Answer to Question #278807 in Abstract Algebra for eubs

Question

a*b=-a-b-2ab; a⊕b=3a+3b

Answer 1

Let the operations * and ⊕ be defined on the set of integers.

The operation * on $\Z$ is associative, if for every $a, b, c, ∈ \Z,$ we have

$(a ^* b) ^* c = a^* (b^*c).$

We have

$$(a ^* b) ^* c=(-a-b-2ab) ^* c$$

$$=a+b+2ab-c+2(a+b+2ab)c$$

$$=a+b-c+2ab+2ac+2bc+4abc$$

$$a^* (b^*c)=a^* (-b-c-2bc)$$

$$=-a+b+c+2bc+2a(b+c+2bc)$$

$$=-a+b+c+2ab+2ac+2bc+4abc$$

Suppose $(a ^* b) ^* c = a^* (b^*c), a, b, c \in \Z.$ Then

$$a+b-c+2ab+2ac+2bc+4abc$$

$$=-a+b+c+2ab+2ac+2bc+4abc$$

$$=>a=c$$

Hence the statement $(a ^* b) ^* c = a^* (b^*c)$ is False for $a\not=c, a, b,c \in \Z.$

The operation $^*$ is not associative.

The operation $^*$ on $\Z$ is commutative, if for every $a, b, ∈ \Z,$ we have

$a ^* b = b^* a$

We have

$$a ^* b =-a-b-2ab$$

$$b ^* a =-b-a-2ba=-a-b-2ab$$

Hence the statement $a ^* b = b^* a$ is True for $a, b \in \Z.$

The operation $^*$ is commutative.

The operation $^*$ on $\Z$ is left-distributive over $⊕$ , if for every $a, b, c∈ \Z,$ we have $a ^*( b⊕c) =(a^*b)⊕(a^*c)$

We have

$$a ^*( b⊕c)=-a-(3b+3c)-2a(3b+3c)$$

$$=-a-3b-3c-6ab-6ac$$

$$(a^*b)⊕(a^*c) =3(-a-b-2ab)+3(-a-c-2ac)$$

$$=-6a-3b-3c-6ab-6ac$$

Suppose $a ^*( b⊕c) =(a^*b)⊕(a^*c) , a, b, c \in \Z.$ Then

$$-a-3b-3c-6ab-6ac$$

$$=-6a-3b-3c-6ab-6ac$$

$$=>a=0$$

Hence the statement $a ^*( b⊕c) =(a^*b)⊕(a^*c)$ is False for $a\not=0, a, b,c \in \Z.$

The operation $^*$ is not left-distributive over $⊕.$

The operation $^*$ on $\Z$ is right-distributive over $⊕$ , if for every $a, b, c∈ \Z,$ we have $( b⊕c)^*a =(b^*a)⊕(c^*a)$

We have

$$( b⊕c)^*a =-(3b+3c)-a-2a(3b+3c)$$

$$=-a-3b-3c-6ab-6ac$$

$$(b^*a)⊕(c^*a) =3(-b-a-2ba)+3(-c-a-2ca)$$

$$=-6a-3b-3c-6ab-6ac$$

Suppose $( b⊕c)^*a =(b^*a)⊕(c^*a) , a, b, c \in \Z.$ Then

$$-a-3b-3c-6ab-6ac$$

$$=-6a-3b-3c-6ab-6ac$$

$$=>a=0$$

Hence the statement $( b⊕c)^*a =(b^*a)⊕(c^*a)$ is False for $a\not=0, a, b,c \in \Z.$

The operation $^*$ is not right-distributive over $⊕.$

The operation $⊕$ on $\Z$ is associative, if for every $a, b, c, ∈ \Z,$ we have

$(a ⊕ b) ⊕ c = a⊕(b⊕c).$

We have

$$(a⊕ b) ⊕c=(3a+3b) ⊕c=3(3a+3b)+3c$$

$$=9a+9b+3c$$

$$a⊕(b⊕c)=a⊕(3b+3c)=3a+3(3b+3c)$$

$$=3a+9b+9c$$

Suppose $(a ⊕ b) ⊕ c = a⊕(b⊕c) , a, b, c \in \Z.$ Then

$$9a+9b+3c=3a+9b+9c=>a=c$$

Hence the statement $(a ⊕ b) ⊕ c = a⊕(b⊕c)$ is False for $a\not=c, a, b,c \in \Z.$

The operation $⊕$ is not associative

The operation $⊕$ on $\Z$ is commutative, if for every $a, b, ∈ \Z,$ we have

$a⊕ b = b⊕ a$

We have

$$a ⊕b =3a+3b$$

$$b⊕a =3b+3a=3a+3b$$

Then $a ⊕ b =3a+3b=3b+3a= b⊕a, a,b\in\Z.$

The operation $⊕$ is commutative.

The operation $⊕$ on $\Z$ is left-distributive over $^*$ , if for every $a, b, c∈ \Z,$ we have $a ⊕( b^*c) =(a⊕b)^*(a⊕c).$

We have

$$a ⊕( b^*c)=3a+3(-b-c-2bc)$$

$$=3a-3b-3c-6bc$$

$$(a⊕b)^*(a⊕c)=-(3a+3b)-(3a+3c)$$

$$-2(3a+3b)(3a+3c)$$

$$=-6a-3b-3c-18a^2-18ab-18ac-18bc$$

Let $a=1, b=c=0.$ Then

$$a ⊕( b^*c)=1 ⊕( 0^*0)$$

$$=3(1)-3(0)-3(0)-6(0)(0)=3$$

$$(a⊕b)^*(a⊕c)=(1⊕0)^*(1⊕0)$$

$$=-6(1)-3(0)-3(0)-18(1)^2$$

$$-18(1)(0)-18(1)(0)-18(0)(0)=-24$$

Since $3\not=-24,$ the operation $⊕$ is not left-distributive over $^*.$

The operation $⊕$ on $\Z$ is right-distributive over $^*$ , if for every $a, b, c∈ \Z,$ we have $( b^*c) ⊕a=(b⊕a)^*(c⊕a).$

We have

$$( b^*c) ⊕a=3(-b-c-2bc)+3a$$

$$=3a-3b-3c-6bc$$

$$(b⊕a)^*(c⊕a)=-(3b+3a)-(3c+3a)$$

$$-2(3b+3a)(3c+3a)$$

$$=-6a-3b-3c-18bc-18ac-18ab-18a^2$$

Let $a=1, b=c=0.$ Then

$$( b^*c) ⊕a=( 0^*0) ⊕1=3(1)-3(0)-3(0)$$

$$-6(0)(0)=3$$

$$(b⊕a)^*(c⊕a)=(0⊕1)^*(0⊕1)$$

$$=-6(1)-3(0)-3(0)-18(0)(0)-18(1)(0)$$

$$-18(1)(0)-18(1)^2=-24$$

Since $3\not=-24,$ the operation $⊕$ is not right-distributive over $^*.$