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Chapter 3: Problem 13

Prove the following proposition: If \(p, q \in \mathbb{Q}\) with \(p

Short Answer

Expert verified
In short, we can find a rational number \(x\) between two given rational numbers \(p\) and \(q\) (with \(p < q\)) by taking their arithmetic mean: \(x = \frac{p+q}{2}\). Since the sum and division of rational numbers are always rational, x is also a rational number. Moreover, we can prove that \(p < x < q\) by showing that \(p < \frac{p+q}{2} < q\).

Step by step solution

01

Find the arithmetic mean

To find the arithmetic mean of p and q, we'll take the average of both numbers: \(x = \frac{p + q}{2}\) Since p and q are rational numbers, x is also a rational number because the sum and division of rational numbers are always rational.
02

Prove that p < x < q

Now we want to prove that p < x < q. For that, let's break into two comparisons: First, we compare p to x: \(x = \frac{p + q}{2} > \frac{p + p}{2} = p\) So, we have proven that \(p < x\). Next, we compare x to q: \(x = \frac{p + q}{2} < \frac{q + q}{2} = q\) And we have proven that \(x < q\). Since we proved that \(p < x\) and \(x < q\), we can conclude that there exists a rational number x such that \(p < x < q \).

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Key Concepts

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

Understanding Rational Numbers
Rational numbers are figures that can be expressed as a fraction where both the numerator and the denominator are integers, and the denominator is not equal to zero. In simpler terms, any number that can be represented as the division of two whole numbers resides in the realm of rational numbers. For example, \frac{1}{2}, \frac{4}{5}, and \frac{-7}{3} are all rational numbers.

Importantly, the set of rational numbers, denoted by \( \mathbb{Q} \) in mathematics, includes all integers since any integer 'a' can be expressed as \frac{a}{1}, which fits the definition of a rational number. Rational numbers encompass a wide range of numerical values, including positive, negative numbers, and zero. When dealing with rational numbers, mathematical operations like addition, subtraction, multiplication, and division (except by zero) will always yield another rational number. This property is utilized in various proofs and is essential in our exercise related to finding a number between two rational numbers.
The Arithmetic Mean
The arithmetic mean, often simply called the 'average', is a concept that might seem straightforward but is a powerful tool in mathematics. It is computed by adding together a set of numbers and then dividing by the count of those numbers. For any two numbers, the arithmetic mean is exactly midway between them.

The formula for the arithmetic mean of two numbers, p and q, is: \( x = \frac{p + q}{2} \). When p and q are rational numbers, their mean, x, is also a rational number due to the fact that the sum and division of rational numbers result in a rational number. The arithmetic mean has a special role in proving inequalities between numbers, which is used in the problem we're discussing. It helps establish a value that we can confidently say lies between two given numbers. When we're given two rational numbers, p and q where p is less than q, the arithmetic mean will always be a number that exists between them.
Navigating Inequalities
Inequalities are mathematical expressions that demonstrate a relationship of non-equality between two values. They make use of symbols such as < (less than), > (greater than), \( \leq \) (less than or equal to), and \( \geq \) (greater than or equal to). When we say that \( p < q \), we are indicating that p is less than q.

In the context of our exercise, we are not just stating that one number is larger or smaller; we are also proving that there is a continuum of rational numbers between any two distinct rational numbers, p and q. This is where the concept of inequality becomes powerful. By using mathematical operations and the properties of rational numbers and arithmetic means, we're capable of demonstrating that there exists at least one rational number that lies between any two distinct rational numbers. As seen in the solution, by finding the arithmetic mean and comparing it to the given numbers, we can create and solve two inequalities that prove the proposition.

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Most popular questions from this chapter

Is the following proposition true or false? Justify your conclusion with a counterexample or a proof. For each integer \(a, 3\) divides \(a^{3}+23 a\).

(a) Use the result in Proposition 3.33 to help prove that the integer \(m=\) 5,344,580,232,468,953,153 is not a perfect square. Recall that an integer \(n\) is a perfect square provided that there exists an integer \(k\) such that \(n=k^{2} .\) Hint: Use a proof by contradiction. (b) Is the integer \(n=782,456,231,189,002,288,438\) a perfect square? Justify your conclusion.

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) For all nonzero integers \(a\) and \(b,\) if \(a+2 b \neq 3\) and \(9 a+2 b \neq 1,\) then the equation \(a x^{3}+2 b x=3\) does not have a solution that is a natural number. \mathrm{P} \text { Proof } . We will prove the contrapositive, which is For all nonzero integers \(a\) and \(b\), if the equation \(a x^{3}+2 b x=3\) has a solution that is a natural number, then \(a+2 b=3\) or \(9 a+2 b=1\). So we let \(a\) and \(b\) be nonzero integers and assume that the natural number \(n\) is a solution of the equation \(a x^{3}+2 b x=3\). So we have $$ \begin{aligned} a n^{3}+2 b n &=3 \quad \text { or } \\ n\left(a n^{2}+2 b\right) &=3 \end{aligned} $$ So we can conclude that \(n=3\) and \(a n^{2}+2 b=1\). Since we now have the value of \(n\), we can substitute it in the equation \(a n^{3}+2 b n=3\) and obtain \(27 a+6 b=3\). Dividing both sides of this equation by 3 shows that \(9 a+2 b=1\). So there is no need for us to go any further, and this concludes the proof of the contrapositive of the proposition. (b) For all nonzero integers \(a\) and \(b,\) if \(a+2 b \neq 3\) and \(9 a+2 b \neq 1,\) then the equation \(a x^{3}+2 b x=3\) does not have a solution that is a natural number. We will use a proof by contradiction. Let us assume that there exist nonzero integers \(a\) and \(b\) such that \(a+2 b=3\) and \(9 a+2 b=1\) and \(a n^{3}+2 b n=3,\) where \(n\) is a natural number. First, we will solve one equation for \(2 b\); doing this, we obtain $$ \begin{aligned} a+2 b &=3 \\ 2 b &=3-a . \end{aligned} $$ We can now substitute for \(2 b\) in \(a n^{3}+2 b n=3\). This gives $$ \begin{array}{l} a n^{3}+(3-a) n=3 \\ a n^{3}+3 n-a n=3 \\ n\left(a n^{2}+3-a\right)=3 . \end{array} $$ By the closure properties of the integers, \(\left(a n^{2}+3-a\right)\) is an integer and, hence, equation (2) implies that \(n\) divides \(3 .\) So \(n=1\) or \(n=3\). When we substitute \(n=1\) into the equation \(a n^{3}+2 b n=3,\) we obtain \(a+2 b=3\). This is a contradiction since we are told in the proposition that \(a+2 b \neq 3\). This proves that the negation of the proposition is false and, hence, the proposition is true.

Prove that for each real number \(x\) and each irrational number \(q,(x+q)\) is irrational or \((x-q)\) is irrational.

A real number \(x\) is defined to be a rational number provided there exist integers \(m\) and \(n\) with \(n \neq 0\) such that \(x=\frac{m}{n}\). A real number that is not a rational number is called an irrational number. It is known that if \(x\) is a positive rational number, then there exist positive integers \(m\) and \(n\) with \(n \neq 0\) such that \(x=\frac{m}{n}\). Is the following proposition true or false? Explain. For each positive real number \(x,\) if \(x\) is irrational, then \(\sqrt{x}\) is irrational.
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