Show that the operation $*: R \times R \rightarrow R$ defined by $a * b = a + 2b$ is not associative.
(N/A) An operation $*$ is associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in R$.
First,calculate $(a * b) * c$:
$(a * b) * c = (a + 2b) * c = (a + 2b) + 2c = a + 2b + 2c$.
Next,calculate $a * (b * c)$:
$a * (b * c) = a * (b + 2c) = a + 2(b + 2c) = a + 2b + 4c$.
Since $a + 2b + 2c \neq a + 2b + 4c$ for all $a, b, c \in R$,the operation is not associative.
For example,let $a = 8, b = 5, c = 3$:
$(8 * 5) * 3 = (8 + 2(5)) * 3 = 18 * 3 = 18 + 2(3) = 18 + 6 = 24$.
$8 * (5 * 3) = 8 * (5 + 2(3)) = 8 * (5 + 6) = 8 * 11 = 8 + 2(11) = 8 + 22 = 30$.
Since $24 \neq 30$,the operation is not associative.
First,calculate $(a * b) * c$:
$(a * b) * c = (a + 2b) * c = (a + 2b) + 2c = a + 2b + 2c$.
Next,calculate $a * (b * c)$:
$a * (b * c) = a * (b + 2c) = a + 2(b + 2c) = a + 2b + 4c$.
Since $a + 2b + 2c \neq a + 2b + 4c$ for all $a, b, c \in R$,the operation is not associative.
For example,let $a = 8, b = 5, c = 3$:
$(8 * 5) * 3 = (8 + 2(5)) * 3 = 18 * 3 = 18 + 2(3) = 18 + 6 = 24$.
$8 * (5 * 3) = 8 * (5 + 2(3)) = 8 * (5 + 6) = 8 * 11 = 8 + 2(11) = 8 + 22 = 30$.
Since $24 \neq 30$,the operation is not associative.
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