# Positive Real Numbers forming a subring

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.

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