Do the real numbers form a division ring with operations $a\oplus b=a+b+\frac12$ and $a\odot b=a+b+2ab$?

Question

Is $(\Bbb R,\oplus,\odot)$ a division ring, where $$a\oplus b = a+b+\frac12$$ and $$a\odot b = a+b+ 2ab?$$

I have only issues with $\odot$. It doesn't work for inverse of $$a=\frac{-1}{2}$$ since $$a^{-1}=\frac{-a}{2a+1}$$ and in the definition of division rings there is condition that every nonzero element has a left inverse.

Answer 1

Note that, since $-1/2$ is the identity element under $(\mathbb{R},⊕)$, it is considered the "zero" element. This is consistent with the fact that $-1/2⊙a=a⊙-1/2=-1/2$, and so $-1/2$ has no inverse. For a ring, you should be considering $(\mathbb{R}\backslash \{-1/2\},⊙)$

Answer 2

What's the problem with a single element not being invertible? After all, all division rings have a single element that isn't invertible, namely the zero element.

Look at $\frac{-1}{2}$: $a\oplus\frac{-1}{2}=a$ and $a\odot \frac{-1}{2}=\frac{-1}{2}$ for all $a$. About now you should be beginning to see that maybe you have to rethink what you mean by "nonzero."

A handy thing to do in situations like this is to see what the multiplicative idempotents are. In this case, $a\odot a=a$ only happens if $a=0$ or $a=\frac{-1}{2}$. This can usually help you quickly find candidates for identities.

Answer 3

If $R$ is indeed a ring (with unit) then $0_{R}=-\frac{1}{2}$ and $1_{R}=0$. For every $a\neq0_{R}$ there must be a $b$ with $a\odot b=a+b+2ab=1_{R}$ wich is indeed the case: $b=\frac{-a}{1+2a}$. Now time to check things like distributivity, associativity and - for addition - commutativity.