# Let * be a binary operation on the set of real numbers R defined as follows: a*b=a+b-3(ab)^2

Let * be a binary operation on the set of real numbers R defined as follows:
$a\cdot b=a+b-3{\left(ab\right)}^{2}$, where $a,b\in \mathbb{R}$
- Prove that * is commutative but not associative algebraic operation on R.
- Find the identity element for * .
- Show that 1 has two inverses with respect to *.
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Donald Cheek
Step 1
The statement is commutative if $a\cdot b=b\cdot$a and associative if $\left(a\cdot b\right)\cdot c=a\cdot \left(b\cdot c\right)$ Step 2
Calculate $a\cdot b$ and $b\cdot a$
$a\cdot b=a+b-3{\left(ab\right)}^{2}$
$b\cdot a=b+a-3{\left(ba\right)}^{2}$
$=a+b-3{\left(ab\right)}^{2}$
$=a\cdot b$
Step 3
Calculate $\left(a\cdot b\right)\cdot c$ and $a\cdot \left(b\cdot c\right)$.
$\left(a\cdot b\right)\cdot c=\left(a+b-3{\left(ab\right)}^{2}\cdot c$
$=\left(a+b-3{\left(ab\right)}^{2}\right)+c-3{\left[\left(a+b-3{\left(ab\right)}^{2}\right)c\right]}^{2}$
$=a+b+c-3{a}^{2}{b}^{2}-3{c}^{2}{\left(a+b-3{\left(ab\right)}^{2}\right)}^{2}$
$a\cdot \left(b\cdot c\right)=a\cdot \left(b+c-3{\left(bc\right)}^{2}\right)$
$=a+\left(b+c-3{\left(bc\right)}^{2}\right)-3{\left[a\left(b+c-3{\left(bc\right)}^{2}\right)\right]}^{2}$
$=a+b+c-3{b}^{2}{c}^{2}-3{a}^{2}{\left(b+c-3\left(b{c}^{2}\right)\right)}^{2}$
$\ne \left(a\cdot b\right)\cdot c$
Thus, not associative.
Step 5
Determine identity.
$a\cdot e=a$
$a+e-3{\left(ae\right)}^{2}=a$
$e-3{e}^{2}{a}^{2}=0$
$e\left(1-3{e}^{2}{a}^{2}\right)=0$
$e=0,1-3{e}^{2}{a}^{2}=0$
$e=0,{e}^{2}=\frac{1}{3{a}^{2}}$
$e=0,e=\frac{1}{\sqrt{3}a}$
Step 6
For inverse ${a}^{-1}$ following condition needs to be satisfied.