Answer to Question #97340 in Abstract Algebra for Brendan

Question #97340
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.
1
Expert's answer
2019-10-27T13:38:57-0400

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.


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