Page:Lagrange - Œuvres (1867) vol. 1.djvu/760

donnent

${\frac {p^{2}}{q^{2}}}=a+{\frac {1}{q^{2}}},\qquad {\frac {p'^{2}}{q'^{2}}}=a+{\frac {1}{q'^{2}}}\,;$

donc

${\frac {p^{2}}{q^{2}}}{\frac {p'^{2}}{q'^{2}}}=a^{2}+a\left({\frac {1}{q^{2}}}+{\frac {1}{q'^{2}}}\right)+{\frac {1}{q^{2}q'^{2}}},$

et tirant la racine carrée,

${\frac {p}{q}}{\frac {p'}{q'}}={\sqrt {a^{2}+a\left({\frac {1}{q^{2}}}+{\frac {1}{q'^{2}}}\right)+{\frac {1}{q^{2}q'^{2}}}}},$

d’où l’on voit que

${\frac {p}{q}}{\frac {p'}{q'}}>a\,;$

de sorte que ${\frac {p}{q}}{\frac {p'}{q'}}-a$ sera toujours une quantité positive : par conséquent $z$ sera aussi un nombre positif.

2^o Soit $pp'-aqq'=p'+u,$ on aura

${\frac {p-1}{q}}{\frac {p'}{q'}}-a={\frac {u}{qq'}}\,;$

or

${\frac {p-1}{q}}={\sqrt {a+{\frac {1}{q^{2}}}}}-{\frac {1}{q}}={\frac {a}{{\sqrt {a+{\cfrac {1}{q^{2}}}}}+{\cfrac {1}{q}}}}\quad {\text{et}}\quad {\frac {p'}{q'}}={\sqrt {a+{\frac {1}{q'^{2}}}}}\,;$

donc

${\frac {p-1}{q}}{\frac {p'}{q'}}=a{\frac {\sqrt {a+{\cfrac {1}{q'^{2}}}}}{{\sqrt {a+{\cfrac {1}{q^{2}}}}}+{\cfrac {1}{q}}}}\,;$

mais $q'>q,$ donc ${\sqrt {a+{\frac {1}{q'^{2}}}}}<{\sqrt {a+{\frac {1}{q^{2}}}}}$ et à plus forte raison $<{\sqrt {a+{\frac {1}{q^{2}}}}}+{\frac {1}{q}}\,;$ donc

${\frac {p-1}{q}}{\frac {p'}{q'}}<a\,;$

donc ${\frac {p-1}{q}}{\frac {p'}{q'}}-a$ sera nécessairement une quantité négative ; par con-