Association, Commutation and Identity Elements on Binary Operations?

Question

Is the following closed, associative or commutative?

f(a, b) = (a+b)/2, where a, b ∈ Z.

I found that it is not closed but I am not sure how to find whether or not it is associative (I was confused about what c would be), commutative or has an identity element.

Also, does the fact that it is not closed mean that none of the other properties matter (associative, commutative and identity element).

Answer 1

If we define, for $a,b \in \mathbb Z$ :

$$f(a,b) = \frac {(a+b)}{2}$$

in order to test for associativity we have to check if:

$$f(f(a,b),c)=f(a,f(b,c))$$

We have :

$$f(f(a,b),c)= \frac {1}{2} [\frac {(a+b)}{2} + c] = \frac {1}{4} (a + b + 2c)$$

while :

$$f(a,f(b,c))= \frac {1}{2} [a +\frac {(b+c)}{2}] = \frac {1}{4} (2a + b + c)$$

Let $a=0$ and $b=c=1$.

Then $f(f(0,1),1)=\frac {1}{4}(0+1+2)=\frac {3}{4}$ and $f(0,f(1,1))=\frac {1}{4}(0+1 +1)=\frac {2}{4}$.

Thus :

$f(f(a,b),c) \ne f(a,f(b,c))$.

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It is of course commutative, because :

$$\frac {(a+b)}{2} = \frac {(b+a)}{2}$$

and thus :

$f(a,b) = f(b,a)$.

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An identity element must be an $i \in \mathbb Z$ such that :

$f(a,i) = a$, for all $a \in \mathbb Z$;

this means that we must have :

$$\frac {(a+i)}{2}=a$$

i.e. $a+i=2a$, i.e $i=a$, for all $a$, and this is impossible.