# Is $(\mathbb R^{+}, \oplus, \otimes)$ a field

abstract-algebrafield-theory

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.

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.