數論的題目，高斯符號的題目，解聯立方程式
設 \(a\) 為任意數，\([a]\) 表高斯符號，\((a )\) 表 \(a-[a]\) ，求下列聯立方程的解：
\[\left\{ {\begin{array}{*{20}{c}}
{x + [y] + (z) = 1.5} \;.............. (1)\\
{y + [z] + (x) = 7.7} \;.............. (2)\\
{z + [x] + (y) = 2.6} \;.............. (3)\\
\end{array}} \right.\]
解答：
由三式相加，可得
\[ x + y+ z + [x] + [y] + [z] + ( x) + ( y) + ( z) = 11.8\]
\[\Rightarrow 2(x+y+z) = 11.8 \Rightarrow x+y+z=5.9\; ....... (4)\]
由 \((4)-(1),\, (4)-(2),\,(4)-(3)\)，得
\[\left\{\begin{array}{*{20}{c}}
y- [y] + z - (z ) = 4.4\\
z- [z] + x- (x ) = -1.8\\
x- [x] + y- (y ) = 3.3\\
\end{array}\right.\]
\[\Rightarrow \left\{\begin{array}{*{20}{c}}
(y) + [z] = 0.4+4\\
(z) + [x] = 0.2+\left(-2\right)\\
(x) + [y] = 0.3+3\\
\end{array}\right.\]
\[\Rightarrow \left\{\begin{array}{*{20}{cc}}
(y)=0.4,& [z] = 4\\
(z)=0.2,& [x] = \left(-2\right)\\
(x)=0.3,& [y] = 3\\
\end{array}\right.\]
\[\Rightarrow \left\{\begin{array}{*{20}{c}}
x= (x) + [x] = -1.7 \\
y=(y) + [y] = 3.4\\
z=(z) + [z] = 4.2\\
\end{array}\right. .\]
