Show that if $n$ and $k$ are positive integers, then $\lceil \frac{n}{k} \rceil = \lfloor \frac{n - 1}{k} \rfloor + 1$ This is answer in the back of the book but it doesn't make sense to me:


*

*There is some $b$ with $(b-1)k < n \leqslant bk$.

*Hence, $(b-1)k \leqslant n-1 < bk$.

*Divide by $k$ to obtain $b-1 < \frac{n}{k} \leqslant b$ and $b-1 \leqslant \frac{n - 1}{k} < b$.

*Hence, $\lceil \frac{n}{k} \rceil = b$ and $\lfloor \frac{n - 1}{k} \rfloor = b-1$.


Where did step one come from? What happened in step two? How did step three turn into step 4?
 A: 1) $n$ is an integer, so you divide the positive real line into disjoint intervals
$$
 (0, k], (k, 2k], (2k, 3k], \cdots,
 $$
then $n$ must fall into one of them. In fact, this shows the existence of $\lceil \frac{n}{k} \rceil$.
2) it meant to say that
$$
 (b - 1)k \leq n - 1 < bk,
 $$
since all of them are integers.
3) Now you divide all these inequalities by $k$.
4) That is the definition of ceil and floor.
$$
\lfloor a \rfloor = \text{the greatest integer that is smaller than }a
$$
$$
\lceil a \rceil = \text{the least integer that is greater than }a
$$

Since you seem to be really confused about this answer from the book, I will write a little bit more about it.
Here is how you should have worked out this proof: 
1) Suppose $\lceil \frac{n}{k} \rceil = b$ some integer, then it suffices to show that $\lfloor \frac{n - 1}{k} \rfloor = b - 1$;
2) What conditions do I get from my assumption $\lceil \frac{n}{k} \rceil = b$?
3) Will conditions I get from 2) imply that $\lfloor \frac{n - 1}{k} \rfloor = b - 1$?
