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Prove that any real number lies between exactly one pair of consecutive integers.

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


First, we prove a lemma.

Lemma: 1 is the smallest element of \mathbb{Z}^+.
Proof. Consider the set S = \{ x \in \mathbb{R} \mid x \geq 1 \}. Then, S is an inductive set since 1 \in S and for all x \in S, x+1 \in S. Thus, \mathbb{Z}^+ \subseteq S (since, by definition, \mathbb{Z}^+ is the set of elements common to every inductive set). But, for any n \in \mathbb{Z}, if n < 1, then n \notin S and hence n \notin \mathbb{Z}^+. Thus, 1 is the least element of \mathbb{Z}^+.

Now, for the theorem of the exercise.
Proof. First, we prove existence.
Let S = \{ n \in \mathbb{Z} \mid n \leq x \}. We know S is non-empty since by I.3.12, Exercise #2 we know there exist integers m and n such that m < x < n. This m must be in our set S. We also know S is bounded above since x is an upper bound by our definition of S. Therefore, by the least upper bound axiom (Axiom 10, p. 25), we know \sup S exists.
Then, by Theorem I.32, we know that for any positive real number h, there is some n \in S such that

    \[ n > \sup S - h. \]

Letting h = 1, we have

    \[ n > \sup S - 1 \implies \sup S < n+1. \]

Therefore, since the supremum of S is an upper bound, we know n+1 \notin S; therefore, x < n+1 by definition of S. But we already know n \in S, and so by definition of S we must have n \leq x. Therefore, we have found an integer n such that

    \[ n \leq x < n+1. \]

Now, for uniqueness. Suppose there are integers n, n' \in \mathbb{Z} with this property. Then we have

    \begin{alignat*}{3}   &n &\leq x &< n+1  \qquad &\text{and} \qquad &n' &\leq x &< n'+1.\\   \implies& -n &\geq -x &> -n-1 \qquad &\text{and} \qquad &n'+1 &> x &\geq n. \end{alignat*}\end{align*}

Then, adding the terms of these inequalities we have,

    \[ n' + 1 -n > 0 > n'-n-1 \quad \implies \quad 1 > n'-n > -1. \]

Without loss of generality, assume n' \geq n. Then this inequality implies, 0 \leq n' - n < 1. But since 1 is the smallest element of \mathbb{Z}^+ (from the lemma above) and \mathbb{Z} is closed under subtraction, we must have n'-n = 0, i.e., n = n'. Thus, any such n is unique. \qquad \blacksquare

4 comments

  1. Pankaj Bhanu says:

    The statement `Then, by Theorem I.32, we know that for any positive real number h, there is some n in S such that n > sup S – h` is incorrect since S only contains integers and is not dense. consider the following example. take x = 4.7 and h = 0.1. sup S – h = 4.6. No n bigger than 4.6 exists

    • Daniel Nagase says:

      I think it’s also unnecessary, since if n' is any other integer less than x, then n' \in S, so n' \leq \mathrm{sup}S.

      It’s also possible to eliminate the appeal to the lemma: suppose m is another integer such that m \leq x  0, then m+1 \leq m+y = n \leq x, so m+1 \leq x, contradicting the hypothesis. Therefore, y = 0 and m = n, as required.

      • Daniel Nagase says:

        Hmm, the text of my comment got a little truncated. Here’s what I wanted to say:

        Suppose m is another integer such that m \leq x < m+1. Then m \in S, so m \leq n (since n is the supremum). Hence, 0 \leq n-m; set y = n-m (such a y exists by Theorem I.2). Then 0 \leq y. If 0 < y, then m+1 \leq m+y = n \leq x, so m+1 \leq x, contradicting the hypothesis. So 0=y and n=m.

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