# Archimedean Property of Real Numbers: Statement, Proof, Example

The Archimedean property of real numbers states that for positive real numbers x and y, there is an integer n>0 such that nx>y. This principle is named after the ancient Greek mathematician Archimedes. In this post, we will learn about the Archimedean property of real numbers along with its proof and applications.

## Proof of Archimedean Property of Real Numbers

We will prove the Archimedean property using the completeness axiom of real numbers. Consider the set

S = {nx : n ∈ ℕ}.

For a contradiction, assume that nx ≤ y for all n ∈ ℕ. This makes y is an upper bound of S. As S is non-empty and bounded above, by the completeness axiom of real numbers Sup(S) exists.

Let M = Sup(S).

As x>0 given, M-x cannot be an upper bound of S.

So M-x < mx for some mx ∈ S.

⇒ M < (m+1)x

As (m+1)x ∈ S, we conclude that M = Sup(S) is not true, which is a contradiction. Thus, our assumption nx ≤ y for all n ∈ ℕ is wrong.

In other words, nx>y for some positive integer n. This completes the proof of the Archimedean property of ℝ. ♣

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## Archimedean Property of Rational Numbers

Question: The set Q of rational numbers satisfies Archimedean property.

To prove the set Q of rational numbers satisfies Archimedean property, we need to show that if x is a positive rational and y is any rational number, then there exists n ∈ ℕ such that

nx > y.

Let us prove the case where y>0 and x<y. For a contradiction, assume that nx ≤ y for all n ∈ ℕ.

⇒ $nx \times \dfrac{1}{y}$ ≤ 1 < m where m ∈ ℕ and m>1.

⇒ $\dfrac{n}{m} \leq \dfrac{y}{x} \cdot \dfrac{n}{m}$ is any rational number. Note that for a fixed x and y, the number y/x is a fixed rational number. So this result is not true always.

That is, nx ≤ y is not true always.

In other words, we have some integer n ∈ ℕ such that nx > y. This proves that the set of rational numbers satisfies the Archimedean property of real numbers.

## Example

Proof:

As the set ℕ of natural numbers is unbounded above, ∃ n ∈ ℕ such that x<n. So it remains to show that n-1 ≤ x.

Let S = {m>x : m ∈ ℕ}.

Note that S is non-empty. Because, n ∈ S and n is the least element of S.

So n-1 ∉ S.

⇒ n-1 ≤ x.

Thus, we have shown that n-1 ≤ x < n for some natural number n.

## FAQs

Q1: State and prove Archimedean property of real numbers.

Statement: If x>0 and y be an arbitrary real number, then the Archimedean property of real number states that ∃ a positive integer n such that nx>y.
Proof: We can prove Archimedean property using the unboundedness property of ℕ. Observe that y/x is a real number. As ℕ is unbounded above, ∃ an integer n>0 such that n> y/x ⇒ nx>y.

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