Math 115. Proof of the Existence of the Greatest-Integer Function

This handout will prove the following lemma mentioned in class:

Lemma. For each real number x there is a unique integer n such that

n ≤ x < n + 1 .

Proof. We first show existence of n. This part will rely on the following axioms.

Archimedean Property of the Real Numbers. For each x ∈ ℝ there is an integer n with n > x.

Well-Ordering Property of the Natural Numbers. Every nonempty subset of ℕ has a smallest element.

Let x be any real number. By the Archimedean Property, there is an m ∈ ℤ with m > x. Let

S = {k ∈ ℤ : k ≥ m − x} .

Note that actually S ⊆ ℕ because m − x > 0 . Also, by the Archimedean Property applied to the real number m − x, S is nonempty.

Therefore we may apply the Well-Ordering Property to S. Let k be the smallest element of S. Since k ∈ S, m − k ≤ x.

Since k is the smallest element of S, we have k−1 ∉ S. Therefore k−1 < m−x, so m − k + 1 > x.

Thus, letting n = m − k, we have

n ≤ x < n + 1 ,

as was to be shown.

Now consider the question of uniqueness. Suppose n, n′ are integers such that n ≤ x < n + 1 and n′ ≤ x < n′ + 1 . Then

n ≤ x < n + 1  and
−n′ − 1 < −x ≤ −n′ ,

where the second line comes from negating all terms of n′ ≤ x < n′ + 1 .

Adding these two double-inequalities gives

n − n′ − 1 < 0 < n − n′ + 1 ,

so adding n′ − n to all three parts gives −1 < n′ − n < 1 . Since n′ − n is an integer in the range (−1, 1) , it must be zero. This gives n′ = n, as was to be shown. □