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Chapter 3: Problem 9

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.

Short Answer

Expert verified
The proposition "For each positive real number x, if x is irrational, then 鈭歺 is irrational" is true. This is because no counterexamples can be found to disprove the proposition and the attempted counterexample led to a contradiction.

Step by step solution

01

Analyzing the given proposition

: We are given a positive real number x, and the aim is to find out if x being irrational implies that 鈭歺 is also irrational. We know that if x is a rational number, then there exist integers m and n such that \(x = \frac{m}{n}\). Furthermore, if x is irrational, then there are no such integers that form this relationship.
02

Finding relationship between x and 鈭歺 in case x is rational

: Let's assume x is a rational number, which means that \(x = \frac{m}{n}\) for some integers m and n with \(n \neq 0\). We want to find the relationship between x and 鈭歺. Since x is rational, is there any guarantee that 鈭歺 will also be rational? We can rewrite x as \(x = \frac{m}{n} = (\frac{\sqrt{m}}{\sqrt{n}})^2\). If both \(\sqrt{m}\) and \(\sqrt{n}\) are rational, then 鈭歺 will also be rational. However, this is not always the case, as there can be examples where x is rational, but 鈭歺 is irrational. For example, let x = 2, which is a rational number. However, 鈭�2 is irrational as there are no integers p and q such that \(\sqrt{2} = \frac{p}{q}\). Therefore, having x as rational doesn't necessarily imply that 鈭歺 will be rational. Now we want to check the proposition in case x is irrational. If x is irrational, does it always imply that 鈭歺 is also irrational?
03

Finding a counterexample, if possible

: Let's try to find a counterexample to disprove the given proposition. Suppose, for the sake of contradiction, that there exists a positive irrational number x for which \(\sqrt{x}\) is rational. In that case, there should exist integers p and q such that \(\sqrt{x} = \frac{p}{q}\). Squaring both sides, we get \(x = \frac{p^2}{q^2}\), which implies that x is a rational number, since p and q are integers. However, this contradicts our assumption that x is irrational. Since we cannot find a counterexample to disprove the proposition, it seems that the proposition is true. So we can conclude:
04

Conclusion

: The proposition "For each positive real number x, if x is irrational, then 鈭歺 is irrational" is true. This is because we cannot find any counterexamples to disprove the proposition and the attempted counterexample led to a contradiction.

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Key Concepts

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

Rational Numbers
A rational number can be thought of as any number that can be represented as a fraction or a ratio. More specifically, these are numbers that can be expressed as the quotient of two integers, where the denominator is not zero. In mathematical terms, a number is rational if it can be written as \(x = \frac{m}{n}\), where \(m\) and \(n\) are integers, and \(n eq 0\).

Some common examples of rational numbers include:
  • Whole numbers like 3 (which can be written as \(\frac{3}{1}\))
  • Fractions like \(\frac{1}{2}\) or \(-\frac{5}{3}\)
  • Decimals that terminate or repeat, like 0.75 (which can be written as \(\frac{3}{4}\))

Rational numbers are dense; between any two rational numbers, another rational number can always be found. This property allows for extensive flexibility when dealing with equations and inequalities that involve rational numbers.
Irrational Numbers
Unlike rational numbers, irrational numbers cannot be expressed as a simple fraction of two integers. These are numbers that have non-terminating, non-repeating decimal expansions. In other words, there aren't two integers \(m\) and \(n\) for which an irrational number can be precisely expressed as \(x = \frac{m}{n}\).

Famous examples of irrational numbers include:
  • \(\pi\), the ratio of a circle鈥檚 circumference to its diameter
  • The number \(e\), the base of the natural logarithm
  • Square roots of non-perfect squares, such as \(\sqrt{2}\)

Unlike rational numbers, irrational numbers do not have a finite or predictable decimal pattern. This unique characteristic gives them a special place in mathematics, as they often arise in various mathematical problems, indicating solutions that can't be neatly expressed as fractions.
Proof by Contradiction
Proof by contradiction is a powerful method of mathematical reasoning used when direct proof is difficult or impossible. The idea behind this approach is to assume that the proposition we're trying to prove is false, and then show that this assumption leads to a contradiction. When a contradiction emerges, it implies that the assumption is false, thereby proving the original proposition true.

Here's a simple breakdown of how proof by contradiction works:
  • Assume the opposite of what you're trying to prove.
  • Use logical reasoning and existing mathematical properties or theorems to explore the consequences of this assumption.
  • Identify a contradiction - usually an inconsistency with established facts or theorems.
  • Conclude that the original assumption must be incorrect, thereby validating the truth of the proposition.

This method was used in the original solution to demonstrate why if \(x\) is irrational, then \(\sqrt{x}\) must also be irrational. By assuming \(\sqrt{x}\) was rational, it led to \(x\) being rational as well, which contradicted the fact that \(x\) was irrational. Therefore, it verified the truth of the proposition.

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Most popular questions from this chapter

The Last Two Digits of a Large Integer. Notice that \(7,381,272 \equiv 72(\) mod 100\()\) since \(7,381,272-72=7,381,200\). which is divisible by 100 . In general, if we start with an integer whose decimal representation has more than two digits and subtract the integer formed by the last two digits, the result will be an integer whose last two digits are \(00 .\) This result will be divisible by 100 . Hence, any integer with more than 2 digits is congruent modulo 100 to the integer formed by its last two digits. (a) Start by squaring both sides of the congruence \(3^{4} \equiv 81\) (mod 100 ) to prove that \(3^{8}=61\) (mod 100 ) and then prove that \(3^{16}=21\) (mod 100 ). What does this tell you about the last two digits in the decimal representation of \(3^{16} ?\) (b) Use the two congruences in Part (24a) and laws of exponents to determine \(r\) where \(3^{20} \equiv r(\bmod 100)\) and \(r \in \mathbb{Z}\) with \(0 \leq r<100 .\) What does this tell you about the last two digits in the decimal representation of \(3^{20} ?\) (c) Determine the last two digits in the decimal representation of \(3^{400}\). (d) Determine the last two digits in the decimal representation of \(4^{804}\). Hint: One way is to determine the "mod 100 values" for \(4^{2}, 4^{4}, 4^{8}\), \(4^{16} \cdot 4^{32}, 4^{64},\) and so on. Then use these values and laws of exponents to determine \(r,\) where \(4^{804} \equiv r(\) mod 100\()\) and \(r \in \mathbb{Z}\) with \(0 \leq r<\) \(100 .\) (e) Determine the last two digits in the decimal representation of \(3^{3356}\). (f) Determine the last two digits in the decimal representation of \(7^{403}\).

Prove that the cube root of 2 is an irrational number. That is, prove that if \(r\) is a real number such that \(r^{3}=2,\) then \(r\) is an irrational number.

One of the most famous unsolved problems in mathematics is a conjecture made by Christian Goldbach in a letter to Leonhard Euler in 1742. The conjecture made in this letter is now known as Goldbach's Conjecture. The conjecture is as follows: Every even integer greater than 2 can be expressed as the sum of two (not necessarily distinct) prime numbers. Currently, it is not known if this conjecture is true or false. (a) Write \(50,142,\) and 150 as a sum of two prime numbers. (b) Prove the following: If Goldbach's Conjecture is true, then every integer greater than 5 can be written as a sum of three prime numbers. (c) Prove the following: If Goldbach's Conjecture is true, then every odd integer greater than 7 can be written as a sum of three odd prime numbers.

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

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