Answer to Question #278805 in Abstract Algebra for Ebs

Question #278805

x*y=x2+2x+y2; x⊕y=x+y


1
Expert's answer
2021-12-13T15:07:29-0500

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

The operation * on "\\Z" is associative, if for every "a, b, c, \u2208 \\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, \u2208 \\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 "\u2295" , if for every "a, b, c\u2208 \\Z," we have "a ^*( b\u2295c) =(a^*b)\u2295(a^*c)"

We have


"a ^*( b\u2295c)=a^2+2a+(b+c)^2"

"(a^*b)\u2295(a^*c) =a^2+2a+b^2+a^2+2a+c^2"

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


"a ^*( b\u2295c)=1 ^*( 0\u22950)=1^2+2(1)+(0+0)^2=3"

"(a^*b)\u2295(a^*c)=(1^*0)\u2295(1^*0 )"

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

Since "3\\not=6," the operation "^*" is not left-distributive over "\u2295."


The operation "^*" on "\\Z" is right-distributive over "\u2295" , if for every "a, b, c\u2208 \\Z," we have "( b\u2295c)^*a =(b^*a)\u2295(c^*a)"

We have


"( b\u2295c)^*a =(b+c)^2+2(b+c)+a^2"

"(b^*a)\u2295(c^*a) =b^2+2b+a^2+c^2+2c+a^2"

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


"( b\u2295c)^*a =( 0\u22950)^*1"

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

"(b^*a)\u2295(c^*a )=(0^*1)\u2295(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 "\u2295."


The operation "\u2295" on "\\Z" is associative, if for every "a, b, c, \u2208 \\Z," we have

"(a \u2295 b) \u2295 c = a\u2295(b\u2295c)."

We have


"(a\u2295 b) \u2295c=(a+b) \u2295c=a+b+c"

"a\u2295(b\u2295c)=a\u2295(b+c)=a+b+c"

Then "(a \u2295 b) \u2295 c =a+b+c= a\u2295(b\u2295c), a,b,c\\in\\Z."

The operation "\u2295" is associative.


The operation "\u2295" on "\\Z" is commutative, if for every "a, b, \u2208 \\Z," we have

"a\u2295 b = b\u2295 a"

We have


"a \u2295b =a+b"

"b\u2295a =b+a=a+b"

Then "a \u2295 b =a+b=b+a= b\u2295a, a,b\\in\\Z."

The operation "\u2295" is commutative.


The operation "\u2295" on "\\Z" is left-distributive over "^*" , if for every "a, b, c\u2208 \\Z," we have "a \u2295( b^*c) =(a\u2295b)^*(a\u2295c)."

We have


"a \u2295( b^*c)=a+b^2+2b+c^2"

"(a\u2295b)^*(a\u2295c)=(a+b)^2+2(a+b)+(a+c)^2"

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


"a \u2295( b^*c)=1 \u2295( 0^*0)=1+0^2+2(0)+0^2=1"

"(a\u2295b)^*(a\u2295c)=(1\u22950)^*(1\u22950)"

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

Since "1\\not=4," the operation "\u2295" is not left-distributive over "^*."


The operation "\u2295" on "\\Z" isright-distributive over "^*" , if for every "a, b, c\u2208 \\Z," we have "( b^*c) \u2295a=(b\u2295a)^*(c\u2295a)."

We have


"( b^*c) \u2295a=b^2+2b+c^2+a"

"(b\u2295a)^*(c\u2295a)=(b+a)^2+2(b+a)+(c+a)^2"

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


"( b^*c) \u2295a=( 0^*0) \u22951=0^2+2(0)+0^2+1=1"

"(b\u2295a)^*(c\u2295a)=(0\u22951)^*(0\u22951)"

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

Since "1\\not=4," the operation "\u2295" is not right-distributive over "^*."



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