Exercise 1, § 8 General Topology, IV If xxx and yyy are real, we have [x+y]=[x]+[y]+εwhere ε=0 or ε=1, [x + y] = [x] + [y] + \varepsilon \quad \text{where } \varepsilon = 0 \text{ or } \varepsilon = 1, [x+y]=[x]+[y]+εwhere ε=0 or ε=1, [x−y]=[x]−[y]−εwhere ε=0 or ε=1, [x - y] = [x] - [y] - \varepsilon \quad \text{where } \varepsilon = 0 \text{ or } \varepsilon = 1, [x−y]=[x]−[y]−εwhere ε=0 or ε=1, [x]+[x+y]+[y]⩽[2x]+[2y], [x] + [x + y] + [y] \leqslant [2x] + [2y], [x]+[x+y]+[y]⩽[2x]+[2y], [x]+[x+1n]+[x+2n]+⋯+[x+n−1n]=[nx], [x] + \left[ x + \frac{1}{n} \right] + \left[ x + \frac{2}{n} \right] + \cdots + \left[ x + \frac{n-1}{n} \right] = [nx], [x]+[x+n1]+[x+n2]+⋯+[x+nn−1]=[nx], [[nx]n]=[x] \left[ \frac{[nx]}{n} \right] = [x] [n[nx]]=[x] for any integer n>0n > 0n>0. Solution No solution has been written yet.