The Integer Part of a Real Number

Locating a Real Number Between Consecutive Integers

The Archimedean Property says that positive natural numbers exceed any specified real number. It therefore provides integer bounds for real numbers, but a more precise question is often useful: which two consecutive integers surround a given real number? The integer part answers this question. Its definition applies to negative numbers as well as positive ones, where careful attention to the direction of the inequalities is essential.

Throughout, \(\mathbb{N}_0=\{0,1,2,\ldots\}\), and \(\mathbb{Z}\) denotes the integers. We use the Archimedean Property from the previous tutorial, along with the fact that every nonempty subset of \(\mathbb{N}_0\) has a least element. The latter follows from the Principle of Mathematical Induction: given a nonempty subset \(A\subseteq\mathbb{N}_0\), choose \(a\in A\); the nonempty finite set \(A\cap\{0,1,\ldots,a\}\) has a least element, which is also least in \(A\).

The inequalities are deliberately non-strict on the left and strict on the right. If \(x\) is itself an integer, its integer part is \(x\), not the integer immediately below it. For a noninteger \(x\), the floor is the integer immediately below \(x\).

Every Real Number Has a Unique Integer Part

To prove existence, we handle nonnegative and negative real numbers separately. For a nonnegative \(x\), the Archimedean Property supplies natural numbers larger than \(x\). The least such number gives the next integer above \(x\). For a negative \(x\), we instead find the least nonnegative integer at least as large as \(-x\), then reverse the resulting inequalities.

Proof. First suppose \(x\geq0\). Define $$ A=\{n\in\mathbb{N}_0:n>x\}. $$ By the Archimedean Property, \(A\) is nonempty. Let \(n\) be its least element. Since \(x\geq0\), \(0\notin A\), so \(n\geq1\). By minimality, \(n-1\notin A\); hence \(n-1\leq x\). Also \(n\in A\), so \(x<n\). Setting \(m=n-1\in\mathbb{Z}\) gives $$ m\leq x<m+1. $$

Now suppose \(x<0\), so \(y=-x>0\). Define $$ B=\{n\in\mathbb{N}_0:n\geq y\}. $$ The Archimedean Property gives a positive natural number greater than \(y\), so \(B\) is nonempty. Let \(n\) be its least element. Since \(y>0\), \(0\notin B\), and therefore \(n\geq1\). Minimality gives \(n-1\notin B\), so \(n-1<y\). The fact that \(n\in B\) gives \(y\leq n\). Negating these inequalities reverses their order: $$ -n\leq x<-(n-1). $$ With \(m=-n\in\mathbb{Z}\), this is exactly \(m\leq x<m+1\). Thus an integer part exists in both cases.

It remains to prove uniqueness. Suppose integers \(m\) and \(p\) both satisfy the defining inequalities. If \(m<p\), then \(p-m\) is a positive integer, so \(p-m\geq1\) and \(m+1\leq p\). But then $$ x<p\leq x $$ because \(p\leq x\) by the defining inequality for \(p\). This is impossible. The same argument with \(m\) and \(p\) exchanged rules out \(p<m\). Therefore \(m=p\), proving uniqueness. \(\square\)

The strict inequality at the upper endpoint is important in both parts of the proof. In the nonnegative case, minimality ensures that the least integer strictly greater than \(x\) is the next integer after its floor. In the negative case, minimality gives a strict inequality after negation. Replacing either strict inequality without justification could invalidate the conclusion.

Fractional Parts and Basic Floor Properties

Subtracting the integer part from a real number leaves a remainder between zero and one. This remainder is called the fractional part. It need not be a nonzero decimal expansion; the definition works uniformly for rational and irrational numbers.

Since \(\lfloor x\rfloor\leq x<\lfloor x\rfloor+1\), subtracting \(\lfloor x\rfloor\) gives $$ 0\leq\{x\}<1. $$ Conversely, the identity \(x=\lfloor x\rfloor+\{x\}\) expresses every real number as an integer plus a number in the half-open interval \([0,1)\). For example, \(\{-17/5\}=-17/5-(-4)=3/5\), which is positive even though the original number is negative.

Proof. Write \(m=\lfloor x\rfloor\) and \(n=\lfloor y\rfloor\). Suppose, for contradiction, that \(m>n\). Since \(m-n\) is a positive integer, \(m-n\geq1\), so \(n+1\leq m\). The defining inequalities and the assumption \(x\leq y\) then give $$ y<n+1\leq m\leq x\leq y, $$ which is impossible. Therefore \(m\leq n\), as claimed. \(\square\)

Monotonicity says that the floor cannot decrease when its input increases. The values need not increase strictly: many distinct real numbers have the same floor. For instance, every \(x\) satisfying \(2\leq x<3\) has floor \(2\).

Proof. Let \(m=\lfloor x\rfloor\), so \(m\leq x<m+1\). Adding the integer \(k\) to all three parts gives $$ m+k\leq x+k<m+k+1. $$ Since \(m+k\) is an integer, it satisfies the defining inequalities for the integer part of \(x+k\). Uniqueness of the integer part therefore gives \(\lfloor x+k\rfloor=m+k=\lfloor x\rfloor+k\). \(\square\)

Why the Integer Part Matters

The integer part converts a real-valued location into an integer index while preserving a precise bound on the remainder. This is useful whenever the real line is divided into intervals of unit length: \(\lfloor x\rfloor=m\) exactly when \(x\in[m,m+1)\). The half-open convention assigns each real number to one interval rather than leaving integer endpoints ambiguous or counting them twice.

A frequent error is to use “truncate toward zero” as a definition of the floor. That agrees with the floor for nonnegative numbers, but not for negative nonintegers. For example, truncating \(-3.4\) toward zero gives \(-3\), whereas the floor is \(-4\), because \(-4\leq-3.4<-3\). Always check the defining inequalities rather than relying on a rounding convention.

The existence proof also illustrates a useful role for the Archimedean Property. It ensures that the relevant set of natural numbers is nonempty; the least-element principle then selects a first integer crossing the chosen bound. Together these ideas produce a precise integer location for every real number, no matter how large, small, or close to an integer it is.