Question

x*y=x²+2x+y²; x⊕y=x+y

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

$$(a ^* b) ^* c=(a^2+2a+b^2) ^* c$$

$$=(a^2+2a+b^2)^2+2(a^2+2a+b^2)+c^2$$

$$a^* (b^*c)= a^* (b^2+2b+c^2)$$

$$=a^2+2a+(b^2+2b+c^2)^2$$

Let $a=b=0, c=2.$ Then

$$(a ^* b) ^* c=(0 ^* 0) ^* 2=$$

$$=(0+0+0)^2+2(0+0+0)+2^2 =4$$

$$a^* (b^*c)= 0^* (0^*2)$$

$$=0+0+(0+0+2^2)^2=16$$

Since $4\not=16,$ then $(a ^* b) ^* c \not= a^* (b^*c).$

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

$$a ^* b =a^2+2a+b^2$$

$$b ^* a =b^2+2b+a^2$$

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

$$a^2+2a+b^2=b^2+2b+a^2=>a=b$$

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

The operation $^*$ is not 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

$$a ^*( b⊕c)=a^2+2a+(b+c)^2$$

$$(a^*b)⊕(a^*c) =a^2+2a+b^2+a^2+2a+c^2$$

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

$$a ^*( b⊕c)=1 ^*( 0⊕0)=1^2+2(1)+(0+0)^2=3$$

$$(a^*b)⊕(a^*c)=(1^*0)⊕(1^*0 )$$

$$=1^2+2(1)+0+1^2+2(1)+0=6$$

Since $3\not=6,$ 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

$$( b⊕c)^*a =(b+c)^2+2(b+c)+a^2$$

$$(b^*a)⊕(c^*a) =b^2+2b+a^2+c^2+2c+a^2$$

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

$$( b⊕c)^*a =( 0⊕0)^*1$$

$$=(0+0)^2+2(0+0)+1^2=1$$

$$(b^*a)⊕(c^*a )=(0^*1)⊕(0^*1)$$

$$=0^2+2(0)+1^2+0^2+2(0)+1^2=2$$

Since $1\not=2,$ 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

$$(a⊕ b) ⊕c=(a+b) ⊕c=a+b+c$$

$$a⊕(b⊕c)=a⊕(b+c)=a+b+c$$

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

The operation $⊕$ is associative.

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

$a⊕ b = b⊕ a$

We have

$$a ⊕b =a+b$$

$$b⊕a =b+a=a+b$$

Then $a ⊕ b =a+b=b+a= b⊕a, a,b\in\Z.$

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

$$a ⊕( b^*c)=a+b^2+2b+c^2$$

$$(a⊕b)^*(a⊕c)=(a+b)^2+2(a+b)+(a+c)^2$$

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

$$a ⊕( b^*c)=1 ⊕( 0^*0)=1+0^2+2(0)+0^2=1$$

$$(a⊕b)^*(a⊕c)=(1⊕0)^*(1⊕0)$$

$$=(1+0)^2+2(1+0)+(1+0)^2=4$$

Since $1\not=4,$ the operation $⊕$ is not left-distributive over $^*.$

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

We have

$$( b^*c) ⊕a=b^2+2b+c^2+a$$

$$(b⊕a)^*(c⊕a)=(b+a)^2+2(b+a)+(c+a)^2$$

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

$$( b^*c) ⊕a=( 0^*0) ⊕1=0^2+2(0)+0^2+1=1$$

$$(b⊕a)^*(c⊕a)=(0⊕1)^*(0⊕1)$$

$$=(0+1)^2+2(1+0)+(0+1)^2=4$$

Since $1\not=4,$ the operation $⊕$ is not right-distributive over $^*.$