1

I am trying to prove: $$\lceil{\frac{\lceil{\frac{x}{a}}\rceil}{b}}\rceil=\lceil{\frac{x}{ab}}\rceil$$ where $x \in R$ and $a,b$ are integers greater than zero.

Clearly, when $a$ divides $x$ the result holds true. Now we assume $a$ doesn't divide x, i.e. $x=ka+u$ where $k$ is an integer but $u$ could be real with $u<k$.

Now working on L.H.S : $$\lceil{\frac{\lceil{\frac{ka+u}{a}}\rceil}{b}}\rceil$$

which gives : $$\lceil{\frac{\lceil{{k+\frac{u}{a}}}\rceil}{b}}\rceil$$ with $0<\frac{u}{a}<1$ and $k \in Z$.

Thus, it gives: $\lceil{\frac{{{k+1}}}{b}}\rceil$

Now considering R.H.S : $\lceil{\frac{x}{ab}}\rceil=\lceil{\frac{ka+u}{ab}}\rceil=\lceil\frac{k}{b}+\frac{u}{ab}\rceil=\lceil\frac{k}{b}+\frac{u/a}{b}\rceil$

Here, let $0<c=\frac{u}{a}<1$

Hence, R.H.S becomes $\lceil\frac{k}{b}+\frac{c}{b}\rceil$

Now I am comparing L.H.S and R.H.S -> $\lceil\frac{k+1}{b}\rceil:\lceil\frac{k+c}{b}\rceil$

Again if $b$ divides $k$ we are done, else take $k=tb+v$, hence L.H.S : R.H.S = $\lceil\frac{tb+v+1}{b}\rceil:\lceil\frac{tb+v+c}{b}\rceil=\lceil t+\frac{v+1}{b}\rceil:\lceil t+\frac{v+c}{b}\rceil=t+\lceil\frac{v+1}{b}\rceil:t+\lceil\frac{v+c}{b}\rceil$

Now I claim that $\lceil\frac{v+1}{b}\rceil=\lceil\frac{v+c}{b}\rceil=1$, because $v$ being an integer can't exceed $b$ so, maximum, it could be $b-1$ and substituting $v=b-1$ above yields the equality.

Is it correct ?

Or could there be a better way to prove this ?

User9523
  • 2,134

1 Answers1

1

I didn't read through your proof, but I would do it this way: $x$ can be expressed as

$$x = m \cdot ab + n \cdot a + r$$

where $m, n \in \mathbb{Z}$, $r \in \mathbb{R}$ with $0 \leqslant n < b$ and $0 \leqslant r < a$. There are 3 cases:

  1. $n = 0$ and $r = 0$: then both sides are clearly equal to $m$.

  2. $n > 0$ and $r = 0$:

$$\left\lceil \frac{ \lceil\frac{x}{a}\rceil}{b} \right\rceil = \left\lceil \frac{mb+n}{b} \right\rceil = m+1 = \left\lceil \frac{x}{ab} \right\rceil$$

  1. $r > 0$:

$$\left\lceil \frac{ \lceil\frac{x}{a}\rceil}{b} \right\rceil = \left\lceil \frac{mb+n+1}{b} \right\rceil = m+1 = \left\lceil \frac{x}{ab} \right\rceil$$

Adayah
  • 13,487