Suppose $a > b$, let $a - b = \varepsilon > 0$. Take $m > \frac{1}{\varepsilon} + 1$ and we will have \[ \lfloor bm + b + \varepsilon \rfloor + \lfloor (m - 1)\varepsilon \rfloor \le \lfloor bm + b + \varepsilon m \rfloor \le \lfloor bm + b + \varepsilon \rfloor \]Then, we have $\lfloor (m - 1)\varepsilon \rfloor \le 0$, a contradiction. The other case is similar.

$ \lfloor an + b \rfloor \ge \lfloor a + bn \rfloor $ $an+ b + 1 \ge a + bn -1$ $an-a+b-bn \ge -2$ $a(n-1) + b(1-n) \ge -2$ $(n-1)(a-b) \ge -2$ So we have $(1-n)(a-b) \leq 2$. Similarly, we can obtain $(1-m)(b-a) \leq 2$. Both of these implies $a=b$.

I stumped on this problem

$[an+b] \geq [a+bn]$ Let's assume $a>b$ 1.$[an+b] \le an+b+1$ 2.$[a+bn] \geq a+bn-1$ Then, $an+b+1 \geq a+bn-1$ $\implies{(n-1)(a-b) \geq -2}$ Same as for $[a+bm] \geq [am+b]$ $[a+bm] \le a+bm+1$ $[am+b] \geq am+b-1$ Then, $(m-1)(b-a) \geq -2$ Okay these cases give us $b-a \le 0$ which is contradiction. Then take $b>a$ contradiction again. And we are done hence $a=b$ $\blacksquare$