بين الخاصيات التالية :
$\displaystyle{\displaylines{(\forall x \in \mathbb{R}) \, (\forall n \in \mathbb{Z}) \, : \quad \text{E}(x + n) = n + \text{E}(x)}}$
$\displaystyle{\displaylines{(\forall x,y \in \mathbb{R}) \, : \quad \text{E}(x) + \text{E}(y) \leq \text{E}(x + y) \leq \text{E}(x) + \text{E}(y) + 1}}$
$\displaystyle{\displaylines{(\forall x \in \mathbb{R}) \, (\forall n \in \mathbb{N}^{*}) \, : \quad 0 \leq \text{E}(n x) - n \text{E}(x) \leq n - 1}}$
راجع الدرس :
دالة الجزء الصحيح
ليكن
$\displaystyle{\displaylines{n \in \mathbb{Z}}}$ و
$\displaystyle{\displaylines{x \in \mathbb{R}}}$ :
نعلم أن
$\displaystyle{\displaylines{\text{E}(x) \leq x < \text{E}(x) + 1}}$إذن
$\displaystyle{\displaylines{n + \text{E}(x) \leq x + n < n + \text{E}(x) + 1}}$نضع
$\displaystyle{\displaylines{N = n + \text{E}(x)}}$لدينا
$\displaystyle{\displaylines{N \in \mathbb{Z}}}$ بحيث
$\displaystyle{\displaylines{N \leq x + n < N + 1}}$إذن
$\displaystyle{\displaylines{\text{E}(x+n) = N = n + \text{E}(x)}}$خلاصة :
$\displaystyle{\displaylines{(\forall x \in \mathbb{R}) \, (\forall n \in \mathbb{Z}) \, : \quad \text{E}(x + n) = n + \text{E}(x)}}$
ليكن
$\displaystyle{\displaylines{x,y \in \mathbb{R}}}$, لدينا :
$\displaystyle{\displaylines{\left\{ \begin{array}{cl}x - 1 < \text{E}(x) \leq x \\y - 1 < \text{E}(y) \leq y\end{array} \right. \implies x + y - 2 < \text{E}(x) + \text{E}(y) \leq x + y}}$نضع
$\displaystyle{\displaylines{N = \text{E}(x) + \text{E}(y)}}$, لدينا
$\displaystyle{\displaylines{N \leq x + y}}$إذن
$\displaystyle{\displaylines{N \leq \text{E}(x + y)}}$خلاصة
$\displaystyle{\displaylines{\text{E}(x) + \text{E}(y) \leq \text{E}(x + y)}}$ولدينا :
$\displaystyle{\displaylines{x + y - 2 < N}}$إذن
$\displaystyle{\displaylines{\text{E}(x + y - 2) < N}}$وكما بينا سابقا :
$\displaystyle{\displaylines{\text{E}(x + y - 2) = \text{E}(x + y) - 2}}$وبالتالي
$\displaystyle{\displaylines{\text{E}(x + y) < N + 2}}$إذن
$\displaystyle{\displaylines{\text{E}(x + y) \leq N + 1}}$ (لأن
$\displaystyle{\displaylines{N}}$ و
$\displaystyle{\displaylines{\text{E}(x + y)}}$ أعداد صحيحة)
خلاصة
$\displaystyle{\displaylines{\text{E}(x + y) \leq \text{E}(x) + \text{E}(y)}}$$\displaystyle{\displaylines{(\forall x,y \in \mathbb{R}) \, : \quad \text{E}(x) + \text{E}(y) \leq \text{E}(x + y) \leq \text{E}(x) + \text{E}(y) + 1}}$
ليكن
$\displaystyle{\displaylines{n \in \mathbb{N}^{*}}}$ و
$\displaystyle{\displaylines{x \in \mathbb{R}}}$ :
نعلم ان
$\displaystyle{\displaylines{n x - 1 < \text{E}(nx) \leq n x}}$ونعلم :
$\displaystyle{\displaylines{x - 1 < \text{E}(x) \leq x}}$إذن
$\displaystyle{\displaylines{-n x \leq -n \text{E}(x) < n - n x}}$إذن
$\displaystyle{\displaylines{ - 1 < \text{E}(n x) - n \text{E}(x) < n}}$وبالتالي
$\displaystyle{\displaylines{0 \leq \text{E}(n x) - n \text{E}(x) \leq n-1}}$خلاصة :
$\displaystyle{\displaylines{(\forall x \in \mathbb{R}) \, (\forall n \in \mathbb{N}^{*}) \, : \quad 0 \leq \text{E}(n x) - n \text{E}(x) \leq n - 1}}$
