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Chapter 3: Problem 6

Are the following statements true or false? Justify each conclusion. (a) For each positive real number \(x,\) if \(x\) is irrational, then \(x^{2}\) is irrational. (b) For each positive real number \(x\), if \(x\) is irrational, then \(\sqrt{x}\) is irrational. (c) For every pair of real numbers \(x\) and \(y\), if \(x+y\) is irrational, then \(x\) is irrational and \(y\) is irrational. (d) For every pair of real numbers \(x\) and \(y\), if \(x+y\) is irrational, then \(x\) is irrational or \(y\) is irrational.

Short Answer

Expert verified
(a) True. If we assume \(x^2\) is rational, we get a contradiction that x must be rational as well. Thus, \(x^2\) must be irrational. (b) True. If we assume that \(\sqrt{x}\) is rational, we get a contradiction that x must be rational as well. Thus, \(\sqrt{x}\) must be irrational. (c) False. A counterexample is when x=2 (rational) and y=\(\pi\) (irrational), their sum is irrational, but x is not irrational. (d) True. The contrapositive states "if both x and y are rational, then x+y is rational" is true. Thus, the original statement is true as well.

Step by step solution

01

Statement (a): For each positive real number x, if x is irrational, then \(x^2\) is irrational.

In this statement, consider a positive irrational number x. We need to find out if \(x^2\) is irrational as well. Let's assume the opposite: Suppose \(x^2\) is rational. Then, \(x^2\) could be written as \(\frac{p}{q}\) with p and q being integers with no common factors other than 1, and q not equal to 0. Now, if we take the square root of \(x^2\), we have $$x = \sqrt{x^2} = \sqrt{\frac{p}{q}} = \frac{\sqrt{p}}{\sqrt{q}}.$$ Since \(x\) is irrational, either sqrt(p) or sqrt(q) must be irrational. However, since the square root of a rational number is either rational or irrational, we have a contradiction: our assumption that \(x^2\) is rational implies that x must be rational as well. Thus, x must be irrational. Therefore, the statement is true.
02

Statement (b): For each positive real number x, if x is irrational, then \(\sqrt{x}\) is irrational.

Consider again a positive irrational number x. We need to find out if the square root of x is irrational as well. This time, let us prove the statement by contradiction. Assume the opposite: that x is irrational, but \(\sqrt{x}\) is rational. Thus, \(\sqrt{x} = \frac{p}{q}\), where p and q are integers with no common factors other than 1, and q not equal to 0. Now, if we square both sides of the equation and simplify, we have $$x = (\frac{p}{q})^2 = \frac{p^2}{q^2}.$$ Since \(x\) is irrational and both p and q are integers, we have a contradiction: our assumption that \(\sqrt{x}\) is rational implies that x must be rational as well. Therefore, the statement is true.
03

Statement (c): For every pair of real numbers x and y, if x+y is irrational, then x is irrational and y is irrational.

Let us consider a pair of real numbers x and y such that x+y is irrational. We want to see if x is irrational and y is irrational. This statement can be disproven by using a counterexample. Suppose for example, we have x as a rational number, say x = 2, and y as the irrational number \(y = \pi\). We get their sum as $$x + y = 2 + \pi.$$ Since 蟺 is irrational, the sum is also irrational. In this case, x+y is irrational, but x is not irrational. Hence, the statement is false.
04

Statement (d): For every pair of real numbers x and y, if x+y is irrational, then x is irrational or y is irrational.

Once more, we need to investigate if either x or y is irrational when their sum is irrational. We can prove this statement by contrapositive. The contrapositive is: "If both x and y are rational, then x+y is rational." Let x and y be rational numbers. By definition, rational numbers can be represented as fractions. Let $$x = \frac{p_1}{q_1}$$ and $$y = \frac{p_2}{q_2}$$, where \(p_1\), \(p_2\), \(q_1\), and \(q_2\) are integers and \(q_1\) and \(q_2\) are not equal to 0. The sum x+y can now be expressed as $$x + y = \frac{p_1}{q_1} + \frac{p_2}{q_2} = \frac{p_1q_2 + p_2q_1}{q_1q_2}.$$ Now, since the product and sum of integers results in an integer, the numerator and denominator of the fraction are also integers. Thus, x+y is a rational number. Therefore, the contrapositive is true, which means the original statement 鈥� "if x+y is irrational, then x is irrational or y is irrational" 鈥� is also true.

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Key Concepts

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

Real Number Properties
Understanding the properties of real numbers is essential in all fields of mathematics, including the realm of irrational numbers. Real numbers consist of all the possible numbers on the number line, encompassing both rational and irrational numbers. One key property is that real numbers are closed under addition, subtraction, multiplication, and non-zero division. This means that performing these operations on any two real numbers will always result in another real number.

Another essential property is the distributive property, which allows us to multiply a number by a sum of numbers and distribute the multiplication over the sum. For example, for any real numbers a, b, and c, the distributive property dictates that \(a \times (b + c) = a \times b + a \times c\). It is through these foundational properties that we explore the behavior of irrational numbers within the real number system.
Proof by Contradiction
Proof by contradiction is a powerful method of mathematical proof, particularly useful when dealing with irrational numbers. In this approach, we start by assuming the opposite of what we want to prove. Then, through a series of logical deductions, we demonstrate that this assumption leads to an impossibility or contradiction. This contradiction shows that our initial assumption must be false, and hence the statement we wanted to prove must be true.

For instance, when proving that the square of an irrational number is irrational (as seen in statement (a) of the exercise), assuming the square is rational leads to a contradiction that the original number must be both rational and irrational, which is impossible. Hence our initial assumption is false, confirming that the square of an irrational number is indeed irrational.
Rational and Irrational Numbers
Rational numbers are numbers that can be expressed as the ratio of two integers, where the denominator is not zero. Examples include \(\frac{1}{2}\), \(3\), and \(-\frac{5}{4}\). In contrast, irrational numbers cannot be written as such a ratio; they have endless, non-repeating decimal expansions. Classic examples of irrational numbers are \(\frac{p^2}{q^2}\).



