Is $(\mathbb R^{+}, \oplus, \otimes)$ a field? Note that $x \oplus y = xy$ and $x \otimes y = x^{\ln y}$ for $x,y \in\mathbb R^{+}$.

I have already proved that it is a ring and that it is commutative.

I also found out that the multiplicative identity is Euler's number $e.$

I do not know how to proceed in showing it is a field though.

My first thought is that it should be something like $x^{\ln z} = e = z^{\ln x}$ for $x \in \mathbb R^{+}$, where $z$ is the multiplicative inverse. But I'm struggling to find $z.$

Any help would be appreciated.

## Best Answer

To prove it's a field you additionally need to prove that each non-zero(non-additive identity) element has a multiplicative inverse.

$$x\otimes y = x^{\ln(y)}=e$$

So $\ln(y)\ln(x)=1\implies y = e^{\frac{1}{\ln(x)}}$

Also $ e^{\frac{1}{\ln(x)}}\otimes x=e^{(\frac{1}{\ln(x)})\ln(x)}=e$.

So each non-zero(additive identity) element $x\in R$ has inverse $e^{\frac{1}{\ln(x)}}$.

Now note that the additive identity is $1$. As $x\oplus 1 =1\oplus x = x$.

Now as in a field , you cannot invert the additive identity. Here the inverse of $1$ does not exist as $e^{\frac{1}{\ln(x)}}$ does not make sense at $x=1$ ( Compare this situation to when $\frac{1}{0}$ does not make sense in $\mathbb{R}$ with usual addition and multiplication. Here $1$ is playing the role of $0$ and $e$ is playing the role of $1$.

Also I forgot to mention. In order to complete the proof . You should also verify the associativity and distributive laws.