= Solution
If $S'$ moves at $+v$ relative to $S$, the <Lorentz transformations> are
$$
\boxed{x'=\gamma(x-vt),\quad t'=\gamma(t-vx/c^2),\quad \gamma=(1-v^2/c^2)^{-1/2}.}
$$
A rod at rest in $S$ has length $L$. To measure it in $S'$, select endpoint events simultaneous there: $\Delta t'=0$, so $\Delta t=vL/c^2$ and $\Delta x'=\gamma(L-v\Delta t)=L/\gamma$. This is <length contraction>.
Now let a clock at rest in $S'$ cross the two rod endpoints. These are different events, with $\Delta x=L$, $\Delta t=L/v$ and $\Delta x'=0$. Their transformed time interval is
$$
\Delta t'=\gamma\left(\frac Lv-\frac{vL}{c^2}\right)=\frac{L}{\gamma v}=\frac{L'}v.
$$
Thus
$$
\boxed{L'=L/\gamma,\qquad \Delta t=\gamma\Delta t'.}
$$
The moving clock measures the <proper time> $\Delta t'$. The rod's contracted length divided by its <speed> in that clock's frame gives exactly the same crossing <time>; <time dilation> and <length contraction> are consistent consequences of the same transformation. Measuring a length and measuring a traversal interval require different simultaneity conditions.
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