determine whether the binary operation * under $ℝ$ is associative and commutative for $a * b =$ $\sqrt[3]{a^3+b^3}$

Question

Let $\ast$ be a binary operation on $\mathbb{R}$ defined by $$ a\ast b := \sqrt[3]{a^3 + b^3}. $$ Is $\ast$ associative and commutative?

I tried to apply the concepts of commutative and associative groups, but the answers are probably all wrong. My attempt is as follows:

Commutativity:

\begin{align} &a * b =\sqrt[3]{a^3+b^3} \quad\text{and}\quad b * a =\sqrt[3]{b^3+a^3} \\ &\qquad \implies a*b = b*a. \end{align}

Associativity:

$$ (a * b)*c = \sqrt[3]{\sqrt[3]{a^3+b^3}+c^3}\quad\text{and}\quad a*(b*c) = \sqrt[3]{\sqrt[3]{b^3+c^3}+a^3} $$

Answer 1

Your approach to associativity is wrong - you've missed an exponent. Your operation is defined by $$x*y=\sqrt[3]{x^3+y^3}.$$ So setting $x=a*b$ and $y=c$, we get

$$(a*b)*c=\sqrt[3]{(a*b)^{\color{red}{3}}+c^3}=\sqrt[3]{(\sqrt[3]{a^3+b^3})^{\color{red}{3}}+c^3}.$$ Note the red exponents, and contrast that with what you've written.

Now we can simplify things a lot - $(\sqrt[3]{a^3+b^3})^{\color{red}{3}}$ is just $a^3+b^3$! So in fact we have $$(a*b)*c=\sqrt[3]{a^3+b^3+c^3}.$$ Now, compute $a*(b*c)$ (paying attention to the exponent issue above) and compare!

Answer 2

More generally, if $\star$ is a binary operation on $A$, and $f:B\to A$ is a bijection, then we define binary operation $\star_f$ on $B$ as:

$$b_1\star_f b_2 = f^{-1}(f(b_1)\star f(b_2))$$

Then $\star$ is commutative (associative) if and only if $\star_f$ is commutative (respectively, associative.)

In your case, $A=B=\mathbb R$, $\star$ is addition, and $f(x)=x^3$ is the bijection.

Answer 3

Your proof of commutativity is correct.

Edit:

I have made the same mistake as you did. $*$ is indeed associative, as you can see:

\begin{align}(a * b) * c &= \sqrt[3]{(a*b)^3 + c^3} \\ &= \sqrt[3]{\left(\sqrt[3]{a^3+b^3}\right)^3 + c^3} \\ &= \sqrt[3]{a^3+b^3+c^3}\\ &=\sqrt[3]{a^3+\left(\sqrt[3]{b^3+c^3}\right)^3} \\ &= \sqrt[3]{a^3 + (b*c)^3} \\ &= a*(b*c) \end{align}

Answer 4

It's associative...

$(a*b)*c=\sqrt[3]{(a*b)^{{3}}+c^3}=\sqrt[3]{(\sqrt[3]{a^3+b^3})^{3}+c^3}=\sqrt[3]{a^3+b^3+c^3}$

It's easier and better posing $d=a*b$

$(a*b)*c=d*c$

Then we got

$d*c=\sqrt[3]{(d)^{{3}}+c^3}=\sqrt[3]{(a*b)^{{3}}+c^3}$

$\sqrt[3]{(a*b)^{{3}}+c^3}=\sqrt[3]{(\sqrt[3] {a^3+b^3)})^3+c^3}=\sqrt[3]{a^3+b^3+c^3}$