MHB Prove Relations: $e,b,d\in \mathbb{Z},d\neq 0$

[evinda]

Hello! (Wave)Let $e,b \in \mathbb{Z}, d \neq 0$.
How could we prove the following? Could you maybe give me a hint?

• If $d>0$ then $e \text{ div } d = \lfloor \frac{e}{d} \rfloor$
  $$$$
• If $d<0$ then $e \text{ div } d = \lceil \frac{e}{d} \rceil $

Could we show the above, using the definitions? (Thinking)

$$\lfloor x \rfloor =max \{ m \in \mathbb{Z}: m \leq x \}$$

$$\lceil x \rceil=\min \{ l \in \mathbb{Z}: l \geq x\}$$

 

[Evgeny.Makarov]

What are the definitions of $e\text{ div }d$, $\lfloor x\rfloor$ and $\lceil x\rceil$?

 

[evinda]

What are the definitions of $e\text{ div }d$, $\lfloor x\rfloor$ and $\lceil x\rceil$?

If $d>0$ it is $e=k \cdot d+r, r<d$ where $k=e\text{ div }d=\lfloor \frac{e}{d} \rfloor$, right? (Thinking)

And if $d<0$, is it then like that? (Thinking)

$e=k \cdot d+r, r<d$ where $k=e\text{ div }d=\lceil \frac{e}{d} \rceil$

 

[Evgeny.Makarov]

If $d>0$ it is $e=k \cdot d+r, r<d$ where $k=e\text{ div }d=\lfloor \frac{e}{d} \rfloor$, right? (Thinking)

And if $d<0$, is it then like that? (Thinking)

$e=k \cdot d+r, r<d$ where $k=e\text{ div }d=\lceil \frac{e}{d} \rceil$

I don't know because $e\text{ div }d$ does not have a universally accepted definition. You have to go with the one used in your book or course. In contrast, $\lfloor x\rfloor$ is pretty unambiguous, but even then there are variations: for example, W|A rounds negative values up instead of down even though "integer part" is usually considered a synonym of "floor function".

Also, I assume that $\lfloor \frac{e}{d} \rfloor$ and $\lceil \frac{e}{d} \rceil$ are not part of the definition of $e\text{ div }d$. Then what is the difference between the two clauses for $d>0$ and $d<0$ in your post? And are there are no other restrictions on $r$, such as $0\le r$? In short, it would be nice if you wrote complete and precise definitions.

 

[johng1]

Here's some information on "a div b". I think in mathematics, method 2 or 3 below is used most often, while in CS, method 1 is usually the case.