Let $a$ and $b$ be positive integers. Prove that
$$ \gcd (a, b) = \gcd (a + b, [a, b]) $$
Hint: You may find the result of Exercises 1.3, question 16 useful: If $\gcd (x, y) = 1$ then $\gcd (x + y, xy) = 1$.
Let $g=\gcd(a,b)$. This means, for some integers $x,y$
$$ a = gx $$
$$ b = gy $$
Substituting,
$$ \begin{align} \gcd (a + b, [a, b]) & = \gcd( gx + gy, [gx, gy] ) \\ \\ & = \gcd( g(x+y), g[x,y] ) \quad \text{ex (2.4).7} \\ \\ & = g \times \gcd(x+y, [x,y]) \quad \text{Proposition 1.11} \\ \\ & = g \times \gcd \left ( x+y, \frac{xy}{\gcd(x,y)} \right ) \quad \text{Proposition 2.22} \end{align} $$
By Proposition 1.5, $\gcd(\frac{a}{g}, \frac{b}{g})=1=\gcd(x,y)$. This gives us
$$ \begin{align} \gcd (a + b, [a, b]) & = g \times \gcd(x+y, xy) \\ \\ & = g \quad \text{using hint} \; \gcd (x, y) = 1 \implies \gcd (x + y, xy) = 1 \end{align}$$
We have shown $ \gcd (a, b) = \gcd (a + b, [a, b]) $.