Verify associativity for the following three mappings: $f: N \rightarrow Z_{0}$ (the set of non-zero integers), $g: Z_{0} \rightarrow Q$ and $h: Q \rightarrow R$ given by $f(x)=2 x, g(x)=1 / x$ and $h(x)=e^{x}$.
Given that $f: N \rightarrow Z_{0}, g: Z_{0} \rightarrow Q$ and $h: Q \rightarrow R$.
$g \circ f: N \rightarrow Q$ and $h o g: Z_{0} \rightarrow R$
$\Rightarrow h \circ(g o f): N \rightarrow R$ and (hog) of $N \rightarrow R$
So, both have the same domains.
$(g o f)(x)=g(f(x))=g(2 x)=\frac{1}{2 x} \quad \ldots(1)$
$($ hog $)(x)=h(g(x))=h\left(\frac{1}{x}\right)=e^{\frac{1}{x}} \quad \ldots(2)$
Now,
$(h o(g o f))(x)=h((g o f)(x))=h\left(\frac{1}{2 x}\right)=e^{\frac{1}{2 x}}$ $[$ from (1) $]$
$((\operatorname{hog}) \circ f)(x)=(\operatorname{hog})(f(x))=(\operatorname{hog})(2 x)=e^{\frac{1}{2 x}}$ $[$ from (2) $]$
$\Rightarrow(h o(g o f))(x)=((h o g) o f)(x), \forall x \in N$
So, $h o(g o f)=(h o g) o f$
Hence, the associative property has been verified.
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