Proving Floor and Ceiling of a Rational Number Suppose x,y $ \in \mathbb{Z}^+ $
Prove $\lceil x/y \rceil = \lfloor (x-1)/y \rfloor + 1$
I was considering using the definition of floor and ceiling to prove this. But this does not seem like a valid proof to me as I assume the right hand side is already equal to the left hand side.  
This is what I have so far:

Let n$ \in \mathbb{Z} $  
$\lceil x/y \rceil = n \iff n-1 < x/y \leq n $  
$\lfloor (x-1)/y \rfloor = n -1 \iff n-1 \leq (x-1)/y < n $
  Then $\lfloor (x-1)/y \rfloor +1 = n \iff n-1 \leq (x-1)/y < n + 1 $  
Then both sides are equivalent.  

Would a better way be to prove using a contradiction by assuming both sides are not equal?
 A: If $y$ divides $x$ then:


*

*$\lceil\frac{x}{y}\rceil=\frac{x}{y}$

*$\lfloor\frac{x-1}{y}\rfloor=\frac{x}{y}-1$


Hence $\lceil\frac{x}{y}\rceil=\lfloor\frac{x-1}{y}\rfloor+1$

If $y$ does not divide $x$ then:


*

*$\lceil\frac{x}{y}\rceil=\lfloor\frac{x}{y}\rfloor+1$

*$\lfloor\frac{x-1}{y}\rfloor=\lfloor\frac{x}{y}\rfloor$


Hence $\lceil\frac{x}{y}\rceil=\lfloor\frac{x-1}{y}\rfloor+1$
A: starting from line 2, the inequality can be rewritten as
\begin{equation*}
n-1+1/y
\leq
x/y
<
n+1/y
\end{equation*}
since x,y are integers, the left inequality is equivalent to
\begin{equation*}
n-1
<
x/y
\end{equation*}
and the right inequality is equivalent to
\begin{equation*}
x/y
\leq 
n
\end{equation*}
A: I might want to use a property of integer arithmetic $a \lt b \iff a+1 \le b$, as in:
$\qquad\;\;\lceil x/y \rceil = n $
$ \iff n-1 \lt x/y \leq n $
$ \iff (n-1)y \lt x \leq ny $
$\iff (n-1)y \le x-1 \lt ny$
$\iff n-1 \le (x-1)/y \lt n $
$\iff  \lfloor(x-1)/y\rfloor =n-1$  
$\iff  \lfloor(x-1)/y\rfloor +1=n$  
so $\lceil x/y \rceil = \lfloor(x-1)/y\rfloor$ + 1
