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.
Reflexive: For m $\in \mathbb{Z}$, we have m + 4m =5m, and therefore 5 divides m + 4m and (m,m) $\in$ R. For this, I believe that because both m and n are in the relation R then |m|=|n| and so m can be substituted for n giving (m,m). Is this correct?
Symmetric: If (m,n) $\in R$, then 5|(m + 4n). Thus, 5k = m + 4n for some k $\in \mathbb Z$. We want to show that 5|(n + 4m). Hence,
(Line 1)
$$(4)5k = 20k = 4(m+4n) = 4m + 16n,$$
(Line 2) $$n + 4m = n + 20k - 16n = 5(4k - 3n)$$
(Line 3) $$and 5|(n + 4m).$$
In line one, it appears that both side of the equation are being multiplied by 4 but I don't understand why. Could someone explain why this is the case?
In line two, I don't understand whe n + 4m = n + 20k - 16n
I understand the rest but not these two part. Any help is welcome.
Thanks,
Tony
I think the following is a cleaner approach. We are using the fact that if a number divides two numbers, then it also divides their sum and their difference. Using this, we would like to sum or subtract appropriate numbers to get from $5\mid n+4m$ to $5\mid 4n+m$.
Now, we know $5\mid n+4m$, and trivially $5\mid 5m+5n$. By the first comment $5\mid n+4m\color{red}{-5m}-\color{blue}{5n}$. This equals $-4n-m$, we so know $5\mid -4n-m$. But of course, if a number divides $-a$ it also divides $a$, so we get $5\mid 4n+m$ a we wanted.