Integer division that rounds up: Difference between revisions

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'''Part 1'''. If <math>y</math> divides <math>x</math> we have <math>x=ay</math> for some <math>a\in\mathbb N_+</math>. Thus we have
'''Part 1'''. If <math>y</math> divides <math>x</math> we have <math>x=ay</math> for some <math>a\in\mathbb N_+</math>. Thus we have
<math>
<math>
\begin{align}
\begin{align}
\left \lceil  \frac{x+y-1}{y} \right \rceil  
\left \lceil  \frac{x+y-1}{y} \right \rceil  
=
&=
\left \lceil  \frac{x}{y} + \frac{y-1}y \right \rceil  \\
\left \lceil  \frac{x}{y} + \frac{y-1}y \right \rceil  \\
=
&=
\left \lceil  \frac{ay}{y} + \frac{y-1}y \right \rceil  \\
\left \lceil  \frac{ay}{y} + \frac{y-1}y \right \rceil  \\
=
&=
\frac{ay}{y} \\
\frac{ay}{y} \\
=
&=
\frac{x}{y}
\frac{x}{y}
\end{align}
\end{align}
</math>
</math>
because <math>0 \leq \frac{y-1}y < 1</math>. This is ok.
because <math>0 \leq \frac{y-1}y < 1</math>. This is ok.

Revision as of 11:08, 8 July 2024

Introduction

Usual integer division rounds down: for . To round up (if overflow is not an issue), you can use following algorithm with the usual roundig down division:

Proof

Proof is in two parts; 1st if divides , and if not. Note that usual integer division rounds down.

Part 1. If divides we have for some . Thus we have

because . This is ok.