Calculus

Set Theory

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Floor and Ceiling Functions

Definitions

Let \(x\) be a real number. The floor function of \(x,\) denoted by \(\lfloor {x} \rfloor\) or \(\text{floor}\left( x \right),\) is defined to be the greatest integer that is less than or equal to \(x.\)

The ceiling function of \(x,\) denoted by \(\lceil {x} \rceil\) or \(\text{ceil}\left( x \right),\) is defined to be the least integer that is greater than or equal to \(x.\)

For example,

\[{\lfloor{\pi}\rfloor = 3,\;\;}{\lceil{\pi}\rceil = 4,\;\;}\kern0pt{\lfloor{5}\rfloor = 5,\;\;}\kern0pt{\lceil{5}\rceil = 5.}\]

\[{\lfloor{- e}\rfloor = -3,\;\;}{\lceil{-e}\rceil = -2,\;\;}\kern0pt{\lfloor{-1}\rfloor = -1,\;\;}\kern0pt{\lceil{-1}\rceil = -1.}\]

It follows from the definitions that the floor and ceiling functions have type \(\mathbb{R} \to \mathbb{Z}.\) Formally, for any \(x \in \mathbb{R},\) they can be defined as

\[\begin{array}{*{20}{l}} \text{floor:} & {\lfloor {x} \rfloor = \max \left\{ {n \in \mathbb{Z}:n \le x} \right\}}\\ \text{ceiling:} & {\lceil {x} \rceil = \max \left\{ {n \in \mathbb{Z}:n \le x} \right\}} \end{array}\]

Graphs of the Floor and Ceiling Functions

The floor and ceiling functions look like a staircase and have a jump discontinuity at each integer point.

The graph of the floor function.
Figure 1.
The graph of the ceiling function.
Figure 2.

Properties of the Floor and Ceiling Functions

There are many interesting and useful properties involving the floor and ceiling functions, some of which are listed below. The number \(n\) is assumed to be an integer.

  1. \(\left\lfloor x \right\rfloor = n \;\text{ iff }\; n \le x \lt n + 1\)
  2. \(\left\lceil x \right\rceil = n \;\text{ iff }\; n – 1 \lt x \le n\)
  3. \(\left\lfloor x \right\rfloor = n \;\text{ iff }\; x – 1 \lt n \le x\)
  4. \(\left\lceil x \right\rceil = n \;\text{ iff }\; x \le n \lt x + 1\)
  5. \(\left\lfloor { – x} \right\rfloor = – \left\lceil x \right\rceil \)
  6. \(\left\lceil { – x} \right\rceil = – \left\lfloor x \right\rfloor \)
  7. \(\left\lfloor x \right\rfloor + \left\lfloor { – x} \right\rfloor \) \(= \left\{ {\begin{array}{*{20}{l}} 0 &{\text{ if } x \in \mathbb{Z}}\\ { – 1} &{\text{ if } x \notin \mathbb{Z}} \end{array}} \right.\)
  8. \(\left\lceil x \right\rceil + \left\lceil { – x} \right\rceil \) \(= \left\{ {\begin{array}{*{20}{l}} 0 &{\text{ if } x \in \mathbb{Z}}\\ 1 &{\text{ if } x \notin \mathbb{Z}} \end{array}} \right.\)
  9. \(\left\lfloor {x + n} \right\rfloor = \left\lfloor x \right\rfloor + n\)
  10. \(\left\lceil {x + n} \right\rceil = \left\lceil x \right\rceil + n\)

Fractional Part Function

The fractional part of a number \(x \in \mathbb{R}\) is the difference between \(x\) and the floor of \(x:\)

\[\left\{ x \right\} = x – \left\lfloor x \right\rfloor .\]

For example,

\[{\left\{ 2 \right\} = 2 – \left\lfloor 2 \right\rfloor }={ 2 – 2 }={ 0}\]

\[{\left\{ {3.51} \right\} = 3.51 – \left\lfloor {3.51} \right\rfloor }={ 3.51 – 3 }={ 0.51}\]

\[{\left\{ {\frac{7}{3}} \right\} = \frac{7}{3} – \left\lfloor {\frac{7}{3}} \right\rfloor }={ \frac{7}{3} – 2 }={ \frac{1}{3}}\]

\[{\left\{ { – 5.98} \right\} }={ – 5.98 – \left\lfloor { – 5.98} \right\rfloor }={ – 5.98 – \left( { – 6} \right) }={ – 5.98 + 6 }={ 0.02}\]

The graph of the fractional part function looks like a sawtooth wave, with a period of \(1.\)

Graph of the fractional part function.
Figure 3.

The range of fractional part function is the half-open interval \(\left[ {0,1} \right).\)

Some other properties of the fractional part are

  1. \(\left\{ x \right\} = 0 \;\text{ iff }\; x \in \mathbb{Z}\)
  2. \(\left\{ {x + n} \right\} = \left\{ x \right\}, n \in \mathbb{Z}\)
  3. \(\left\{ x \right\} + \left\{ { – x} \right\} \) \(= \left\{ {\begin{array}{*{20}{l}} 0 &{\text{if } x \in \mathbb{Z}}\\ 1 &{\text{if } x \notin \mathbb{Z}} \end{array}} \right.\)

Solved Problems

Click or tap a problem to see the solution.

