= Solution
Use signed <velocities> in the <shock frame>; the <Mach number> depends on their squares. For a genuine nonradiative <perfect gas> <shock wave>, the <Rankine-Hugoniot conditions for a perfect gas> give a common <mass flux> $j=\rho_Lu_L=\rho_Ru_R$, a common <momentum flux> $p+\rho u^2$, and a common specific stagnation <enthalpy> $\gamma p/[(\gamma-1)\rho]+u^2/2$. Here $e$ in the given <equation of state> is <specific internal energy>, despite the source's informal “energy density” wording.
Put $r=\rho_R/\rho_L$, $P=p_R/p_L$ and $M^2=\rho_Lu_L^2/(\gamma p_L)$. Mass and momentum give $u_R=u_L/r$ and $P=1+\gamma M^2(1-1/r)$. Substituting these into the energy relation and clearing denominators gives
$$
(r-1)\left[((\gamma-1)M^2+2)r-(\gamma+1)M^2\right]=0.
$$
The $r=1$ factor is the no-jump solution. On the nontrivial <shock wave> branch,
$$
\boxed{r=\frac{(\gamma+1)M^2}{(\gamma-1)M^2+2},\qquad
P=\frac{2\gamma M^2-(\gamma-1)}{\gamma+1}.}
$$
In particular $r-1=2(M^2-1)/[(\gamma-1)M^2+2]$. With $\gamma>1$, compression is therefore equivalent to $M_L^2>1$. The other <Mach number> is
$$
M_R^2=\frac{M_L^2}{rP}
=\frac{(\gamma-1)M_L^2+2}{2\gamma M_L^2-(\gamma-1)}.
$$
Positive <pressure> makes its denominator positive, and $M_R^2-1$ has the opposite sign to $M_L^2-1$. Thus a nontrivial physical <shock wave> has a supersonic side and a subsonic side; the entropy-admissible flow goes from the former to the latter.
The <shock compression ratio> also yields $u_Lu_R=(\gamma-1)u_L^2/(\gamma+1)+2\gamma p_L/[(\gamma+1)\rho_L]$. Rearranging produces the first <velocity> identity in the question. The <pressure> relation gives the second identity in the equivalent form
$$
p_R=\frac{2\rho_Lu_L^2-(\gamma-1)p_L}{\gamma+1}.
$$
The numerator here contains <mass density>, as the original PDF shows; the converted TeX incorrectly replaces that <mass density> by <pressure>.
For reflection, the once-shocked gas is the common left state in both applications. Before rebound, its <velocity> in the <shock frame> is $a=u_s-U_+<0$ and the gas ahead has <velocity> $-U_+$; after rebound, it is $b=u_s+U_->0$ and the gas next to the wall has <velocity> $U_-$. In both cases the right <velocity> minus the left <velocity> equals $-u_s$. Applying the first identity therefore gives the same <quadratic equation>,
$$
v^2-\frac{\gamma+1}{2}u_sv-\frac{\gamma p_s}{\rho_s}=0,
$$
with roots $a$ and $b$. Their opposite signs distinguish the two roots, so \b[$\boxed{(u_s-U_+)(u_s+U_-)=-\gamma p_s/\rho_s}$].
Write $q_0=p_0/p_s$, $q_1=p_1/p_s$. The <pressure> identity applied to each <shock wave> gives
$$
a^2=\frac{p_s}{2\rho_s}[(\gamma+1)q_0+\gamma-1],\qquad
b^2=\frac{p_s}{2\rho_s}[(\gamma+1)q_1+\gamma-1].
$$
Squaring the root-product relation and eliminating the <velocities> yields
$$
[(\gamma+1)q_0+\gamma-1][(\gamma+1)q_1+\gamma-1]=4\gamma^2,
$$
and hence the <normal shock reflection at a rigid wall> relation
$$
\boxed{\frac{p_1}{p_s}=\frac{3\gamma-1-(\gamma-1)p_0/p_s}{\gamma-1+(\gamma+1)p_0/p_s}.}
$$
Taking $p_0/p_s\to0$ gives \b[$\boxed{p_1/p_s\to(3\gamma-1)/(\gamma-1)}$]; for a <monatomic gas> this limiting amplification is six.
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