Bounding the Fractional Part

Idea · Intermediate

Algebra: Inequalities

[x]:\ \text{Integer part} \qquad\qquad \{x\}:\ \text{Fractional part}

We can write $x$ as:

x=k+r, \qquad k\in\mathbb Z,\quad r\in\mathbb R,\quad 0\le r<1.

The key idea is to use the bound

0\le r<1.

r\ge0 \qquad\text{and}\qquad r<1.

This gives two inequalities that restrict the possible values of $k$ . These inequalities are often quadratic inequalities.

Thus, the general strategy is

x=k+r \quad\longrightarrow\quad \text{express }r\text{ in terms of }k \quad\longrightarrow\quad 0\le r<1 \quad\longrightarrow\quad \text{solve the resulting inequalities}.

Find all real numbers $x$ such that

x^2-2[x]-2=0.

x=k+r,

where $k$ and $r$ are the integer and fractional parts, respectively.

x^2-2[x]-2=0

(k+r)^2-2k-2=0

k^2+2kr+r^2-2k-2=0

r^2+2kr+k^2-2k-2=0.

The roots are

\begin{aligned} r &= \frac{-2k\pm\sqrt{(2k)^2-4(1)(k^2-2k-2)}}{2} \\[4pt] &= \frac{-2k\pm\sqrt{8k+8}}{2} \\[4pt] &= -k\pm\sqrt{2k+2}. \end{aligned}

2k+2\ge0 \quad\Longrightarrow\quad k\ge-1.

Case 1: $k=-1$

r = -(-1)+\sqrt{2(-1)+2} = 1 \qquad \text{\(\times\)}

which is not possible because $r<1$ .

Case 2: $k\ge0$

0 \le r = -k+\sqrt{2k+2} < 1 \qquad\qquad\text{(1)}

The left-side inequality gives

0\le-k+\sqrt{2k+2}

k\le\sqrt{2k+2}

k^2\le2k+2

k^2-2k-2\le0.

By Quadratic Inequalities, the roots are

\begin{aligned} k &= \frac{2\pm\sqrt{2^2+4(1)(2)}}{2} \\[4pt] &= \frac{2\pm\sqrt{12}}{2} \\[4pt] &= 1\pm\sqrt3. \end{aligned}

k\in[1-\sqrt3,\ 1+\sqrt3].

$k$ should be a nonnegative integer, so

k=0,1,2 \qquad\qquad\text{(2)}

The right-side inequality of (1) gives

-k+\sqrt{2k+2}<1

\sqrt{2k+2}<k+1

2k+2<k^2+2k+1

0<k^2-1=(k-1)(k+1).

By Quadratic Inequalities, the roots are $-1$ and $1$ , so

k\in(-\infty,-1)\cup(1,+\infty) \qquad\qquad\text{(3)}

By (2) and (3),

k=2.

By (1),

r = -k+\sqrt{2k+2} = -2+\sqrt{2(2)+2} = -2+\sqrt6.

x = k+r = 2+(-2+\sqrt6) = \sqrt6.

\boxed{x=\sqrt6}.