Example 1

Solve the equation \[{x^3} – \left\lfloor x \right\rfloor = 5.\]

Example 2

Solve the equation \[{\left\lfloor {x – \frac{1}{2}} \right\rfloor + \left\lfloor {x – \frac{5}{2}} \right\rfloor + \left\lfloor {x – \frac{9}{2}} \right\rfloor }={ 3.}\]

Example 3

Solve the equation \[{x^2} + 2{\left\{ x \right\}^2} = 12.\]

Example 4

Find all \(x\) satisfying the equation \[\left\{ {3x} \right\} = \frac{1}{3}.\]

Example 1.

Solve the equation \[{x^3} – \left\lfloor x \right\rfloor = 5.\]

Solution.

We represent the number \(x\) as

\[x = \left\lfloor x \right\rfloor + \left\{ x \right\},\]

where the fractional part \(\left\{ x \right\}\) satisfies the condition \(0 \le \left\{ x \right\} \lt 1.\)

Substitute \(\left\lfloor x \right\rfloor = x – \left\{ x \right\}\) in the equation to get

\[{{x^3} – \left\lfloor x \right\rfloor = 5,}\;\; \Rightarrow {{x^3} – \left( {x – \left\{ x \right\}} \right) = 5,}\;\; \Rightarrow {{x^3} – x + \left\{ x \right\} = 5,}\;\; \Rightarrow {\left\{ x \right\} = x – {x^3} + 5.}\]

It is clear that

\[{0 \le x – {x^3} + 5 \lt 1,}\;\; \Rightarrow {- 5 \le x – {x^3} \lt – 4,}\;\; \Rightarrow {4 \lt {x^3} – x \le 5.}\]

By taking the derivative of the function \(f\left( x \right) = {x^3} – x\) we find its local extrema:

\[{f^{\prime}\left( x \right) = \left( {{x^3} – x} \right)^{\prime} }={ 3{x^2} – 1 }={ 0,}\;\; \Rightarrow {{x_{1,2}} = \pm \frac{1}{{\sqrt 3 }}.}\]

The function \(f\left( x \right)\) is increasing on the intervals \(\left( { – \infty , – \large{\frac{1}{{\sqrt 3 }}}\normalsize} \right)\) and \(\left( {\large{\frac{1}{{\sqrt 3 }}}\normalsize,\infty } \right),\) and is decreasing on the interval \(\left( { – \large{\frac{1}{{\sqrt 3 }}}\normalsize,\large{\frac{1}{{\sqrt 3 }}}\normalsize} \right).\) The maximum value on the interval \(\left( { – \infty ,\large{\frac{1}{{\sqrt 3 }}}\normalsize} \right)\) is attained at \({{x_1} = – \frac{1}{{\sqrt 3 }}}:\)

\[{f\left( {{x_1}} \right) = f\left( { – \frac{1}{{\sqrt 3 }}} \right) }={ {\left( { – \frac{1}{{\sqrt 3 }}} \right)^3} – \left( { – \frac{1}{{\sqrt 3 }}} \right) }={ – \frac{1}{{3\sqrt 3 }} + \frac{1}{{\sqrt 3 }} }={ \frac{2}{{3\sqrt 3 }} \lt 4.}\]

Thus, the function takes the values from \(4\) to \(5\) on the last interval \(\left( {\large{\frac{1}{{\sqrt 3 }}}\normalsize,\infty } \right)\) where it is strictly increasing. Note that

\[{f\left( 1 \right) = {1^3} – 1 = 0,\;\;}\kern0pt{f\left( 2 \right) = {2^3} – 2 = 6.}\]

So, the solution of the equation must be in the range \(1 \lt x \lt 2,\) that is, \(\left\lfloor x \right\rfloor = 1.\) This yields:

\[{{x^3} – \left\lfloor x \right\rfloor = 5,}\;\; \Rightarrow {{x^3} – 1 = 5,}\;\; \Rightarrow {{x^3} = 6,}\;\; \Rightarrow {x = \sqrt[3]{6}.}\]

Example 2.

Solve the equation \[{\left\lfloor {x – \frac{1}{2}} \right\rfloor + \left\lfloor {x – \frac{5}{2}} \right\rfloor + \left\lfloor {x – \frac{9}{2}} \right\rfloor }={ 3.}\]

Solution.

