Pagina:Werkecarlf02gausrich.djvu/17

5.

Theorema. Sit ${\textstyle x}$ quantitas positiva non integra, inter cuius multipla ${\textstyle x}$, ${\textstyle 2x}$, ${\textstyle 3x\ldots }$ usque ad ${\textstyle nx}$ nullum fiat integer; ponatur ${\textstyle [nx]=h}$, unde facile concluditur, etiam inter multipla quantitatis reciprocae ${\textstyle {\frac {1}{x}}}$, ${\textstyle {\frac {2}{x}}}$, ${\textstyle {\frac {3}{x}}\ldots }$ usque ad ${\textstyle {\frac {h}{x}}}$ integrum non reperiri. Tum dico fore

$$\left.{\begin{array}{l}{\phantom {+}}[x]+[2x]+[3x]\dots +[nx]\\+\left[{\frac {1}{x}}\right]+\left[{\frac {2}{x}}\right]+\left[{\frac {3}{x}}\right]\ldots +\left[{\frac {h}{x}}\right]\end{array}}\right\}=nh$$

Dem. Seriei ${\textstyle [x]+[2x]+[3x]\ldots +[nx]}$, quam ponemus ${\textstyle =\Omega }$, membra prima usque ad ${\textstyle \left[{\frac {1}{x}}\right]^{\text{tum}}}$ inclus. manifesto omnia erunt ${\textstyle =0}$; sequentia usque ad ${\textstyle \left[{\frac {2}{x}}\right]^{\text{tum}}}$ cuncta ${\textstyle =1}$; sequentia usque ad ${\textstyle \left[{\frac {3}{x}}\right]^{\text{tum}}}$ cuncta ${\textstyle =2}$ et sic porro. Hinc fit

$$\left.{\begin{array}{rl}{\Omega }&=0\times \left[{\frac {1}{x}}\right]\\&+1\times \left\{\left[{\frac {2}{x}}\right]-\left[{\frac {1}{x}}\right]\right\}\\&+2\times \left\{\left[{\frac {3}{x}}\right]-\left[{\frac {2}{x}}\right]\right\}\\&+3\times \left\{\left[{\frac {4}{x}}\right]-\left[{\frac {3}{x}}\right]\right\}\\&\qquad {\text{etc.}}\\&+(h-1)\left\{\left[{\frac {h}{x}}\right]-\left[{\frac {h-1}{x}}\right]\right\}\\&+h\left\{n-\left[{\frac {h}{x}}\right]\right\}\end{array}}\right\}=hn-\left[{\frac {1}{x}}\right]-\left[{\frac {2}{x}}\right]-\left[{\frac {3}{x}}\right]\ldots -\left[{\frac {h}{x}}\right]$$

Q. E. D.

6.

Theorema. Designantibus ${\textstyle k}$, ${\textstyle p}$ numeros positivos impares inter se primos quoscunque, erit

$$\left.{\begin{array}{l}{\phantom {+}}{\left[{\frac {k}{p}}\right]+\left[{\frac {2k}{p}}\right]+\left[{\frac {3k}{p}}\right]\ldots +\left[{\frac {{\frac {1}{2}}(p-1)k}{p}}\right]}\\+{\left[{\frac {p}{k}}\right]+\left[{\frac {2p}{k}}\right]+\left[{\frac {3p}{k}}\right]\ldots +\left[{\frac {{\frac {1}{2}}(k-1)p}{k}}\right]}\end{array}}\right\}={\frac {1}{4}}(k-1)(p-1).$$

Demonstr. Supponendo, quod licet, ${\textstyle k<p}$, erit ${\textstyle {\frac {{\frac {1}{2}}(p-1)k}{p}}}$ minor quam ${\textstyle {\frac {1}{2}}k}$, sed maior quam ${\textstyle {\frac {1}{2}}(k-1)}$, adeoque ${\textstyle \left[{\frac {{\frac {1}{2}}(p-1)k}{p}}\right]={\frac {1}{2}}(k-1)}$. Hinc patet, theorema praesens ex praec. protinus sequi, statuendo illic ${\textstyle {\frac {k}{p}}=x}$, ${\textstyle {\frac {1}{2}}(p-1)=n}$, adeoque ${\textstyle {\frac {1}{2}}(k-1)=h}$.