Natural Numbers Have No Real Upper Bound
The completeness results in “Why Completeness Matters” guarantee suprema and infima for suitable sets of real numbers. A basic consequence is that the natural numbers cannot have a real upper bound. In other words, however large a real number is, some natural number exceeds it. This fact is called the Archimedean Property.
Here we use \(\mathbb{N}_0=\{0,1,2,\ldots\}\), and call an element of \(\mathbb{N}_0\) positive if it is at least \(1\). Using positive natural numbers rather than \(\mathbb{N}_0\) makes no difference to the unboundedness claim, since \(0\) adds only one element.
The quantifiers matter. The property says that for each real number \(x\), a natural number can be found that exceeds that particular \(x\). It does not say there is one natural number exceeding all real numbers. Indeed, no real number can be an upper bound for the positive natural numbers.
Completeness Proves the Archimedean Property
Suppose, for contradiction, that the positive natural numbers are bounded above in \(\mathbb{R}\). They form a nonempty set, so the Least Upper Bound Property gives them a supremum. The defining approximation property of a supremum then produces a natural number close enough to that supremum to contradict its being an upper bound.
Proof. Let \(P=\{1,2,3,\ldots\}\), viewed as a subset of \(\mathbb{R}\). Suppose that \(P\) is bounded above. Since \(P\) is nonempty, the Least Upper Bound Property gives \(s=\sup P\).
Because \(s-1<s\), the number \(s-1\) is not an upper bound for \(P\). Otherwise it would be an upper bound smaller than the least upper bound \(s\). Therefore there is some \(n\in P\) such that $$ s-1<n. $$ Adding \(1\) gives \(s<n+1\). But \(n+1\in P\), and \(s\), as an upper bound for \(P\), must satisfy \(n+1\leq s\). These inequalities contradict one another. Thus \(P\) is not bounded above, which means that for every \(x\in\mathbb{R}\), some \(n\in P\) satisfies \(n>x\). \(\square\)
This argument shows precisely how completeness enters: it supplies a supremum under the assumption that the natural numbers are bounded above. The successor \(n+1\) then contradicts the supremum’s upper-bound property. The proof relies on the fact that adding \(1\) to a positive natural number gives another positive natural number.
Worked Example: Finding a Natural Number Above a Specific Real
Consider \(x=\sqrt{7}+2\). Since both \(\sqrt{7}\) and \(3\) are positive, comparing their squares shows that \(\sqrt{7}<3\): indeed, \(7<9\). Adding \(2\) gives $$ \sqrt{7}+2<3+2=5. $$ Thus the positive natural number \(n=5\) exceeds \(x\). The Archimedean Property guarantees that such an \(n\) exists for every real \(x\); this example verifies one particular choice directly.
Equivalent Form: Reciprocals Can Be Arbitrarily Small
The same property can be expressed in terms of positive reciprocals. It says that no matter how small a positive real number \(\varepsilon\) is, some reciprocal \(1/n\) of a positive natural number is smaller than \(\varepsilon\). The reciprocal comparison follows from the order rule that reciprocals reverse the order of positive numbers.
Proof. First suppose the Archimedean Property holds, and let \(\varepsilon>0\). Its reciprocal \(\varepsilon^{-1}\) is positive. Choose a positive natural number \(n>\varepsilon^{-1}\). Since \(n\) and \(\varepsilon^{-1}\) are positive, reciprocal comparison gives $$ \frac{1}{n}<\frac{1}{\varepsilon^{-1}}=\varepsilon. $$ This proves the reciprocal statement.
