donnent
![{\displaystyle {\frac {p^{2}}{q^{2}}}=a+{\frac {1}{q^{2}}},\qquad {\frac {p'^{2}}{q'^{2}}}=a+{\frac {1}{q'^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84024dd49c499e4129a41f7805e1edb0c379f414)
donc
![{\displaystyle {\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}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91a464bdad261e9dadfd3b4efc65ab4b373813b3)
et tirant la racine carrée,
![{\displaystyle {\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}}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/372c8c32c31fe9a4d2e6b6041f8d0b4f55536b6e)
d’où l’on voit que
![{\displaystyle {\frac {p}{q}}{\frac {p'}{q'}}>a\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5aa5e788b42f402e4517a556665fa9177fb4070a)
de sorte que
sera toujours une quantité positive : par conséquent
sera aussi un nombre positif.
2o Soit
on aura
![{\displaystyle {\frac {p-1}{q}}{\frac {p'}{q'}}-a={\frac {u}{qq'}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/414cd6008bdca594db91fe5a5ac17f44286e9bc5)
or
![{\displaystyle {\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}}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/536d2691c9f58df010e8fabd8c0ff82af77224b9)
donc
![{\displaystyle {\frac {p-1}{q}}{\frac {p'}{q'}}=a{\frac {\sqrt {a+{\cfrac {1}{q'^{2}}}}}{{\sqrt {a+{\cfrac {1}{q^{2}}}}}+{\cfrac {1}{q}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e54e09424cca75057bb0e03d95add54f24b5f90)
mais
donc
et à plus forte raison
donc
![{\displaystyle {\frac {p-1}{q}}{\frac {p'}{q'}}<a\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22645b19327f237e9052c415070cd5e85862edb3)
donc
sera nécessairement une quantité négative ; par con-