It's important to distinguish between the two because it affects the overall understanding of statements like those in the exercise. For example, the square root of a positive irrational number is itself irrational (statement (b)), which defies the property we see with rational numbers, where the square root may or may not be rational. This highlights how the mutual exclusivity of rational and irrational numbers underpins the inherent structure of the real numbers.

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Most popular questions from this chapter

Using a Contradiction to Prove a Case Is Not Possible. Explore the statements in Parts (a) and (b) by considering several examples where the hypothesis is true. (a) If an integer \(a\) is divisible by both 4 and \(6,\) then it divisible by \(24 .\) (b) If an integer \(a\) is divisible by both 2 and 3 , then it divisible by 6 . (c) What can you conclude from the examples in Part (a)? (d) What can you conclude from the examples in Part (b)? The proof of the following proposition based on Part (b) uses cases. In this proof, however, we use cases and a proof by contradiction to prove that a certain integer cannot be odd. Hence, it must be even. Complete the proof of the proposition. Proposition. Let \(a \in \mathbb{Z}\). If 2 divides \(a\) and 3 divides \(a\), then 6 divides \(a\). Proof: Let \(a \in \mathbb{Z}\) and assume that 2 divides \(a\) and 3 divides \(a\). We will prove that 6 divides \(a\). Since 3 divides \(a\), there exists an integer \(n\) such that $$ a=3 n $$ The integer \(n\) is either even or it is odd. We will show that it must be even by obtaining a contradiction if it assumed to be odd. So, assume that \(n\) is odd. (Now complete the proof.)

Prove each of the following propositions: (a) For each real number \(\theta\), if \(0<\theta<\frac{\pi}{2},\) then \([\sin (\theta)+\cos (\theta)]>1\). (b) For all real numbers \(a\) and \(b,\) if \(a \neq 0\) and \(b \neq 0,\) then \(\sqrt{a^{2}+b^{2}} \neq\) \(a+b\) (c) If \(n\) is an integer greater than \(2,\) then for all integers \(m, n\) does not divide \(m\) or \(n+m \neq n m\). (d) For all real numbers \(a\) and \(b,\) if \(a>0\) and \(b>0,\) then $$ \frac{2}{a}+\frac{2}{b} \neq \frac{4}{a+b} $$

Evaluation of Proofs See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Proposition. If \(m\) is an odd integer, then \((m+6)\) is an odd integer. Proof. For \(m+6\) to be an odd integer, there must exist an integer \(n\) such that $$ m+6=2 n+1 $$ By subtracting 6 from both sides of this equation, we obtain $$ \begin{aligned} m &=2 n-6+1 \\ &=2(n-3)+1 . \end{aligned} $$ By the closure properties of the integers, \((n-3)\) is an integer, and hence, the last equation implies that \(m\) is an odd integer. This proves that if \(m\) is an odd integer, then \(m+6\) is an odd integer. (b) Proposition. For all integers \(m\) and \(n,\) if \(m n\) is an even integer, then \(m\) is even or \(n\) is even. \mathrm{\\{} \text { Proof } . For either \(m\) or \(n\) to be even, there exists an integer \(k\) such that \(m=2 k\) or \(n=2 k .\) So if we multiply \(m\) and \(n,\) the product will contain a factor of 2 and, hence, \(m n\) will be even.

(a) Use cases based on congruence modulo 3 and properties of congruence to prove that for each integer \(n, n^{3}=n(\mathrm{mod} 3)\). (b) Explain why the result in Part (a) proves that for each integer \(n, 3\) divides \(\left(n^{3}-n\right) .\) Compare this to the proof of the same result in Proposition 3.27

Using a Logical Equivalency. Consider the following proposition: Proposition. For all integers \(a\) and \(b,\) if 3 does not divide \(a\) and 3 does not divide \(b,\) then 3 does not divide the product \(a \cdot b\). (a) Notice that the hypothesis of the proposition is stated as a conjunction of two negations (" 3 does not divide \(a\) and 3 does not divide \(b\) "). Also, the conclusion is stated as the negation of a sentence ("3 does not divide the product \(a \cdot b\) "). This often indicates that we should consider using a proof of the contrapositive. If we use the symbolic form \((\neg Q \wedge \neg R) \rightarrow \neg P\) as a model for this proposition, what is \(P,\) what is \(Q,\) and what is \(R ?\) (b) Write a symbolic form for the contrapositive of \((\neg Q \wedge \neg R) \rightarrow \neg P\). (c) Write the contrapositive of the proposition as a conditional statement in English. We do not yet have all the tools needed to prove the proposition or its contrapositive. However, later in the text, we will learn that the following proposition is true. Proposition \(\mathbf{X}\). Let \(a\) be an integer. If 3 does not divide \(a\), then there exist integers \(x\) and \(y\) such that \(3 x+a y=1\) (d) i. Find integers \(x\) and \(y\) guaranteed by Proposition \(\mathrm{X}\) when \(a=5\). ii. Find integers \(x\) and \(y\) guaranteed by Proposition \(\mathrm{X}\) when \(a=2\). iii. Find integers \(x\) and \(y\) guaranteed by Proposition \(\mathrm{X}\) when \(a=-2\). (e) Assume that Proposition \(\mathrm{X}\) is true and use it to help construct a proof of the contrapositive of the given proposition. In doing so, you will most likely have to use the logical equivalency \(P \rightarrow(Q \vee R) \equiv\) \((P \wedge \neg Q) \rightarrow R\)
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