= Solution
The <exponential distribution> has survival <probability> $e^{-u}$ for $u\geq0$. <Conditional probability> therefore gives
$$
\mathbb P(Z>s+t\mid Z>s)=\frac{e^{-(s+t)}}{e^{-s}}=e^{-t},
$$
which proves the <memoryless property>. Let $K=\lfloor Z\rfloor$ and $U=Z-K$. With $q=e^{-1}$,
$$
\boxed{\mathbb P(K=m)=(1-q)q^m,\quad m=0,1,2,\ldots}.
$$
The <geometric distribution> here starts at zero. Summing its first moment gives
$$
\boxed{\mathbb E[K]=\frac q{1-q}=\frac1{e-1}}.
$$
For $0\leq u<1$, sum the density over all integer translates:
$$
\boxed{f_U(u)=\sum_{m=0}^\infty e^{-m-u}=\frac{e^{-u}}{1-e^{-1}}},
$$
with zero density outside that interval. The <integer and fractional parts of an exponential variable> are independent, because for every integer $m\geq0$ and measurable $A\subseteq[0,1)$,
$$
\mathbb P(K=m,U\in A)=\int_Ae^{-m-u}\,du
=\mathbb P(K=m)\int_A f_U(u)\,du.
$$
This factorization proves <independence> of the discrete and continuous components, beyond just checking their separate marginals.
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