MHB Can $B \subseteq A$ Be Proven with Elements Defined as $2b-2$ and $2a$?

[tmt1]

let $A = \left\{ m \in \Bbb{Z} | m = 2a \right\}$ and let $B = \left\{ n \in \Bbb{Z} | n = 2b - 2 \right\}$

Prove that$ B \subseteq A$

My proof is:
let $x = 2b - 2$ for some int $b$ I need to show that $x = 2a$ for some int $a$.

let $a = b - 1$, $b$ is some integer therefore $a$ in this case is some integer. And $2a = 2b - 2$. since $x = 2b - 2$ then , then $x = 2a$ which is what was to be shown.Is this sufficient proof?

 

[Dethrone]

Hi tmt,

The proof is correct, but I think there's a more straightforward way of presenting the same proof.

Given $n = 2b - 2 $, then $n=2(b-1)$. Let $a=b-1$, then $n=2a$. Hence, $B \subseteq A$.

 

[Evgeny.Makarov]

let $A = \left\{ m \in \Bbb{Z} | m = 2a \right\}$ and let $B = \left\{ n \in \Bbb{Z} | n = 2b - 2 \right\}$

It is also better to define the sets $A$ and $B$ as follows.
\[
A = \{2a \mid a\in \Bbb{Z} \},\quad B = \{ 2b - 2 \mid b \in \Bbb{Z} \}
\]
In the original definition it is not clear what $a$ and $b$ are. In particular, it is not clear whether $m=2a$ for all $a$ (well, this is clearly not the case), $m=2a$ for some $a$ or whether $a$ comes from the surrounding context. Also, it is not clear what set $a$ and $b$ range over. If they range over reals, for example, then $1\in A$ since $1=2\cdot(1/2)$ and $1\in\Bbb Z$.

And finally, the vertical bar in the set-builder notation is best typeset with [m]\mid[/m]: it creates correct spaced on both sides.