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Let $m \in \mathbb{R}$ and $n \in \mathbb{N}$. Prove the following facts:

$\lfloor \hspace2mm\rfloor$ Means Floor function: And $\lfloor \lfloor m\rfloor/n\rfloor$ mean the floor of $m$ and then the floor of the floor $m$ divides $n$.

(i) $\lfloor m+n \rfloor = \lfloor m \rfloor+n$

(ii) $\lfloor \lfloor m \rfloor / n \rfloor = \lfloor m/n \rfloor$

(iii) $\lfloor m \rfloor + \lfloor m + 1/n \rfloor + \lfloor m + 2/n \rfloor + \ldots + \lfloor m + (n-1)/n \rfloor = \lfloor m n \rfloor$ (induction?)

Lemma: for $x = nθ$, where $θ = m−\lfloor m \rfloor$.

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3 Answers 3

up vote 3 down vote accepted

For all the problems, the main thing you need is to write $$x = \lfloor x \rfloor + \{x\}$$ where $\lfloor x \rfloor \in \mathbb{Z}$ and $\{x\} \in [0,1)$.

For instance, for the first one

  1. $\lfloor m + n \rfloor = \lfloor \lfloor m \rfloor +\{m\} + n \rfloor = \lfloor \lfloor m \rfloor + n +\{m\} \rfloor = \lfloor m \rfloor + n$.
  2. If $$\lfloor \lfloor m \rfloor/n \rfloor = a \in \mathbb{Z}$$ then we have that $$\lfloor m \rfloor/n = a + b$$ where $b \in \left \{0,\dfrac1n, \dfrac2n, \ldots, \dfrac{(n-1)}n \right\}$. Hence, $$\lfloor m \rfloor = an + bn \implies m = an + bn + c$$ where $c \in [0,1)$. Hence, $$m/n = a + b + \dfrac{c}n$$ Note that $b + \dfrac{c}n \in \left\{ \dfrac{c}n,\dfrac{c+1}n, \ldots, \dfrac{c+(n-1)}n\right\} \subseteq [0,1)$. Hence, $$\lfloor m/n \rfloor = a = \lfloor \lfloor m \rfloor/n \rfloor$$
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The first two are straightforward using the universal property of the floor function, viz. $$\rm k\le \lfloor r \rfloor \iff k\le r,\ \ \ for\ \ \ k\in \mathbb Z,\ r\in \mathbb R$$ Therefore for $\rm\:0 < n\in \mathbb Z,\ r\in \mathbb R,\ $ $$\rm\begin{eqnarray} &\rm k &\le&\:\rm\ \lfloor \lfloor r \rfloor / n\rfloor \\ \iff& \rm k &\le&\ \ \rm \lfloor r \rfloor / n \\ \iff& \rm nk &\le&\ \ \rm \lfloor r \rfloor \\ \iff& \rm nk &\le&\ \ \rm r \\ \iff& \rm k &\le&\ \ \rm r/n \\ \iff& \rm k &\le&\ \ \rm \lfloor r/n \rfloor \\ \\ \Rightarrow\ \ \rm \lfloor \lfloor r\!\!&\rm \rfloor / n\rfloor\ &=&\rm\ \ \lfloor r/n\rfloor \end{eqnarray}$$

since, having equal predecessors, these integers are equal.

If you know a little category theory you can view this universal property of floor as a right adjoint to inclusion, e.g. see Arturo's answer here, or see most any textbook on category theory. But, of course, one need not know any category theory to understand the above proof. Indeed, I've had success explaining this (and similar universal-inspired proofs) to bright high-school students.

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The floor function (also known as the entier function) is defined as having its value the largest integer which does not exceed its argument. When applied to any positive argument it represents the integer part of the argument obtained by suppressing the fractional part.

From this definition it is clear that for all positive real numbers x, y [x + y] > [x] + [y] since [x] and [y] are the integer parts of x and y. It's also obvious that equality is obtained when x and y are integers since they then have zero fractional parts.

let {x} = x - [x] and {y} = y - [y], so they are the decimal parts of x and y.

Then x + y = [x] + [y] + {x} + {y}

Case 1: {x} + {y} < 1 then {x + y} = {[x] + [y] +{x} + {y}} = {{x} + {y}} = {x} + {y} (we can remove from inside the {}s anything to the left of the decimal point!)

so [x+y] = x + y - {x + y} = x + y - {x} - {y} = [x] + [y]

Case 2: {x} + {y} >= 1 (but clearly must be <2) then {x + y} = {[x] + [y] + {x} + {y}} = {{x} + {y}} = {x} + {y} - 1

so [x + y] = x + y - {x + y} = x + y - {x} - {y} + 1 = [x] + [y] + 1

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