Abstract Algebra - showing commutativity and associativity

In summary, the conversation discusses determining if the operation * is both commutative and associative. It is concluded that * is commutative but not associative, and the definitions of commutativity and associativity are provided. The use of the operation in equations is discussed, with the conclusion that the operation is not associative. A more accurate method of determining associativity is suggested.
  • #1
kathrynag
598
0

Homework Statement



I need to determine if a*b=ab+1 is commutative and associative.
* is any arbitrary operation

Homework Equations





The Attempt at a Solution


a*b=ab+1 is commutative, but not associative.
I'm getting stuck in showing why.
b*a=ba+1
a*(b*c)=a(b+1)
?
 
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  • #2
What are you definitions of associativity and commutivity?
 
  • #3
commutative is iff a*b=b*a for all a and b.
associative is iff (a*b)*c=a*(b*c)
 
  • #4
kathrynag said:
commutative is iff a*b=b*a for all a and b.
associative is iff (a*b)*c=a*(b*c)

Ok, so have you tried to use your operation for * in these expressions? If so, what did you get?
 
  • #5
cristo said:
Ok, so have you tried to use your operation for * in these expressions? If so, what did you get?

That's where I get confused in using the operation.
a*b=ab+1
so, b*a=ba+1

a*b=ab+1
so (a*b)*c=a*(b*c)
ab+1*c=a*(bc+1)
(ab+1)c+1=a(bc+1)+1
abc+c+1=abc+a+1
So, not associative.

Is this the right idea?
 
  • #6
kathrynag said:
That's where I get confused in using the operation.
a*b=ab+1
so, b*a=ba+1

Yes, but why? You should state "because multiplication of real numbers in commutative..."

a*b=ab+1
so (a*b)*c=a*(b*c)
ab+1*c=a*(bc+1)
(ab+1)c+1=a(bc+1)+1
abc+c+1=abc+a+1
So, not associative.

This is a very sloppy way of doing things, since the equals signs don't hold: i.e. the thing on the left is not equal to the thing on the right.

Better would be to set it up something like this:

* is associative if(f) (a*b)*c=a*(b*c). Then,

LHS: (a*b)*c=(ab+1)*c=(ab+1)c+1=abc+c+1 ..(#)
RHS: a*(b*c)=a*(bc+1)=a(bc+1)+1=abc+a+1 ..(§)

Since (#)[itex]\neq[/itex](§), then * is not associative.
 

Related to Abstract Algebra - showing commutativity and associativity

1. What is commutativity in abstract algebra?

Commutativity refers to the property of a mathematical operation where the order of the operands does not affect the result. In abstract algebra, it means that for any two elements in a set, the order in which they are multiplied or added does not change the result.

2. How is commutativity shown in abstract algebra?

In order to show commutativity in abstract algebra, we need to prove that for any two elements a and b in a set, a * b = b * a. This can be done using mathematical induction or by providing a counterexample to disprove commutativity.

3. What does associativity mean in abstract algebra?

Associativity is a property of a mathematical operation where the grouping of the operands does not affect the result. In abstract algebra, it means that for any three elements a, b, and c in a set, (a * b) * c = a * (b * c).

4. How is associativity proven in abstract algebra?

To prove associativity in abstract algebra, we need to show that for any three elements a, b, and c in a set, (a * b) * c = a * (b * c). This can be done using mathematical induction or by providing a counterexample to disprove associativity.

5. Why is commutativity and associativity important in abstract algebra?

Commutativity and associativity are important properties in abstract algebra because they allow us to manipulate and simplify mathematical expressions. These properties also hold true in many other branches of mathematics and can be used to solve complex problems and equations.

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