• Jun 16th 2013, 09:58 AM
I-Think
I have a set $S\subset{\mathbb{Q}}$ that is closed under addition and multiplication. Additionally, for any $r\in{\mathbb{Q}}$, one of the three is true
$r\in{S}, -r\in{S}, r=0$

I have to prove that all positive integers are in $S$.
So I assume there exists a positive integer $z\not\in{S}$, hence $-z\in{S}$.
Not sure which direction to go in to get the contradiction.
• Jun 16th 2013, 10:39 AM
Plato
Quote:

Originally Posted by I-Think
I have a set $S\subset{\mathbb{Q}}$ that is closed under addition and multiplication. Additionally, for any $r\in{\mathbb{Q}}$, one of the three is true $r\in{S}, -r\in{S}, r=0$

If you can show that $1\in S$ then you are done because $S$ is closed by addition.

You know that $1\ne 0$. So if $1\notin S$ then $-1\in S$.

What is $(-1)(-1)~?$. What is the contradiction there?
• Jul 9th 2013, 05:55 AM
HallsofIvy
Quote:

Originally Posted by I-Think
I have a set $S\subset{\mathbb{Q}}$ that is closed under addition and multiplication. Additionally, for any $r\in{\mathbb{Q}}$, one of the three is true
$r\in{S}, -r\in{S}, r=0$

I have to prove that all positive integers are in $S$.
So I assume there exists a positive integer $z\not\in{S}$, hence $-z\in{S}$.
Not sure which direction to go in to get the contradiction.

You can't prove it. It is not true.
For example S= {0} satisfies all of the conditions but does not include all of the integers.

You need some additional condition, perhaps the "1 in S" that Plato suggested.
• Jul 9th 2013, 12:15 PM
emakarov
Quote:

Originally Posted by I-Think
Additionally, for any $r\in{\mathbb{Q}}$, one of the three is true
$r\in{S}, -r\in{S}, r=0$

Quote:

Originally Posted by HallsofIvy
For example S= {0} satisfies all of the conditions but does not include all of the integers.

I don't think S = {0} satisfies $\forall r\in\mathbb{Q}.\,r\ne0\to r\in S\lor -r\in S$.
• Jul 9th 2013, 03:12 PM
HallsofIvy
I don't think S = {0} satisfies $\forall r\in\mathbb{Q}.\,r\ne0\to r\in S\lor -r\in S$.
You are right. I was thinking the condition was " $r\in S$".