# Elements of an Equivalence Class

1. Nov 30, 2016

### Numberphile

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

Question:
Let ~ be the equivalence relation on the set ℤ of integers defined by a~b if a-b is divisible by 5. Let k ∈ Em belong to the equivalence class of m, and l ∈ En belong to the equivalence class of n. Prove that the product kl belongs to the equivalence class Emn.

What I've tried:
Since k ∈ Em, then m - k = 5x, for x ∈ ℤ.
Also, since l ∈ En, then n - l = 5y, for y ∈ ℤ.
So, I need to show that m*n - k*l = 5z, for z ∈ ℤ.
I've tried multiplying (m-k)*(n-l), but to no avail.
So I tried one case. Let m=20, k=10, n=30, and l=5.
Well, 20-10=5(2), so 10 ∈ E20, and 30-5=5(5), so 5 ∈ E30
Also, 600-50=5(110), so 50 ∈ E600, which for this case means kl ∈ Emn.

Dilemma:
I'm convinced the product kl belongs to the equivalence class of m*n always, so I'm having trouble actually finding a systematic way of proving it in general, instead of using a specific case.

Last edited: Nov 30, 2016
2. Nov 30, 2016

### Staff: Mentor

Why haven't you calculated what you are actually looking for: $m\cdot n - k \cdot l$?

3. Nov 30, 2016

### Numberphile

I haven't calculated that because I'm not sure what that would represent.
I know an equivalence class is defined as a relation on a set S with x ∈ S, then the equivalence class of x is
Ex = {y ∈ S | x ~ y}

4. Nov 30, 2016

### Staff: Mentor

And here's the way:
$mk-kl = (5x+k)(5y+l)-kl= \dots$

5. Nov 30, 2016

### Numberphile

I may have made a mistake in my "What I've tried" section.
I already corrected it in my original post.

My apologies.

6. Nov 30, 2016

### Staff: Mentor

No, my sorry, I had a typo. Of course it's $mn-kl=(5x-k)(5y-l)-kl= \dots$
The point is: you already have everything you need in your post. What is $mn-kl$?

7. Nov 30, 2016

### Numberphile

I'm not sure that I follow the logic.
mn - kl = (5x+k)(5y+l) - kl

I expanded this, and trivially, I obtain that mn - kl = mn - kl

8. Nov 30, 2016

### Staff: Mentor

You set $m=5x-k$ and $n=5y-l$ which you can do because $k\in E_m$ and $l\in E_n$.
This is correct. Also correct is, that $mn-kl=5z$ for some $z \in \mathbb{Z}$ has to be shown, because it would mean $E_{mn}=E_{kl}$ what we need. Now, simply substitute $m$ and $n$ to get
$mn-kl=(5x-k)(5y-l)-kl=25xy-5xl-5yk+kl-kl=25xy-5xl-5yk=5z$.
This line was all what's been missing.

9. Nov 30, 2016

### Numberphile

Ah, I see it now! (Though I think the signs are a bit off, but nonetheless it should be the same)

Since 25xy + 5xl + 5yk = 5(5xy) + 5(xl) + 5(yk) = 5[5xy + xl +yk], then this is just an 5z, as [5xy + xl + yk] is just some integer.

Perfect! Thank you so much.