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.
Table of Contents
Statement of Archimedean Property of Real Numbers
| Let x and y be two real numbers with x>0. Then there exists a positive integer n such that nx>y. |
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. ♣
Related Topics:
Archimedean Property of Rational Numbers
Question: The set Q of rational numbers satisfies Archimedean property.
Solution:
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
| For every positive real number x, ∃ n ∈ ℕ such that n-1 ≤ x < n. |
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
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.
