Home » Blog » There is exactly one integer between any real number x and x+1.

There is exactly one integer between any real number x and x+1.

by RoRi

Prove that if x \in \mathbb{R}, then there is exactly one n \in \mathbb{Z} such that x \leq n < x+1.

Proof. By I.3.12, Exercise #4 we know that for x \in \mathbb{R} there is exactly one n \in \mathbb{Z} such that

    \[ n \leq x < n+1. \]

If n = x, then we already done since x \leq n < x+1.
If n \neq x, then we have n < x and so n+1 < x+1. But we already have x < n+1; hence,

    \[ x < n+1 < x+1 \quad \implies \quad x \leq n+1 < x+1. \]

So, n+1 is an integer with the desired properties. Uniqueness follows from the uniqueness of our choice of n. \qquad \blacksquare

3 comments

  1. James says:

    Hey, looking at your solution gave me an idea for an alternate solution (please excuse the formatting):
    From problem 4, for an arbitrary real number (-x), there is only one integer (-n) such that:
    (-n) <= (-x) n >= x and x > n – 1 –> x + 1 > n
    Thus, there is exactly one n such that x<= n < x + 1.

    • James says:

      For some reason, the comment didn’t come out right:
      The first line was supposed to be:
      (-n) <= (-x) and (-x) < (-n) + 1.

      • James says:

        And then, multiply both by (-1), and get
        n>=x and x>n-1, the second of which is x+1>n.
        Combined, they are x<=n<x+1.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):