= Solution
Write the standard <Lorentz transformation> as $x'=\gamma(x-vt)$, $ct'=\gamma(ct-vx/c)$, with <Lorentz factor> $\gamma=(1-v^2/c^2)^{-1/2}$ and $|v|<c$. In the <null coordinates in two-dimensional Minkowski spacetime>, addition and subtraction give
$$
x'_+=\gamma(1-v/c)x_+,\qquad x'_-=\gamma(1+v/c)x_-.
$$
Since both factors are positive,
$$
\boxed{x'_+=\lambda(v)x_+,\quad x'_-=\lambda(-v)x_-,\qquad
\lambda(v)=\sqrt{\frac{c-v}{c+v}}.}
$$
Their product is one, so <Lorentz invariance> of the interval follows immediately:
$$
\boxed{x'^2-c^2t'^2=x'_+x'_-=x_+x_-=x^2-c^2t^2.}
$$
Two successive collinear <Lorentz boosts> multiply the positive-coordinate factor by $\lambda(v_2)\lambda(v_1)$ and the negative-coordinate factor by its reciprocal. Thus they have the same form with a velocity $v_3$ determined by
$$
\frac{c-v_3}{c+v_3}
=\frac{(c-v_1)(c-v_2)}{(c+v_1)(c+v_2)}.
$$
Cross multiplication yields the <relativistic velocity-addition formula> formula
$$
\boxed{v_3=\frac{v_1+v_2}{1+v_1v_2/c^2}.}
$$
For $|v_1|,|v_2|<c$, the denominator is positive and
$$
1-\frac{v_3^2}{c^2}
=\frac{(1-v_1^2/c^2)(1-v_2^2/c^2)}{(1+v_1v_2/c^2)^2}>0,
$$
so $v_3$ is admissible. The positive square roots then verify the factor identity itself, not merely its square. In terms of <rapidity> $v/c=\tanh\eta$, this same calculation is addition of rapidities because $\lambda(v)=e^{-\eta}$.
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