# Is this proof that $\lfloor x \rfloor \geq n \left\lfloor \frac{x}{n} \right\rfloor$ correct?

In this text the fractional part of a real $x$ shall be denoted $\{x\}$, such that $x = \lfloor x \rfloor + \{x\}$.

Theorem:

$$\forall x \in \mathbb{R}_{\geq 0} \forall n \in \mathbb{N}_{\geq 1} : \left\lfloor x \right\rfloor \geq n \left\lfloor \frac{x}{n} \right\rfloor$$

Proof:

Let $n$ be a positive integer. Let $x$ be a nonnegative real number.

$$x = n \frac{x}{n} = n \left( \left\lfloor \frac{x}{n} \right\rfloor + \left\{\frac{x}{n}\right\} \right)$$

Keeping in mind that the floor function is increasing:

$$\lfloor x \rfloor = \left\lfloor n \left\lfloor \frac{x}{n} \right\rfloor + n \left\{\frac{x}{n}\right\} \right\rfloor \geq \left\lfloor n \left\lfloor \frac{x}{n} \right\rfloor \right\rfloor = n \left\lfloor \frac{x}{n} \right\rfloor$$

$$\square$$

Having just written it down it seems so clear (and unexpectedly short), but is this proof correct? What could be improved about it?

• Yes, it is correct. – Zarrax Sep 9 '14 at 17:19
• It is ok. Why do you say it is unexpectedly short? – Crostul Sep 9 '14 at 17:34
• @Crostul I suppose it has something to do with the theorem holding true only for integer $n$, a constraint I did not notice earlier. I am very new to this. – Homeward Sep 9 '14 at 17:42

A shorter proof would note that we always have $y \ge \lfloor y \rfloor$, hence ${x \over n} \ge \lfloor {x \over n} \rfloor$. Multiplying by $n$ gives $x \ge n\lfloor {x \over n} \rfloor$. Taking the floor of both sides gives the desired result (noting that the floor of an integer is the integer).