If $\frac{m}{n}$ and $\frac{m'}{n'}$ are neighbours in the Farey sequence, how can I prove that $|mn'-nm'|=1$?
I know a proof by vectors, by I am trying to prove this by induction or other high school method, but I couldn't prove this.
Does anyone know another manner to prove this without vectors?
Thanks for antetion.
You can do it with nothing more than induction and solving a system of two equations in two unknowns.
For the induction step suppose that $\frac{a}b<\frac{c}d$ are neighbors in $F_n$ and that $bc-ad=1$. If $\frac{a}b$ and $\frac{c}d$ are neighbors in $F_{n+1}$, there is nothing to prove. Otherwise there is some $\frac{k}\ell$ between $\frac{a}b$ and $\frac{c}d$ in $F_{n+1}$. Suppose that we can show that $\frac{k}\ell=\frac{a+c}{b+d}$; then
$$bk-a\ell=b(a+c)-a(b+d)=bc-ad=1$$
and
$$\ell c-kd=(b+d)c-(a+c)d=bc-ad=1\;,$$
as desired.
To show that $\frac{k}\ell=\frac{a+c}{b+d}$, note first that
$$\frac{a+c}{b+d}-\frac{a}b=\frac{bc-ad}{b(b+d)}>0$$
and
$$\frac{c}d-\frac{a+c}{b+d}=\frac{bc-ad}{d(b+d)}>0\;,$$
so $\frac{a}b<\frac{a+c}{b+d}<\frac{c}d$.
In any case we can solve the system
$$\begin{align*} &ax+cy=k\\ &bx+dy=\ell\;: \end{align*}$$
multiplying the first equation by $d$ and the second by $c$ and subtracting the first from the second yields the equation
$$x=(bc-ad)x=c\ell-dk>0\;,$$
and similarly we find that $y=bk-a\ell>0$. (The inequalities follow from $\frac{a}b<\frac{k}\ell<\frac{c}d$.)
Thus, there are positive integers $x$ and $y$ such that $\frac{k}\ell=\frac{ax+cy}{bx+dy}$, which implies that $\ell=bx+dy\ge b+d$. In particular, if there is any fraction between $\frac{a}b$ and $\frac{c}d$ with a denominator less than or equal to $n+1$, it is their mediant $\frac{a+c}{b+d}$.