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Let $[\cdot]$ denote the integral part of a real number, that is the only integer $a$ satisfying

$a \leq x < a+1$.

Let $0<r<1$ be a rational number and $m \geq 0$ a positive integer.

How do you prove that the following equality?

$[rm]+[(1-r)m]=m-1$

This should be completely elementary but I don't manage!!

Zev Chonoles
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user84782
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2 Answers2

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The claimed formula is incorrect. If $r=1/2$ and $m=2$ then the two integer parts sum to $m$ rather than $m-1$.

Mikhail Katz
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  • Oh, that's very interesting. Which is the correct formula then? – user84782 Jul 02 '13 at 18:02
  • I am not sure what you are aiming at here, but at any rate of either of the numbers $rm$ and $(1-r)m$ is not an integer, then the sum will be $m-1$ as stated. Therefore one needs to add a condition to ensure that one of them is not an integer. For example, if you assume that $r$ is irrational instead of rational, that should do it. – Mikhail Katz Jul 02 '13 at 18:04
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This will not be true always. Let $r=\frac 12, m=4$, then $[rm]+[(1-r)m]=4\neq m-1$. But this gives a hint: if $rm$ is integral, we have $[rm]+[(1-r)m]=rm+m-rm=m$, while if $rm$ is not integral we have $rm-[rm]+\{rm\}$ and $[rm]+[(1-r)m]=[rm]+[m-[rm]-\{rm\}]=[rm]+m-[rm]-1$ So the formula is correct for $rm$ non-integral, and the $-1$ should be removed for $rm$ integral.

Ross Millikan
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