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Let $L$ be the set of positive real numbers. Two operations are defined: $$ a \oplus b = ab$$ $$a \times b = a^{\log b}.$$

Is L a ring?

1) a $\oplus$ b = ab, ab $\in$ L. 2) a x b = $ a^{\log b}, a^{\log b} \in L$. 3) addition is commutative 4)1 is the additive identity element 5) associative multiplication 6)Distributive laws 7) Associative multiplication.

8) Is 1 the additive inverse?

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    $\begingroup$ Remember that "zero" means the identity relative to $\oplus$ in this context. It will obviously not be the usual $0$ as a real number. Is there a positive real number $x$ such that $ax=a$ for all positive real numbers $a$? Did you try checking all of the other properties? $\endgroup$ – Jonas Meyer Feb 22 '13 at 3:39
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    $\begingroup$ I assumed that it had to be $a^{\log b}$, since $a \log b \not\in L$ when $0<b<1$… $\endgroup$ – Aeolian Feb 22 '13 at 3:43
  • $\begingroup$ @Aeolian: Yes, and it makes more sense in other ways. Thanks. $\endgroup$ – Jonas Meyer Feb 22 '13 at 3:43
  • $\begingroup$ I mean $a^{\log b}$.So the set of positive real numbers under normal addition and multiplication would not be a ring right? Because there is no additive identity? $\endgroup$ – grayQuant Feb 22 '13 at 3:43
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    $\begingroup$ The problem you forsee doesn't exist: the neutral additive element is $\,1\,$. What about the other ring axioms? Is $\,L\,$ an abelian group under that sum? Do we have associativity, distributivity (double sided in case multiplication isn't commutative)...? $\endgroup$ – DonAntonio Feb 22 '13 at 3:50
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You might apply the fact that if $\phi$ is a ring homomorphism (so respecting the addition and multiplication) from a ring $R$ to a set, then its image $\phi(R)$ is again a ring.
Now let $\phi$ be the map from the normal ring of reals $\mathbb{R}$ to $\mathbb{R}_{>0}$ with your addition $\oplus$ and multiplication $\otimes$, defined by $\phi(a)=e^{a}$, where $e$ is the base of the natural logarithm. Obviously this is a well-defined bijection. One can easily check that $\phi$ is a ring homomorphism. Observe that 0 is mapped to 1 (the neutral element w.r.t. $\oplus$), and the unit 1 is mapped to $e$ (the neutral element w.r.t. $\otimes$). Hence, the two rings are even isomorphic!

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You ask:

Is there a zero element in the set of positive real numbers?

What this means in this context is:

Is there a positive real number $a$ such that $a\oplus x=x$ for all positive real numbers $x$?

The defintion of $a\oplus x$ is $ax$, where the latter is ordinary multiplication. Is there a positive real number $a$ such that $ax=x$ for all positive real numbers $x$?

You write:

I see that L is a ring with the given properties but...

I don't see how you can see it is a ring if you don't know whether there is an additive identity. I suggest checking every single one of the properties of a ring to verify whether or not it holds. You can ask for more specific help if you get stuck on any other parts.

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  • $\begingroup$ The additive identity is 1, clear to me now. $\endgroup$ – grayQuant Feb 22 '13 at 3:57
  • $\begingroup$ @grayQuant: So what's left? Have you checked whether every property from the definition of ring holds? $\endgroup$ – Jonas Meyer Feb 22 '13 at 3:58
  • $\begingroup$ Is 1 the additive inverse? Because a $\oplus$ 1 = 1. $\endgroup$ – grayQuant Feb 22 '13 at 4:05
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    $\begingroup$ I think it is 1/a. $\endgroup$ – grayQuant Feb 22 '13 at 4:15
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    $\begingroup$ @grayQuant: Please be careful with notation and think about how the associative property should be written here. It would be $a\times(b\times c)=(a\times b)\times c$. That is not what you have; you seem to have gotten some of the "multiplications" mixed up with ordinary multiplications. For example, $2\times (3\times 4)=2\times (3^{\log 4})=2^{\log(3^{\log 4})}$. $\endgroup$ – Jonas Meyer Feb 22 '13 at 4:29

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