# set of odd integers proof

• Feb 24th 2010, 02:21 PM
james121515
set of odd integers proof
I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?

$\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$

$\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,
$\Longrightarrow x=2k+1+1-1$
$\Longrightarrow x=2k+2-1$
$\Longrightarrow x=2(k+1)-1$

Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$
• Feb 24th 2010, 02:36 PM
Plato
Quote:

Originally Posted by james121515
I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?

$\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$

$\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,
$\Longrightarrow x=2k+1+1-1$
$\Longrightarrow x=2k+2-1$
$\Longrightarrow x=2(k+1)-1$

Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$

That 'way' is correct.
By symmetry you are done.

$x=2k+1=2(k+1)-1$
• Feb 24th 2010, 09:36 PM
james121515

So you are saying that due to symmetry, there is no need to show the other "right to left" containment due to symmetry?

-James
• Feb 25th 2010, 08:21 AM
Plato
Quote:

Originally Posted by james121515
$x=2k+1=2(k+1)-1$