Change the variable: \(x – \large{\frac{1}{2}}\normalsize = z.\) Then the equation becomes

\[\left\lfloor z \right\rfloor + \left\lfloor {z – 2} \right\rfloor + \left\lfloor {z – 4} \right\rfloor = 3.\]

Using the identity \(\left\lfloor {x + n} \right\rfloor = \left\lfloor x \right\rfloor + n,\) we get

\[{\left\lfloor z \right\rfloor + \left\lfloor z \right\rfloor – 2 + \left\lfloor z \right\rfloor – 4 = 3,}\;\; \Rightarrow {3\left\lfloor z \right\rfloor = 9,}\;\; \Rightarrow {\left\lfloor z \right\rfloor = 3.}\]

The last equation means that \(3 \le z \lt 4.\) Returning back to the variable \(x,\) we have

\[{3 \le x – \frac{1}{2} \lt 4,}\;\; \Rightarrow {3\frac{1}{2} \le x \lt 4\frac{1}{2},\;\text{ or }\;}\kern0pt{x \in \left[ {3.5,4.5} \right).}\]

Example 3.

Solve the equation \[{x^2} + 2{\left\{ x \right\}^2} = 12.\]

Solution.

Since, by definition, \(0 \le \left\{ x \right\} \lt 1,\) then

\[{0 \le {\left\{ x \right\}^2} \lt 1,}\;\; \Rightarrow {0 \le 2{\left\{ x \right\}^2} \lt 2.}\]

Hence,

\[10 \lt {x^2} \le 12.\]

Given that \(x^2\) is an increasing function for \(x \gt 0,\) we get

\[\sqrt {10} \lt x \le \sqrt {12} ,\]

that is, the floor of \(x\) is equal to \(3.\) Hence, we can represent the number \(x\) as

\[x = \left\lfloor x \right\rfloor + \left\{ x \right\} = 3 + \theta ,\]

where \(\theta\) is the fractional part of \(x.\)

Substitute this expression into the original equation and solve it for \(\theta.\)

\[{{\left( {3 + \theta } \right)^2} + 2{\theta ^2} = 12,}\;\; \Rightarrow {9 + 6\theta + {\theta ^2} + 2{\theta ^2} = 12,}\;\; \Rightarrow {3{\theta ^2} + 6\theta = 3,}\;\; \Rightarrow {{\theta ^2} + 2\theta = 1,}\;\; \Rightarrow {{\theta ^2} + 2\theta + 1 = 2,}\;\; \Rightarrow {{\left( {\theta + 1} \right)^2} = 2,}\;\; \Rightarrow {\theta + 1 = \sqrt 2 ,}\;\; \Rightarrow {\theta = \sqrt 2 – 1.}\]

Therefore,

\[{x = 3 + \theta }={ 3 + \sqrt 2 – 1 }={ 2 + \sqrt 2 .}\]

Example 4.

Find all \(x\) satisfying the equation \[\left\{ {3x} \right\} = \frac{1}{3}.\]

Solution.

Since \(3x = \left\lfloor {3x} \right\rfloor + \left\{ {3x} \right\},\) we rewrite the equation in the form

\[{\left\{ {3x} \right\} = \frac{1}{3},}\;\; \Rightarrow {3x – \left\lfloor {3x} \right\rfloor = \frac{1}{3},}\;\; \Rightarrow {\left\lfloor {3x} \right\rfloor = 3x – \frac{1}{3}.}\]

The \(RHS\) must be an integer. We denote it by \(n,\) so

\[{3x – \frac{1}{3} = n,}\;\; \Rightarrow {3x = n + \frac{1}{3},}\;\; \Rightarrow {x = \frac{n}{3} + \frac{1}{9},\; n \in \mathbb{Z}.}\]

Then we have

\[{\left\lfloor {3x} \right\rfloor = 3x – \frac{1}{3},}\;\; \Rightarrow {\left\lfloor {3\left( {\frac{n}{3} + \frac{1}{9}} \right)} \right\rfloor = n,}\;\; \Rightarrow {\left\lfloor {n + \frac{1}{3}} \right\rfloor = n.}\]

It follows from the last equation that

\[n \le n + \frac{1}{3} \lt n + 1.\]

This double inequality is valid for any \(n \in \mathbb{Z}.\) Hence, the solution set is given by

\[x = \left\{ {n \in \mathbb{Z} \,\bigg|\, \frac{n}{3} + \frac{1}{9}} \right\}.\]

We can check the answer for a few values of \(n:\)

  1. \(\kern-1pt n = -1:\) \[{x = – \frac{2}{9},}\;\; \Rightarrow {\left\{ {3x} \right\} = \left\{ {3 \cdot \left( { – \frac{2}{9}} \right)} \right\} }={ \left\{ { – \frac{2}{3}} \right\} }={ \left\{ { – 1 + \frac{1}{3}} \right\} }={ \frac{1}{3};}\]
  2. \(\kern-1pt n = 0:\) \[{x = \frac{1}{9},}\;\; \Rightarrow {\left\{ {3x} \right\} = \left\{ {\frac{1}{3}} \right\} = \frac{1}{3};}\]
  3. \(\kern-1pt n = 2:\) \[{x = \frac{7}{9},}\;\; \Rightarrow {\left\{ {3x} \right\} = \left\{ {3 \cdot \frac{7}{9}} \right\} }={ \left\{ {\frac{7}{3}} \right\} }={ \left\{ {2 + \frac{1}{3}} \right\} }={ \frac{1}{3}.}\]