- 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 *.

Brock Brown
2022-01-07
Answered

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 *.

- 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 *.

You can still ask an expert for help

Donald Cheek

Answered 2022-01-08
Author has **41** answers

Step 1

The statement is commutative if$a\cdot b=b\cdot$ a and associative if $(a\cdot b)\cdot c=a\cdot (b\cdot c)$
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$(a\cdot b)\cdot c$ and $a\cdot (b\cdot c)$ .

$(a\cdot b)\cdot c=(a+b-3{\left(ab\right)}^{2}\cdot c$

$=(a+b-3{\left(ab\right)}^{2})+c-3{\left[(a+b-3{\left(ab\right)}^{2})c\right]}^{2}$

$=a+b+c-3{a}^{2}{b}^{2}-3{c}^{2}{(a+b-3{\left(ab\right)}^{2})}^{2}$

$a\cdot (b\cdot c)=a\cdot (b+c-3{\left(bc\right)}^{2})$

$=a+(b+c-3{\left(bc\right)}^{2})-3{\left[a(b+c-3{\left(bc\right)}^{2})\right]}^{2}$

$=a+b+c-3{b}^{2}{c}^{2}-3{a}^{2}{(b+c-3\left(b{c}^{2}\right))}^{2}$

$\ne (a\cdot b)\cdot c$

Thus, not associative.

Step 5

Determine identity.

$a\cdot e=a$

$a+e-3{\left(ae\right)}^{2}=a$

$e-3{e}^{2}{a}^{2}=0$

$e(1-3{e}^{2}{a}^{2})=0$

$e=0,1-3{e}^{2}{a}^{2}=0$

$e=0,{e}^{2}=\frac{1}{3{a}^{2}}$

$e=0,e=\frac{1}{\sqrt{3}a}$

Step 6

For inverse$a}^{-1$ following condition needs to be satisfied.

The statement is commutative if

Calculate

Step 3

Calculate

Thus, not associative.

Step 5

Determine identity.

Step 6

For inverse

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