+10k : k, ∈ℤ

 

[fourier jr]

I thought you had to show that you get a partition of the integers but they just want a description... I think I get what you did for that part but maybe you could clean it up & actually describe in words what all that means. The rest looks ok to me.

 

[HallsofIvy]

Certainly your notation is bad- "n= {m: m= n+10k, k∈ℤ}" you have an integer, n, equal to a set of integers!
Perhaps you meant "[n]' on the left, indicating the equivalence class n is in. Of course, there are only 10 such equivalence classes. You could just write each out:
{0, 10, -10, 20, -20, ...}, {1, 11, -9, 21, -19, ...}, etc.