A={n∈ℤ: 6∣n and 8∣n} and B={n∈ℤ: 48∣n}. Is A⊆B?

Question

A={n∈ℤ: 6∣n and 8∣n} and B={n∈ℤ: 48∣n}.

Is A⊆B? Is B⊆A?

I'm pretty sure that they are subsets of each other, because any n that 6 and 8 would both divide would have to be divisible by 6*8, but I'm not sure how to proof this formally.

Answer 1

You have $$6\mathrel{|}n \wedge 8\mathrel{|}n \iff 24\mathrel{|}n,$$ since $\text{lcm}(6,8) = 24$.

Answer 2

Its the latter that holds since if $48 \mid n \Rightarrow 6 \mid n, 8 \mid n$, and the former does not since $24 \in A$ but not in $B$.

Answer 3

$(A\subseteq B)\!\iff\!(6,8\mid n\,\Rightarrow\, 48\mid n)$ by definition of a subset. $\ \ \ (1)$

Similarly, $(B\subseteq A)\!\iff\!(48\mid n\,\Rightarrow\, 6,8\mid n)$. $\ \ \ (2)$

Since $\text{lcm}(6,8)=24$, we have $\ (6,8\mid n\!\iff\!24\mid n)$.

So in $(1)$ you're asked if $\ 24\mid n\,\Rightarrow\, 48\mid n$, $\ $ which is clearly false: e.g., take $n=24$.

In $(2)$, you're given $n=48k$ for some $k\in\Bbb Z$, so

$\,\Rightarrow\, n=6(8k)=8(6k)\,\Rightarrow\, 6,8\mid n$, so $(2)$ must be true.

Answer 4

Hint $\,\ 6,8\mid n \iff 3,4\mid n/\color{#c00}2 \iff 12\mid n/2 \iff 24\mid n,\,$

$\ \ $ i.e. $\ {\rm lcm}(6,8)\, =\, \color{#c00}2\,{\rm lcm}(3,4) = 24\ $ by factoring out $\,\gcd(6,8) = \color{#c00}2$