Question

Prove that the rational subfield \(R\) of any ordered field is Archimedean. [Hint: If $$ x=\frac{k}{m} \text { and } y=\frac{p}{q} \quad(k, m, p, q \in N), $$ then \(n x>y\) for \(n=m p+1]\)

Short answer

The rational subfield \(R\) is Archimedean because for any rationals \(x,y > 0\), \(nx > y\) with \(n = mp + 1\).

Step-by-step solution

Understanding the Problem

We need to prove that the rational subfield \(R\) of any ordered field is Archimedean. An ordered field is Archimedean if for any positive elements \(x\) and \(y\) in the field, there exists a natural number \(n\) such that \(nx > y\).

Expressing Rational Numbers

Recall that in the ordered field, rational numbers can be represented as fractions. Let \(x = \frac{k}{m}\) and \(y = \frac{p}{q}\) where \(k, m, p, q\) are natural numbers and \(x, y > 0\).

Choosing a Suitable Natural Number

To find \(n\) such that \(nx > y\), we follow the hint: choose \(n = mp + 1\). This is our candidate for the natural number that satisfies the condition \(nx > y\).

Verifying the Condition

Calculate \(nx = n \cdot \frac{k}{m} = (mp + 1) \cdot \frac{k}{m} = pk + \frac{k}{m}\). We need this to be greater than \(y = \frac{p}{q}\). Multiply both sides by \(mq\) to clear the denominators: \ \(mq( pk + \frac{k}{m}) > mq \cdot \frac{p}{q}\) \ \(pqmk + qk > mpk\).

Simplifying the Inequality

Re-arrange the inequality \(pqmk + qk > mpk\) by noting that \(qk > (mpk - pqmk)\). Since \(qk\) is positive and everything on the right is less than \(qk\), the inequality holds. Therefore, \(nx > y\) for \(n = mp + 1\).

Conclusion

We have shown that for \(x = \frac{k}{m}\) and \(y = \frac{p}{q}\), there exists an \(n = mp + 1\) such that \(nx > y\) holds, confirming the Archimedean property for the rational subfield \(R\).

Key concepts

Ordered Field

An ordered field is a mathematical structure that blends two concepts: a field, and an order. A field is a set equipped with two operations, typically addition and multiplication, that behaves similarly to the familiar number systems like the rational numbers, real numbers, or complex numbers. These operations must follow certain rules, such as the existence of additive and multiplicative inverses for every non-zero element.

Meanwhile, an order on this set means that you can compare any two elements to decide which one is greater or smaller. A field is called an ordered field if it has a subset, often denoted as positive elements, and every element can be compared consistently. So, in an ordered field, for any elements \(a\), \(b\), and \(c\), the following hold:

• If \(a < b\), then \(a + c < b + c\).
• If \(0 < a\) and \(0 < b\), then \(0 < a \, b\).
• Transitivity: If \(a < b\) and \(b < c\), then \(a < c\).

Understanding the basic structure of an ordered field helps us explore deeper properties like the Archimedean property.

Rational Subfield

The rational subfield is essentially the set of all rational numbers within an ordered field. The rational numbers are those numbers that can be expressed as the fraction of two natural numbers, like \(\frac{3}{4}\) or \(-\frac{5}{2}\).

When dealing with any ordered field, we often find a copy or a version of the rational numbers inside it. This is because rational numbers follow the same rules of arithmetic and order, acting like a core or a foundation for more complex numbers.

One of the powerful results about rational subfields is the Archimedean property, which states that for any two positive elements of the rational subfield, we can find a natural number that scales one of the elements to eventually exceed the other. This property is a cornerstone in understanding the behavior of numbers in an ordered field and helps in building intuition for limits and continuity within analytical contexts.

Natural Numbers

Natural numbers are the most basic and intuitive set of numbers used for counting and ordering, starting from 1, 2, 3, and so on. They do not include zero or any negative numbers. In the context of mathematics and ordered fields, natural numbers serve as a core building block.

When we talk about scaling a rational number within an ordered field, the natural numbers are the steps or increments by which we multiply. For instance, multiplying a rational number \(x = \frac{k}{m}\) by a natural number \(n\) scales \(x\) consistently, allowing us to reach or exceed another number, \(y\), according to the Archimedean property.

Without natural numbers, we would struggle to express many fundamental mathematical facts. They are the fundamental steps that help in "measuring" within the rational subfield and ensuring that any gaps between elements can be bridged. Their role is critical in proving inequalities, especially in ordered fields.

Inequality Proof

Proving inequalities is a fundamental part of verifying properties in ordered fields, like the Archimedean property. In the context of our original problem, we used inequality proof to show that for any two positive rational numbers \(x = \frac{k}{m}\) and \(y = \frac{p}{q}\), there exists a natural number \(n\) such that \(nx > y\).

The trick to establishing this inequality, as hinted in the exercise, involves selecting a specific natural number \(n = mp + 1\). When we multiply the rational number \(x\) by \(n\), it yields \((mp + 1) \cdot \frac{k}{m}\). Distributing this result and simplifying leads us to \(pk + \frac{k}{m}\).

By rearranging and manipulating this expression, we compare it to \(y = \frac{p}{q}\). Clearing out the denominators by multiplying through by \(mq\) ensures we're comparing like terms. Ultimately, simplifying the inequality shows \(pqmk + qk > mpk\), confirming that \(nx > y\) under this configuration.

This logical process showcases the precision needed in inequality proofs and how choosing the right multiplier can be key to demonstrating a number property in ordered fields.