I am just tying to figure our this example but am having difficulty understand the math being used.
The example state: Let R be a relation on the set $\mathbb{Z}$ defined as $(m,n)\in$ R if and only if $5|(m + 4n)$. We show first that $R$ is an equivalent relation.
(If you notice missing parts it is because I have already asked questions regarding this being reflexive and symmetric in another post.)
The example states: If $(m,n)$ and $(n,u)$ are in $R$, then $5k = m + 4n$ and $5s = n + 4u$ for some $k,s\in u$. Hence,
$$m + 4u = 5k - 4n + 5s - n = 5(k + s - n)$$ and $5|(m + 4u)$. Hence, $(m,u)\in $ R
I understand that the above equation is a combination of the two equations: $5k = m + 4n$ and $5s = n + 4u$ but, I don't understand why I can combine these two equations.
Continuing the example: The equivalent classes are $$[m]=\{{(n \in \mathbb{Z}|(m,n) \in R)}\} = \{n \in \mathbb Z|5|(m + 4n) \}.$$
Verify that $$ [0] = \{n \in \mathbb Z|5|4n\} = \{5k|k \in \mathbb Z\} $$ $$ [1] = \{n \in \mathbb Z|5|1 + 4n\} = \{1,6,11,16,...\} = \{1 + 5k|k \in \mathbb Z\} $$ $$ [2] = \{n \in \mathbb Z|5|2 + 4n\} = \{2,7,12,17,...\} = \{2 + 5k|k \in \mathbb Z\}. $$
I can see how $\{1,6,11,16,...\} = \{1 + 5k|k \in \mathbb Z \}$ is related. If I let k = 0, I get 1; If I let k = 1, I get 6, etc. But I don't see how $\{n \in \mathbb Z|5|1 + 4n\} = \{1,6,11,16,...\}$ is related. I've tried using 0, 1, and 2 for n but that does not work and, I'm not sure what I am missing.
There is another small part to this equation, but I would like to try and figure it out before posting it, if i have trouble with it.
Thanks for any help.
Tony
Thanks for any help.
Have you noticed that $5$ divides $m+4n$ if and only if $5$ divides $m-n$? This will make your calculations much easier. For instance, to show transitivity, if $5$ divides $m-n$ and $n-q$, then $$ 5 \, | \, m-q = (m-n) + (n-q) $$ It also makes it more obvious now that the equivalence class of $1$ is just $1 + 5k$, since $(5k+1)-1$ is divisible by $5$.
Hope that helps,