Mark the tick against the correct answer in the following:
Let $Z^{+}$be the set of all positive integers. Then, the operation $*$ on $Z^{+}$defined bya $* b=a^{b}$ is
A. commutative but not associative
B. associative but not commutative
C. neither commutative nor associative
D. both commutative and associative
According to the question,
$\mathrm{Q}=\{$ All integers $\}$
$\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}^{*} \mathrm{~b}=\mathrm{a}^{\mathrm{b}}\right\}$
Formula
$*$ is commutative if $a * b=b^{*} a$
$*$ is associative if $(a * b) * c=a *(b * c)$
Check for commutative
Consider, $a * b=a^{b}$
And, $b * a=b^{a}$
Both equations are not the same and will not always be true .
Therefore, $*$ is not commutative (1)
Check for associative
Consider, $(a * b) * c=\left(a^{b}\right) * c=\left(a^{b}\right)^{c}$
And, $a^{*}\left(b^{*} c\right)=a^{*}\left(b^{c}\right)=a^{\left(b^{c}\right)}$
$E x a=2 b=3 c=4$
$\left(a^{*} b\right) * c=\left(2^{3}\right) * c=(8)^{4}$
$a^{*}\left(b^{*} c\right)=2 *\left(3^{4}\right)=2^{(81)}$
Both the equation are not the same and therefore will not always be true.
Therefore, $*$ is not associative (2)
Now, according to the equations (1), (2)
Correct option will be (C)
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