set of odd integers proof

**james121515** · February 24th 2010, 02:21 PM

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, $A\subseteq B$

**Plato** · February 24th 2010, 02:36 PM

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.

$x=2k+1=2(k+1)-1$

**james121515** · February 24th 2010, 09:36 PM

Thanks for your response.

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

-James

**Plato** · February 25th 2010, 08:21 AM

Originally Posted by james121515

Thanks for your response.

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

$x=2k+1=2(k+1)-1$
There is the symmetry.

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