Associative Operation/Examples/Non-Associative/xy+1
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Example of Non-Associative Operations
Let $\R$ denote the set of real numbers.
Let $\circ$ denote the operation on $\R$ defined as:
- $\forall x, y \in \R: x \circ y := x y + 1$
Then $\circ$ is not an associative operation, despite being commutative.
Proof
Let $x, y, z \in \R$.
We have:
\(\ds \paren {x \circ y} \circ z\) | \(=\) | \(\ds \paren {x y + 1} z + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x y z + z + 1\) |
\(\ds x \circ \paren {y \circ z}\) | \(=\) | \(\ds x \paren {y z + 1} + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x y z + x + 1\) |
But unless $x = z$ it is not the case that $\paren {x y + 1} z + 1 = x y z + x + 1$.
However, $x y + 1 = y x + 1$ and so $\circ$ is commutative.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $4$. Groups: Exercise $2$