Magma $(\mathbb R,*)$ with a binary operation $\;a*b=a+b-2a^2b^2$

Question

Let $(\mathbb R, *)$ be a magma with a binary operation: $$a*b=a+b-2a^2b^2$$ Prove

$(a)$ the binary operation is commutative, but not associative,

$(b)$ $0$ is a neutral element for that operation,

$(c)$ $\forall x\in\langle-\frac{1}{2},+\infty\rangle$ there are two inverses,

$(d)$ $\forall x\in \langle-\infty,-\frac{1}{2}\rangle$ there is no inverse,

$(e)$ for $x=\frac{1}{2}$ there exists a unique inverse.

My work:

$(a)$ $\forall a,b\in \mathbb R\;\;\;a*b=b*a$ $$a*b=a+b-2a^2b^2=b+a-2a^2b^2=b*a$$ $(b)$ $$a*b=a=a+b-2a^2b^2=a+b(1-2a^2b)\implies b(1-2a^2b)=0$$$$\implies b=0\;\;\;\; (b\; \text{independent of} \;a)$$ $(c)$ Asummption $\exists \;b_{1,2}\in \mathbb R$ $b_{1,2}\;=(a\;\text{inverse})$ $$2b^2a^2-a-b=0$$ $$\implies D=1+8b^3>0\implies b\in\Big\langle-\frac{1}{2},+\infty\rangle$$ $(c)\implies$ in $(d)\;:$ $$D<0\implies\forall a\in\langle-\infty,\frac{1}{2}\Big\rangle \;\nexists \;a\;\text{inverse}$$ $(c) \wedge (d)\implies$ in $(e):$ for$\;a=-\frac{1}{2}\;!\exists b$ such that $b=a$ inverse, $D=0$

What should I correct?

Answer 1

All in all, it would be way cleaner if $a,b$ didn't appear out of nowhere : eg for a), begin with "Let $a,b \in \mathbb{R}$".

• a) you forgot to show that $*$ is not associative. Hint : just take $a=1$, $b=2$ and $c=3$ and compute $(a*b)*c$ and $a*(b*c)$.
• b) You didn't prove that $0$ is a neutral element. What you've proven is that if there is a neutral element, then it must be $0$. The correct way to show that $0$ is a neutral element is simply to compute $a*0$ and to check that $a*0=a$.
• c) What did you prove here? I don't get it. The correct way to go is to take $x\in\langle-\frac{1}{2},+\infty\rangle$ arbitrary, then to show that the equation $x*b=0$ has two solutions.
• d) Do this : take arbitrary $x\in \langle -\infty, \,-\frac{1}{2}\rangle$ and show that the equation $x*b=0$ has no solution.
• e) Show that the equation $\frac{-1}{2}*b=0$ has a unique solution.