Contradiction and rational numbers

I have a set $\displaystyle S\subset{\mathbb{Q}}$ that is closed under addition and multiplication. Additionally, for any $\displaystyle r\in{\mathbb{Q}}$, one of the three is true

$\displaystyle r\in{S}, -r\in{S}, r=0$

I have to prove that all positive integers are in $\displaystyle S$.

So I assume there exists a positive integer $\displaystyle z\not\in{S}$, hence $\displaystyle -z\in{S}$.

Not sure which direction to go in to get the contradiction.

Help please?

Re: Contradiction and rational numbers

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

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

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

What is $\displaystyle (-1)(-1)~?$. What is the contradiction there?

Re: Contradiction and rational numbers

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Originally Posted by

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

$\displaystyle r\in{S}, -r\in{S}, r=0$

I have to prove that all positive integers are in $\displaystyle S$.

So I assume there exists a positive integer $\displaystyle z\not\in{S}$, hence $\displaystyle -z\in{S}$.

Not sure which direction to go in to get the contradiction.

Help please?

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.

Re: Contradiction and rational numbers

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**I-Think** Additionally, for any $\displaystyle r\in{\mathbb{Q}}$, one of the three is true

$\displaystyle r\in{S}, -r\in{S}, r=0$

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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 $\displaystyle \forall r\in\mathbb{Q}.\,r\ne0\to r\in S\lor -r\in S$.

Re: Contradiction and rational numbers

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**emakarov** I don't think S = {0} satisfies $\displaystyle \forall r\in\mathbb{Q}.\,r\ne0\to r\in S\lor -r\in S$.

You are right. I was thinking the condition was "$\displaystyle r\in S$".