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Chapter 3: Problem 19

Given any two rational numbers \(r\) and \(s\) with \(r

Short Answer

Expert verified
Based on the provided step-by-step solution, the short answer is: Given two rational numbers, r and s, with r < s, we can find another rational number between them by calculating the average: \(\frac{r + s}{2}\). By expressing r and s as fractions, calculating the common denominator, and combining the fractions, we find the average to be a rational number. Additionally, we show that r < average < s by comparing the differences between the average and r and the average and s. Overall, we can conclude that there exists a rational number between r and s: \[r < \frac{ad + bc}{2bd} < s\]

Step by step solution

01

Determine the average of r and s

We start by finding the average of the given rational numbers r and s. The average of two numbers can be found using the formula: \[average = \frac{r + s}{2}\]
02

Show that the average is rational

Since r and s are both rational numbers, we can express them as fractions: \[r = \frac{a}{b}\] and \[s = \frac{c}{d}\], where a, b, c, and d are integers (with b and d not equal to 0). Using the given formula for finding the average, we get: \[average = \frac{(\frac{a}{b}) + (\frac{c}{d})}{2}\] Now, find a common denominator for the fractions a/b and c/d: \[average = \frac{(\frac{ad}{bd}) + (\frac{bc}{bd})}{2}\] Combine the fractions: \[average = \frac{ad + bc}{2bd}\] Since the numerators ad and bc are integers (as the product of integers) and the denominator 2bd is also an integer (as the product of integers), the average of r and s is a rational number.
03

Prove the average is between r and s

To prove the average lies between r and s, we need to show that r < average < s. Let's find the difference between the average and r: \[average - r = \frac{ad + bc}{2bd} - \frac{a}{b}\] To find the common denominator, we multiply the second fraction by 2d: \[average - r = \frac{ad + bc}{2bd} - \frac{2ad}{2bd}\] Combine the fractions: \[average - r = \frac{-ad + bc}{2bd}\] We are given that r < s, which means \[a * d < b*c\] (since the fractions are in lowest terms). Since this inequality holds, the numerator of average - r is positive, making the average greater than r. Next, we'll show that the average is less than s. Calculate the difference between s and the average: \[s - average = \frac{c}{d} - \frac{ad + bc}{2bd}\] Find the common denominator and multiply the first fraction by 2b: \[s - average = \frac{2bc}{2bd} - \frac{ad + bc}{2bd}\] Combine the fractions: \[s - average = \frac{bc - ad}{2bd}\] As mentioned earlier, since r < s, we know that \(a * d < b*c\), which means that the numerator of s - average is positive, making the average less than s. Thus, we have shown that the average is a rational number that lies between r and s: \[r < \frac{ad + bc}{2bd} < s\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Number Theory
Number theory is the branch of mathematics dedicated to exploring the fascinating properties of numbers. It often deals with integers, whole numbers, and their relationships and characteristics. Understanding how these numbers interact helps build fundamental mathematical concepts.
Rational numbers, which are a key concept in number theory, are numbers that can be written as a fraction where both the numerator and the denominator are integers, and the denominator is not zero. This means any number that can be expressed as a ratio of two integers falls within this category.
For example, the numbers 1/2, -5/3, and 4 can all be classified as rational because they can be expressed as a fraction or an integer. They are important in number theory because they help in understanding other types of numbers and mathematical operations.
Number theory, in this way, becomes the cornerstone for exploring how rational numbers fit within the wide world of mathematics.
  • Focus on rational numbers as fractions formed by integers.
  • The importance of rational numbers in constructing mathematical operations.
Intermediate Value Theorem
The Intermediate Value Theorem (IVT) is a principle in calculus, albeit slightly adapted in this context to the realm of rational numbers. Its main function is to confirm that for any two values (in this case, rational numbers) the intervals contain a value in-between.
While traditionally focused on continuous functions, the theorem, when applied to rational numbers, helps in affirming that there's always another rational number between any two rational numbers you could pick.
Mathematically, if you pick two rational numbers, say \( r \) and \( s \) with \( r < s\), by averaging them, you end up with a value strictly between them, showcasing the IVT's influence in ensuring no gaps exist between consecutive rational numbers.
  • Understanding the placement of rational numbers within any two selected rational values.
  • Basic extension of the IVT into sequences of rational numbers.
Mathematical Proof
Mathematical proof is essential to establish the validity of mathematical statements and concepts. It serves as a logical verification, ensuring that the conclusions drawn are irrefutably true.
In the context of rational numbers, a proof would firmly establish that no matter how you select two different rational numbers \( r \) and \( s \), there will always be a rational number between them.
This exercise provides a step-by-step proof that shows the validity of this claim by considering the average of \( r \) and \( s \) calculated using \( \frac{r + s}{2} \). Using component-wise operations and properties of rational numbers, the proof ensures that this average is not only rational but also strictly located between \( r \) and \( s \).
  • The role of proofs in confirming mathematical ideas.
  • Using logical steps to show the presence of a number between two rational numbers.
Inequalities
Inequalities are mathematical expressions showing the relationship between two non-equal values. They play a crucial role in illustrating how one number is greater than, less than, or different from another.
Within rational numbers, inequalities are fundamental. When given \( r < s \), it leads to exploring any element \( x \) satisfying \( r < x < s \).
In proving the existence of a rational number between \( r \) and \( s \), inequalities demonstrate that the average \( \frac{r + s}{2} \) satisfies \( r < \frac{r + s}{2} < s \).
This approach not only highlights the importance of choosing \( r \) and \( s \) but also ensures these principles effectively explain the positioning of numbers according to inequalities.
  • Utilizing inequalities to define the range and position of rational numbers.
  • Understanding how inequalities solidify mathematical claims.

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Most popular questions from this chapter

The quotient-remainder theorem says not only that there exist quotients and remainders but also that the quotient and remainder of a division are unique. Prove the uniqueness. That is, prove that if \(a\) and \(d\) are integers with \(d>0\) and if \(q_{1}, r_{1}, q_{2}\), and \(r_{2}\) are integers such that $$ a=d q_{1}+r_{1} \quad \text { where } 0 \leq r_{1}

a. Rewrite the following theorem in the form \(\forall \longrightarrow\) if then b. Fill in the blanks in the proof, Theorem: The sum of any even integer and any odd integer is odd. Proof: Suppose \(m\) is any even integer and \(n\) is (a) By definition of even, \(m=2 r\) for some \((b)\), and by definition of odd, \(n=2 s+1\) for some integer \(s\). By substitution and algebra, \(m+n=\underline{(c)}=2(r+s)+1 .\) Since \(r\) and \(s\) are both integers, so is their sum \(r+s\). Hence \(m+n\) has the form 2 . (some integer) \(+1\), and \(s o\) (d) by definition of odd.

If \(n\) is any odd integer, then \((-1)^{n}=-1\).

For each integer \(n\) with \(1 \leq n \leq 10, n^{2}-n+11\) is a prime number.

Is \(\frac{1}{0}\) an irrational number? Explain.
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