Fractional Part Function

The floor function $\lfloor x \rfloor$ is defined to be the greatest integer less than or equal to the real number $x$. The fractional part function $\{ x \}$ is defined to be the difference between these two:

Let $x$ be a real number. Then the fractional part of $x$ is

$$\{x\}= x -\lfloor x \rfloor. \ _\square$$

This is the graph of the function $y=\{x\}.$

For nonnegative real numbers, the fractional part is just the "part of the number after the decimal," e.g.

$$\{3.64 \} = 3.64 - \lfloor 3.64 \rfloor = 3.64 - 3 = 0.64.$$

But for negative real numbers, this is no longer this case:

$$\{-3.64 \} = -3.64 - \lfloor -3.64 \rfloor = -3.64 - (-4) = 0.36.$$

Note that in both cases $\{ x \}$ is nonnegative.

The following are some examples of how fractional part functions work:

• $\{ 1 \} = 1 - 1 = 0.$
• $\left\{ \sqrt{2} \right\} = \sqrt{2}-1 = 0.4142\ldots.$
• $\{ \pi \} = \pi - 3 = 0.14159\ldots.$
• $\left\{ -\frac{17}5 \right\} = -\frac{17}5 - (-4) = \frac35.$

1. $0 \le \{ x \} < 1$, and $0 = \{ x \}$ if and only if $x$ is an integer.
2. $\{x \} + \{-x\} = \begin{cases} 0 & \text{if } x \text{ is an integer} \\ 1 & \text{otherwise.} \end{cases}$
3. If $a$ and $b$ are integers and $b > 0$, then $\big\{ \frac{a}{b} \big\} = \frac{r}{b}$, where $r$ is the remainder from dividing $a$ by $b$.

$\{ 9 \} + \{-9\} = 0+0 = 0$, but $\{9.01\} + \{-9.01\} = 0.01+0.99 = 1$. The fractional part is always nonnegative.

Since $47 = 13 \cdot 3 + 8$, it follows that $\left\{\frac{47}{13} \right\} = \frac{8}{13}$.

For problems involving the floor function and the fractional part function, it often helps (for ease of notation) to write $x = n+r ,$ where $n = \lfloor x \rfloor$ and $r = \{ x\}$. Then $n$ is an integer and $0 \le r < 1$.

Let $x$ be a positive real number such that

$$x^2+\{x\}^2 = 27.$$

Find $x$.

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Write $x = n+r$ as suggested. Now note that $x^2 \le x^2+\{x\}^2 < x^2+1$, so $26 < x^2 \le 27$. So $n = 5$. Then

$$\begin{aligned} (5+r)^2+r^2 &= 27 \\ 25+10r+2r^2 &=27 \\ 2r^2+10r-2 &= 0, \end{aligned}$$

so $r = \frac{-5+\sqrt{29}}2$ by the quadratic formula.

Therefore, $x = 5 + r = \frac{5+\sqrt{29}}2$. $_\square$

Find the smallest real number $m$ such that for all positive real numbers $x$,

$$\{ x \} + \left\{ \frac1{x} \right\} < m.$$

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If $x < 1$, we can write $x = \frac1y, y \ge 1,$ and $\{ x \} + \big\{ \frac1{x} \big\} = \{ y \} + \big\{ \frac1{y} \big\}.$ So we may assume $x \ge 1$. Then $\big\{ \frac1{x} \big\} = \frac1{x}$. Writing $x = n+r$, with $n$ a positive integer, the left side becomes

$$r + \frac1{n+r}.$$

For fixed $r$, this is maximized when the denominator is minimized, i.e. $n = 1$.

Now, $r + \frac1{1+r}$ is an increasing function for $0 \le r < 1$, its derivative being $1-\frac1{(1+r)^2}$. As $r \to 1^-$, the sum goes to $1+\frac1{1+1} = \frac{3}{2}$. So the answer is $\frac{3}{2}$. $_\square$

$($Exercise: If $x$ is allowed to be negative, the answer is $m=2.)$

There are many interesting integrals involving the fractional part function. A good way to evaluate definite integrals of this type is to break up the interval of integration into intervals on which the greatest integer function is constant; then the original integral is a sum of integrals which are easier to evaluate.

Find

$$\int_0^1 \left\{ \frac1{x} \right\} \, dx.$$

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On the interval $\left(\frac1{n+1},\frac1n \right]$, $\left\lfloor \frac1{x} \right\rfloor = n$. So the integral on that interval becomes

$$\begin{aligned} \int_{1/(n+1)}^{1/n} \left( \frac1{x}-n \right) \, dx &= \ln \frac1n-\ln \frac1{n+1} -n\left(\frac1n-\frac1{n+1}\right) \\ &= \ln(n+1)-\ln n - \frac1{n+1}. \end{aligned}$$

So the integral is

$$\sum_{n=1}^{\infty} \int_{1/(n+1)}^{1/n} \left\{ \frac1{x} \right\} \, dx = \sum_{n=1}^\infty \left(\ln(n+1)-\ln n - \frac1{n+1}\right)$$

and the $k^\text{th}$ partial sum telescopes to $\ln(k+1)-\left( \frac12 + \cdots + \frac1{k+1} \right)$. The limit is $1-\gamma$, where $\gamma$ is the famous Euler-Mascheroni constant. $_\square$