Conversely, suppose that for every \(\varepsilon>0\), some positive natural number \(n\) satisfies \(1/n<\varepsilon\). Let \(x\in\mathbb{R}\). If \(x<0\), then \(n=1>x\), as required. If \(x\geq0\), then \(x+1>0\), so the assumed reciprocal statement applies with \(\varepsilon=1/(x+1)\). Choose a positive natural number \(n\) such that $$ \frac{1}{n}<\frac{1}{x+1}. $$ Both \(n\) and \(x+1\) are positive. Reciprocal comparison therefore gives \(x+1<n\), and hence \(x<n\). Thus the Archimedean Property holds for every real \(x\). \(\square\)
Worked Example: Choosing a Reciprocal Below a Given Bound
Let \(\varepsilon=7/50\). Choose \(n=8\). To verify the strict inequality, note that both denominators are positive and $$ \frac{1}{8}<\frac{7}{50} \quad\Longleftrightarrow\quad 50<56. $$ Since \(50<56\), we have \(1/8<7/50\). This is an explicit instance of the reciprocal form of the property: the chosen reciprocal is positive and lies below the specified positive bound.
A standard application is that the sequence of reciprocals \(1/n\) converges to \(0\). Indeed, let \(\varepsilon>0\). By the reciprocal form, choose a positive natural number \(N\) with \(1/N<\varepsilon\). For every positive natural number \(n\geq N\), reciprocal comparison gives \(0<1/n\leq1/N<\varepsilon\). Thus \(|1/n-0|<\varepsilon\) for all \(n\geq N\), which is exactly convergence to \(0\).
Scaling and Bounding Real Numbers
The Archimedean Property also lets us find a natural number whose positive multiple exceeds any specified real bound. If \(c>0\) and \(M\in\mathbb{R}\), apply the property to \(M/c\). The resulting natural number \(n>M/c\) satisfies \(cn>M\), because multiplication by a positive number preserves strict order.
Proof. Since \(c>0\), the quotient \(M/c\) is defined. By the Archimedean Property, choose a positive natural number \(n>M/c\). Multiplying this inequality by \(c>0\) preserves its direction, so \(nc>M\). \(\square\)
Worked Example: A Multiple Exceeding a Prescribed Bound
Take \(c=3/5\) and \(M=100\). The choice \(n=167\) works, since $$ nc=167\cdot\frac{3}{5}=\frac{501}{5}=100+\frac{1}{5}>100. $$ Thus a positive natural number multiple of \(3/5\) exceeds \(100\). The corollary guarantees an appropriate multiple for every positive \(c\) and every real bound \(M\), not only for these particular values.
Another useful consequence is that any one real number can be bounded in absolute value by a positive natural number. Given \(x\in\mathbb{R}\), apply the Archimedean Property to \(|x|\) to obtain \(n>|x|\). The basic absolute-value bound then gives $$ -n<x<n. $$ This is a finite bound chosen for the particular \(x\). It does not make the whole real line bounded.
Why the Property Matters—and a Common Pitfall
The Archimedean Property connects the order and algebra of the real numbers to their natural-number structure. It justifies choosing natural-number indices large enough to meet a real-valued bound, and its reciprocal form gives a direct way to make reciprocals smaller than any prescribed positive tolerance. These facts are used throughout analysis when estimates require a sufficiently large integer or a sufficiently small reciprocal.
Do not confuse “the natural numbers are unbounded” with “the natural numbers contain every real number.” Unboundedness says that some natural number exceeds any chosen real number; it does not say that a real number itself is natural. For instance, the property supplies a natural number greater than \(\sqrt{7}+2\), but it does not identify that real number as a natural number.
The proof also depends on the completeness of \(\mathbb{R}\). The field and order rules alone do not ensure the existence of a supremum for every nonempty bounded-above set. Here, completeness rules out the possibility that the positive natural numbers have a real upper bound. The Archimedean Property is one concrete way that the completeness of the real numbers constrains their order.
Check Your Understanding
Use the definition and the proved results to answer the following questions.
- What does it mean for the positive natural numbers to be unbounded above in \(\mathbb{R}\)?
- In the proof of the Archimedean Property, why can \(s-1\) not be an upper bound when \(s\) is the supremum of the positive natural numbers?
- How does the inequality \(n>\varepsilon^{-1}\), with \(\varepsilon>0\), give \(1/n<\varepsilon\)?
- Why does \(1/n<1/(x+1)\), for \(x\geq0\), imply \(n>x\)?
- Explain why the choice \(n=167\) gives \(n(3/5)>100\).