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Prove some properties of the greatest integer function

For any x \in \mathbb{R} we denote the greatest integer less than or equal to x by [x]. Prove the following properties of the function [x]:

  1. [x+n] = [x] + n for any integer n.
  2. [-x] = \displaystyle{\begin{cases}  -[x] & \text{if } x \text{ is an integer,} \\ -[x]-1 & \text{otherwise.} \end{cases}}
  3. [x+y] = [x]+[y] or [x]+[y]+1.
  4. [2x] = [x] + \left[ x + \frac{1}{2} \right].
  5. [3x] = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right].

  1. Proof. Let [x+n] = m for some integer m. Then,

        \begin{align*}  m \leq x+n < m+1 &&\implies  && m-n &\leq x < m-n+1 \\ &&\implies && [x] &= m-n \\ &&\implies && [x] + n &=  m. \end{align*}

    But, we defined m=[x+n]. Thus, [x+n] = [x] + n for any n \in \mathbb{Z}. \qquad \blacksquare

  2. Proof. If x \in \mathbb{Z}, then x = n for some n \in \mathbb{Z}. Hence, [x]=n and

        \[ -x=-n \quad \implies \quad [-x]=-n \quad \implies \quad [-x]=-[x]. \]

    On the other hand, if x \notin \mathbb{Z}, then let [x] = n. This gives us,

        \begin{align*}  n \leq x < n+1 &&\implies && -n-1 &< -x < n &(n \neq x \text{ since } x \notin \mathbb{Z}) \\ &&\implies && [-x] &= -n-1 = -[x]-1. \qquad \blacksquare \end{align*}

  3. Proof. Let [x] = m and [y] = n, then we have

        \[ m \leq x < m+1 \qquad \text{and} \qquad n \leq y < n+1. \]

    So, adding, we obtain,

        \[ m+n \leq x+y < m+n+2. \]

    Thus,

        \[ [x+y] = m+n = [x] + [y] \qquad \text{or} \qquad [x+y] = m+n+1 = [x] + [y] + 1. \qquad \blacksquare\]

  4. Proof. By part (c) we have,

        \[ [2x] = [x+x] = [x] + [x] \qquad \text{or} \qquad [x] + [x] + 1. \]

    If [2x] = [x] + [x], then let [x] = n,

        \begin{align*}  && [2x] &= 2n \\ \implies && 2n &\leq 2x < 2n+1 \\ \implies && n &\leq x < n + \frac{1}{2} \\ \implies && n &\leq x + \frac{1}{2} < n+1 \\ \implies && \left[ x + \frac{1}{2} \right] &=n . \end{align*}

    Thus, [2x] = 2n = n+n = [x] + \left[ x + \frac{1}{2} \right].
    On the other hand, if [2x] = [x] + [x] + 1, then let [x] = n, and

        \begin{align*}   && [2x] &= 2n+1 \\  \implies && 2n+1 &\leq 2x < 2n+2 \\  \implies && n+\frac{1}{2} &\leq x < n+1 \\  \implies && n +1 &\leq x + \frac{1}{2} <  n+2 \\  \implies && \left[ x + \frac{1}{2} \right] &= n+1. \end{align*}

    Thus, [2x] = n+n+1 = [x] + \left[ x + \frac{1}{2} \right]. \qquad \blacksquare

  5. Proof. By part (c) we have

        \[ [3x] = [x+x+x] = [x+x] + [x] \qquad \text{or} \qquad [x+x]+[x] + 1. \]

    And,

        \[ [x+x] = [x] + [x] \qquad \text{or} \qquad [x]+[x]+1. \]

    So, putting these together we have,

        \[ [3x] = [x]+[x]+[x], \qquad \text{or} \qquad [x]+[x]+[x]+1, \qquad \text{or} \qquad [x]+[x]+[x]+2. \]

    If [3x] = [x] + [x] + [x], then, let [x] = n, so

        \begin{align*} && 3n &\leq 3x < 3n+1 \\ \implies && n &\leq x < n + \frac{1}{3} \\ \implies && n &\leq x + \frac{1}{3} < n+1 & \text{and}&& n &\leq x+\frac{2}{3} < n+1 \\ \implies && [x] &= \left[ x+ \frac{1}{3} \right] = \left[ x+ \frac{2}{3} \right] = n.  \end{align*}

    Thus, [3x] = 3n = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right].
    Next, if [3x] = [x] + [x] + [x] + 1, then let [x] = n, giving us,

        \begin{align*}  && 3n+1 &\leq 3x < 3n+1 \\ \implies && n + \frac{1}{3} &\leq x < n + \frac{2}{3} \\ \implies && n + \frac{1}{3} & \leq x + \frac{1}{3} < n+1 &\implies \quad \left[ x + \frac{1}{3} \right] &=n \\ \text{and}, \ \implies && n + 1 &\leq x + \frac{2}{3} < n+2 &\implies \quad \left[ x + \frac{2}{3} \right] &= n+1. \end{align*}

    Thus,

        \[ [3x] = 3n+1  \quad \implies \quad [3x] = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right]. \]

    Finally, if [3x] = [x] + [x] + [x] + 2, then let [x] = n, and we have

        \begin{align*} && 3n+2 &\leq 3x < 3n+3 \\ \implies && n+\frac{2}{3} &\leq x < n+1 \\ \implies && n+1 &\leq x+\frac{1}{3} < n+2 & \text{and} && n+1 \leq x+\frac{2}{3} < n+2.  \end{align*}

    So,

        \[ [3x] = 3n+2 = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right]. \qquad \blacksquare \]

7 comments

  1. Anonymous says:

    Hello,

    I think proof (b) is wrong, when you multiplied by (-1) you did not multiply n. Here is they way i proved it:

    [x]<x multiplying by -1

    -[x]>-x>-[x]-1 Now we proved that -[x]-1 is less than -x.

    What remains to be proved is that (-[x] – 1) is both an integer and the maximum integer < -x

    The first one is quite obvious and the second one can be proved by absurd

    BTW, thks for your blog, is helping a lot with my calculus!

  2. Sebastian says:

    On the b proof, on the line that says -n-1<-x<n shouldn't that n be negative since im assuming you multiplied all by -1?

    And can i ask why you pick -n-1 and equate it with the [-x] i dont get that step.

    Thanks in advance

    • Sebastian says:

      Also on c how did you get the m+n+1 from the m+n+2, can you just take that like that?
      I dont think you can because thats a segment on a line and you cant know that x+y is not on the m+n+1 to m+n+2 segment.

      Maybe thats a dumb question but i thought i should ask anyway XD

      • PRADEEP SINGH says:

        @Sebastian: Your query can be dealt within two cases:
        CASE:1 When x+y is in between m+n and m+n+1,then [x+y]=m+n=[x]+[y].
        CASE:2 When x+y is in between m+n+1 and m+n+2,then [x+y]=m+n+1=[x]+[y]+1

  3. Anonymous says:

    Just to point out a typo. In section \emph{e}, right after the line “If [3x] = [x] + [x] + [x] …” where you define the first inequality, the right side of the inequality says 3x + 1 instead of 3n+1, as well as the next line after that one. And after that everything else is fine.

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