Prove by induction floor and ceiling

Question

I couldn't solve below question ... I was able to start the solution .. but I couldn't turn left side to be same as write side in the inductive step:

Answer 1

Hint: you only need to show

$$\left\lceil \frac{n+1}{m}\right\rceil - \left\lceil \frac nm \right\rceil = \left\lfloor \frac{n+m}{m} \right\rfloor - \left\lfloor \frac{n+m-1}{m}\right\rfloor$$

Now the LHS is either $0$ or $1$, usually $0$. So is the RHS.

When is LHS $1$? When is RHS $1$?

When $\frac nm \in \mathbb N$, both LHS and RHS are equal to 1. Otherwise they are both zeros. Therefore, $$\left\lceil \frac{n+1}{m}\right\rceil - \left\lceil \frac nm \right\rceil = \left\lfloor \frac{n+m}{m} \right\rfloor - \left\lfloor \frac{n+m-1}{m}\right\rfloor \tag1$$ $$\left\lceil \frac nm \right\rceil = \left\lfloor \frac{n+m-1}{m}\right\rfloor \text{ (from last step)} \tag 2$$ $(1)+(2)$, then we have $$ \left\lceil \frac{n+1}{m}\right\rceil = \left\lfloor \frac{n+m}{m} \right\rfloor $$