Integer division that rounds up: Difference between revisions
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== Proof == | == Proof == | ||
Proof is in two parts; 1st if <math>y</math> divides <math>x</math>, and if not. | Proof is in two parts; 1st if <math>y</math> divides <math>x</math>, and if not. Note that usual integer division rounds down. | ||
'''Part 1'''. If <math>y</math> divides <math>x</math> we have <math>x=ay</math> for some <math>a\in\mathbb N_+</math>. | '''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> | |||
\begin{align} | |||
F=ma | |||
\end{align} | |||
</math> |
Revision as of 11:04, 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