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I got stucked with this question, anyone please help me. Is this a group or not? If it...

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jolas | (Level 1) eNoter

Posted June 7, 2013 at 10:26 PM via web

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I got stucked with this question, anyone please help me.

Is this a group or not? If it is not, Identify the condition being violated and tell whether it is a monoid or a semigroup.If it is a group, show your complete proof

`(R,*) wherea*b =a^(2) + b`

note R is a set of real numbers

 

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 8, 2013 at 2:35 AM (Answer #1)

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Determine if `{RR,*}` is a group where `a*b=a^2+b` :

A group must follow the following axioms: closure, associativity, identity, and inverses.

(1) The set is closed under `*` since a real number squared plus another real number is a real number.

(2) Let `a,b,c in RR` . We will check if `(a*b)*c=a*(b*c)`

`(a*b)*c=(a^2+b)*c=(a^2+b)^2+c=a^4+2a^2b+b^2+c`

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

The operation is not associative.

(3) Suppose e is the identity element. Let `a in RR, a != 0` . Then `a*e=e*a=a`

`e*a=a ==> e^2+a=a ==> e=0`

`a*e=a ==> a^2+e=a ==> e=a-a^2 !=0` except when a=1. So there is no unique identity element.

(4) With no identity element there can be no inverses.

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This is not a group as it fails the associative axiom, the identity axiom, and the inverse axiom.

This is not a monoid since the binary operation is not associative.

This is not a semigroup as the operation is not associative.

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