I was wondering if the subset of positive real numbers ${\mathbb{R}_{>0}}$ forms a subring of the real numbers $\mathbb{R}$ under the regular operations of addition and multiplication.

My thought so far is that since $1$ is clearly in ${\mathbb{R}_{>0}}$, we know ${\mathbb{R}_{>0}}$ has the multiplicative identity of $\mathbb{R}$.

However, since ${\mathbb{R}_{>0}}$ is not closed under subtraction (take $1, 2$ in ${\mathbb{R}_{>0}}$), this is not a subring.

I was wondering then the difference between addition and subtraction? Why can I not use close under addition, since isn't subtraction the same as addition? Is it because the additive inverse of $1$ and $2$ is not in the set ${\mathbb{R}_{>0}}$?

In other words, is closure under subtraction the same as requiring an abelian group under addition?

Thanks for your help, new to rings.

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    $\begingroup$ Closure under subtraction means: Closure under addition AND under taking inverses. $\endgroup$ – Stefan Perko Jan 27 '15 at 21:18
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    $\begingroup$ Yes, you need additive inverses and don't have them, and that is it. $\endgroup$ – Jonas Meyer Jan 27 '15 at 21:19
  • $\begingroup$ Just wanted to confirm my thoughts were correct. Still new to this stuff, so wasn't sure if my reasoning was right. $\endgroup$ – jstnchng Jan 27 '15 at 21:19
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    $\begingroup$ $(\mathbb{R}^+,+)$ is not a subgroup of $(\mathbb{R},+)$. $\endgroup$ – MattAllegro Jan 27 '15 at 21:19

No it is not a subring, for it is not a subgroup (and every subring is a subgroup)!


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