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For m,n ∈ N, we define: m ≈ n ⇔ number $m^2-n^2$ is a multiple of 3

I'm new to discrete math, and have problems with understanding notations. The actual task asks for finding 4 elements for equivalence class but I only need to understand what this line means.

Murad Davudov
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    This is not an equivalence relation because it is not reflexive. – John Douma Jan 15 '19 at 19:44
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    If it is $m^2-n^2$ it will indeed be an equivalence relation, noting that $0$ is a multiple of three. I highly encourage you to check for yourself that it does satisfy all properties necessary to be called an equivalence relation. The punchline is going to be that every multiple of three is related to every other multiple of three forming an equivalence class, and every non-multiple of three will be related to every other non-multiple of three forming the only other equivalence class. If it were $m^2+n^2$ note that $1^2+1^2$ is not a multiple of $3$, hence not reflexive. – JMoravitz Jan 15 '19 at 19:49
  • @JMoravitz It says that $1\not\approx1$, but $3\approx3$. We can take the subset of $\mathbb N$. – Michael Rozenberg Jan 15 '19 at 20:15
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    $1^2-1^2 = 0 = 0\cdot 3$ is a multiple of three. $1^2+1^2=2$ is not a multiple of three. If "it" is talking about how $1\not\approx 1$ then you must be referring to the relation $m\approx n\iff m^2\color{red}{+} n^2$ is a multiple of three, which as mentioned is not an equivalence relation and is a different relation than the one where it were instead $m^2\color{red}{-}n^2$ is a multiple of three. – JMoravitz Jan 15 '19 at 20:17
  • @JMoravitz It's an equivalence relation on the subset of $\mathbb N$. I really don't see a problem. – Michael Rozenberg Jan 15 '19 at 20:20
  • @JMoravitz, what does it mean to find 4 elements of equivalence class [0]? From what I found, it is set of elements which 0 has relations with. So [0] = (0, 3, 6, 9...) right? – Murad Davudov Jan 15 '19 at 20:25
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    The equivalence class of $x$, notated $[x]$ is the set $[x] = {y~:~y\approx x}$ is the set of all elements which are related to $x$. Note that in an equivalence relation if $x\approx y$ then $[x]=[y]$. Note also, we generally reserve the word "equivalence class" to be used very explicitly only in the situation where the relation being discussed is an equivalence relation. Yes, here in both the $m^2-n^2$ and in the $m^2+n^2$ example, the set of elements which are related to zero will be ${0,3,6,9,\dots}$ the multiples of three, as I mentioned in my first comment. – JMoravitz Jan 15 '19 at 20:29

2 Answers2

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Do a few examples.

If $n = 1$ then $m\approx n$ means $m^2 - 1$ is a multiple of $3$. So such numbers are: $1$ (because $1^2 - 1 = 0$ is a multiple of $3$. $2$ (because $2^2 -1 =3$). $4,5,7...$.

The numbers that are not $\approx 1$ so far seem to be $3$ and $6$ (because $3^2 - 1=8$ and $6^2 - 1=35$ not multiples of three). However $6^2 - 3^2 = 27$ is a multiple of three.

So it seems like we have $\{1,2,4,5,7....\} = \{$ all non-multiples of $3\}$ are $\approx$ to each other and all $\{3,6,...\}=\{$ all multiples of $3\}$ are $\approx$ to each other.

Can we prove that.

1) If $n$ and $m$ are not multiples of $3$ then $n\approx m$. Can we prove that.

Proof: If $n$ and $m$ are not multiples of $3$ then they have a remainder when divided by $3$; either $1$ or $2$. So $n = 3k +\{1,2\}$ for some $k$ and either plus $1$ or $2$ and $m = 3j + \{1, 2\}$ for some $j$ and either pluse $1$ or $2$.

So $n^2 - m^2 = (3k + \{1,2\})^2 - (3j+\{1,2\})^2$

$ = 9k^2 + \{6,12\}k + \{1, 4\} - 9j^2-\{6,12\}j - \{1,4\}$

$= 9k^2 -9j^2 + \{6,12\}k -\{6,12\}j + \{0,-3,3,0\} $

$= 3[k^2 - 3j^2 +\{2,4\}k - \{2,4\}k + \{0,-1, 1\}]$ is a multiple of $3$.

So yes, if $m$ and $n$ are both not multiples of $3$ then $m \approx n$.

2) If $m,n$ are both multiples of $3$ then $m \approx n$. Can we prove that?

Proof: If $m = 3k$ and $n = 3j$ then $m^2 - n^2 = 9k^2 - 9j^2 = 3(3k^2 -3j^2)$ is a multiple of $3$ so $m \approx n$>

3) If $m$ is a multiple of $3$ and $n$ is not (or vice versa) then $m \not \approx n$. Can we prove that?

Pf: If $m =3k$ and $n = 3j + \{1,2\}$ then $m^2 - n^2=$

$9k^2 - 9j^2 -\{6,12\}j - {1,4}$ which is not a mutiple of $3$. So $m \not \approx n$.

.....

So $m \approx n$ if either $m,n$ are both multiples of $3$ or if $m,n$ are both not multiples of $3$.

Now.... we haven't actually proven that is an equivalence relationship.

1) Is it reflexive. Is $m\approx m$ always? Is $m,m$ both either multiples of $3$ or both not? Well, since $m$ is the same thing as itself that is obvious.

2) Is it symmetric? If $m\approx n$ does $n\approx m$? If $m,n$ are either both or both not multiples of $3$ will $n,m$ be either both or both not multiples of $3$? Well, since $m,n$ and $n,m$ are the same pair anything that is true of both of one pair is true of ... the same pair.

3) Is it transitive? If $m \approx n$ and $n\approx p$ does that mean $m \approx p$. Well, if $m,n$ are both the same sort of thing, and $n, p$ are both the same sort of thing, then $m$ is the same sort of thing that $n$ is, and $p$ is also the same sort of thing that $n$ is, then both $m$ and $p$ are the same sort of thing.

So $\approx$ is reflexive, transitive, and symmetric. So $\approx$ is an equivalence relation.

And to to partition all the natural elements into equivalence classes so that all the elements in each class are all the numbers that are equivalent to each other is:

$C_1 = \{1,2,4,5,7,8,10,.....\} = \{n\in \mathbb N| n$ is not a multiple of $3\}$. and $C_2 = \{13,6,9,...\} = \{n\in \mathbb N| n$ is a multiple of $ 3 \}$.

fleablood
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  • Thank you sir. You literally made everything clear. – Murad Davudov Jan 15 '19 at 20:43
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    @Murad Simpler: $,m\approx n!!\overset{\rm\ \ def}\iff 3,$ divides $,m^2-n^2\iff f(m) = f(n)\ $ for $,f(x) = x^2\bmod 3\ $ and such relations are always equivalence relations. – Bill Dubuque Jan 15 '19 at 21:38
  • I'm trying to get the idea to how to even think about these concepts. But that is a simpler (and !shorter!) explanation. But I'm not sure how we can generalize "such relations". – fleablood Jan 15 '19 at 22:19
  • @Murad and fleablood The point is that many equivalence relations have natural definitions of that form (equivalence kernels), i.e. $,m\approx n\iff f(m) = f(n),,$ so it saves a lot of work if we can appeal to that general result. Moreover, proofs are often much simpler from that viewpoint, i.e. pulling back the equivalence properties along a function. – Bill Dubuque Jan 16 '19 at 16:39
  • Not going to disagree with that. – fleablood Jan 16 '19 at 16:51
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For example, $3\approx6$ because $3^2-6^2$ is divisible by $3.$

Michael Rozenberg
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