Answer to Question #97340 in Abstract Algebra for Brendan

Question

Consider the following sets together with binary operations.
Are they user-friendly? Z with binary operation
z1 . z2 = 2z1 - 4z2
Is the set closed under the operation? Is the operation commutative ?
Is the operation associative ? Is there an identity? If there is an identity
element then does every element have an inverse relative to the operation _
Consider R together with x . y = x/y. Ask the same questions as in last
example.

Answer 1

1. $\mathbb{Z}$  with binary operation $z_1\cdot z_2 = 2z_1 - 4z_2$ .

The set $\mathbb{Z}$  is closed under $\cdot$  because if $z_1, z_2\in \mathbb{Z}$ , then $2z_1\in \mathbb{Z}$ , $4z_2\in \mathbb{Z}$ , $2z_1 - 4z_2\in \mathbb{Z}$ .

The operation $\cdot$  is not commutative because

$0\cdot 1 = 2\times 0 - 4\times 1 = -4 \neq 2 = 2\times 1 - 4\times 0 = 1\cdot 0.$

The operation $\cdot$  is not associative because

$(1\cdot 0)\cdot 0 = 2\cdot 0 = 4 \neq 2 = 1\cdot 0 = 1\cdot (0\cdot 0).$

If $z$  is an identity, then $0 = 0\cdot z = -4z$  and $1 = 1\cdot z = 2 -4z$ . Subtracting the first equation from the last gives $1=2$ , contradiction. Therefore, there is no identity.

2. $\mathbb{R}$  with binary operation $x\cdot y = x/y$ .

The set $\mathbb{R}$  is closed under $\cdot$  because if $x, y\in \mathbb{R}$ , then $x/y\in \mathbb{R}$ . Also $\cdot$  is a partial operation because $x\cdot 0$  is undefined for all $x\in \mathbb{R}$ .

The operation $\cdot$  is not commutative because

$1\cdot 2 = 1/2 \neq 2 = 2/1 = 2\cdot 1.$

The operation $\cdot$  is not associative because

$(1\cdot 1)\cdot 2 = (1/1)/2 = 1/2 \neq 2 = 1/(1/2) = 1\cdot (1\cdot 2).$

If $x$  is an identity, then $1 = x\cdot 1 = x$  and $2 = x\cdot 2 = x/2$ . Dividing the last equation by the first gives $2/1 = (x/2)/x = 1/2$ , contradiction. Therefore, there is no identity.