we have
$A =\{m ∈ \mathbb{Z}|m=6r-5,r ∈ \mathbb{Z}\} $ and
$B = \{n ∈ \mathbb{Z}| n= 3s+1, s ∈ \mathbb{Z}\}$
prove $A ⊆ B$

I have
Proof: suppose $A = \{m ∈ \mathbb{Z}|m=6r-5,r ∈ \mathbb{Z}\}$,
$B = \{n ∈ \mathbb{Z}| n= 3s+1, s ∈ \mathbb{Z}\}$
suppose P.B.A.C integers x ∈ m and y ∈ n
by substitution $x=6r-5$ and $y=3s+1$, where $r$ and $s ∈ \mathbb{Z}$
and.. I have no idea where to go. should I set these equal to eachother? I have no clue

  • $\begingroup$ Solve for s in the following equation. $6r - 5 = 3s+1$ then notice that $s$ will be an integer $\endgroup$ Commented Apr 28, 2021 at 0:50
  • 3
    $\begingroup$ What on earth is P.B.A.C.? $\endgroup$
    – Théophile
    Commented Apr 28, 2021 at 1:49
  • $\begingroup$ Particular but arbitrarily chosen. $\endgroup$ Commented Apr 28, 2021 at 20:11
  • $\begingroup$ @Slowly_Learning Ah, interesting. I've never seen that abbreviation before, and I wouldn't consider it standard. I would go so far as to say that it's not useful either: why write, for example, "let $n$ be a particular but arbitrarily chosen integer" when one can instead just write "let $n$ be an integer"? $\endgroup$
    – Théophile
    Commented Apr 28, 2021 at 22:30
  • $\begingroup$ I asked my teacher the same thing (: $\endgroup$ Commented Apr 29, 2021 at 0:06

2 Answers 2


Let $m \in A$, $r \in \mathbb{Z}$ be such that $m=6r-5$. Then $m=6(r-1)+1=3(2(r-1))+1 $, and so $m \in B$.


$x\in A$ iff $(x-1)/6\in \Bbb Z$.

$x\in B$ iff $(x-1)/3\in \Bbb Z$.

Therefore $x\in A\implies (x-1)/6\in \Bbb Z \implies (x-1)/3=2\cdot (x-1)/6\in \Bbb Z \implies x\in B.$


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