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

Short Answer

Expert verified
For any real number \(x\) and irrational number \(q\), either \((x+q)\) or \((x-q)\) is irrational. To prove this, we consider two cases: 1. If \(x\) is rational: Express \(x\) as \(p/q\). Then \((x+q) = (p/q) + q = (p+q^2)/q\), which is irrational, as it is the sum of a rational number and an irrational number. 2. If \(x\) is irrational: If \((x+q)\) is rational, then \((x-q)\) must be irrational, as the sum of a rational number and an irrational number is always irrational. Similarly, if \((x-q)\) is rational, then \((x+q)\) must be irrational. Thus, in both cases, either \((x+q)\) or \((x-q)\) is irrational.

Step by step solution

01

Case 1: x is a rational number

Suppose x is a rational number. We can express x as the ratio of two integers, say x = p/q, where p and q are integers. Now, we analyze the sum and difference of x and q: 1. (x + q) = (p/q) + q 2. (x - q) = (p/q) - q We proceed to look for a contradiction to show that either one of them must be irrational.
02

Working with (x + q)

To work with (p/q) + q, we need to find a common denominator. This will convert q into a fraction as well. 1. (p/q) + q = (p/q) + (q * q / q) 2. (p/q) + (q^2/q) = (p + q^2) / q Now we need to determine if (p + q^2) / q is rational or irrational.
03

Analyzing (x + q) for irrationality

If (x + q) is rational, then (p + q^2) / q can be expressed as a ratio of two integers. But, since q is irrational, it cannot be expressed as a ratio of two integers, and thus q^2 is also irrational. Therefore, (p + q^2) must be irrational as it is the sum of a rational number (p) and an irrational number (q^2). So, we conclude that (x + q) is irrational.
04

Conclusion for Case 1

Since we proved that (x + q) is irrational for x being a rational number, we can conclude that for each real number x and each irrational number q, (x+q) is irrational in this case. There is no need to further analyze (x - q) in this case, as we already found one of the expressions to be irrational.
05

Case 2: x is an irrational number

Now, suppose x is an irrational number. Then, we want to show that either (x + q) is irrational or (x - q) is irrational.
06

Working with (x + q) and (x - q)

First, let us consider the following fact: The sum, or the difference, of a rational number and an irrational number is always irrational. Now we analyze the sum and difference of x and q: 1. (x + q) = x + q 2. (x - q) = x - q Notice that when x is irrational, then (x + q) and (x - q) are sums of an irrational number and an irrational number.
07

Conclusion for Case 2

For the case where x is irrational, we cannot ensure that either (x + q) or (x - q) is irrational, since the sum, or the difference, between two irrational numbers can be either rational or irrational. However, if (x + q) is rational, then (x - q) must be irrational, as the sum of a rational number and an irrational number is always irrational. Similarly, if (x - q) is rational, then (x + q) must be irrational. Thus, we can conclude that for each real number x and each irrational number q, either (x+q) is irrational or (x-q) is irrational in this case as well.

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Key Concepts

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

Irrational Numbers
Irrational numbers are a fascinating and crucial part of the number system. An irrational number is defined as a number that cannot be written as a simple fraction or ratio of two integers, where the numerator and the denominator are both whole numbers. This means that irrational numbers cannot be expressed in the form \( \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \eq 0 \). A classic example of an irrational number is \( \sqrt{2} \), which cannot be precisely expressed as a fraction of two integers.

Another way to look at irrational numbers is through their decimal expansions; these numbers have non-terminating, non-repeating decimal parts. For example, the number \( \pi \) ≈ 3.14159... continues infinitely without a repeating pattern. Because of their non-repeating nature, irrational numbers are crucial in mathematics for various reasons, including the fact that they expand our understanding beyond what can be represented by simple ratios, and present unique properties that are explored through algebra, geometry, and calculus.

When working with irrational numbers in mathematical proofs, such as the one given in the exercise, their unique characteristics must be carefully considered to determine the rationality or irrationality of expressions they're involved in. This helps students understand the nuances of number classification and reinforces a foundation for more complex mathematical concepts.
Rational Numbers
In contrast to irrational numbers, rational numbers can be written as a ratio of two integers. Specifically, a number is rational if you can find two integers \( p \) and \( q \) such that the number equals \( \frac{p}{q} \), with the condition that \( q \eq 0 \). These numbers can be either positive, negative, or even zero.

Rational numbers have a predictable structure either in their fractional form or when written as decimals. They can have a finite number of decimal places, like \( \frac{1}{2} \)= 0.5, or an infinite number of decimal places that repeat in a pattern, like \( \frac{1}{3} \)= 0.333... where the digit '3' repeats indefinitely. This repeatability is key to identifying rational numbers.

Understanding rational numbers is pivotal when considering operations involving both rational and irrational numbers. For instance, adding or multiplying a rational number with an irrational number will always give an irrational result, which is central to solving the provided exercise. Through the clear definition and properties of rational numbers, students can develop problem-solving skills that involve recognizing number types and applying properties that govern their interactions.
Proof by Contradiction
Proof by contradiction is a classical and powerful method to affirm the truth of a statement by first assuming that the opposite of the statement is true and then showing that this assumption leads to a contradiction. If a premise leads to an impossibility, then the premise must be false and consequently, the original statement is true.

The technique often begins with the statement 'Assume for the sake of contradiction that...' followed by the opposite of what we aim to prove. As we work through the implications of this assumption, we reach an illogical end or contradict a known fact, thus invalidating our initial assumption. The beauty of this approach is that it can sometimes prove statements that are difficult to prove directly.

Using proof by contradiction in the context of irrational and rational numbers allows us to tackle claims regarding the nature of certain numbers with certainty. As showcased in the exercise solutions provided, by assuming that both (x + q) and (x - q) could be rational when x is a rational number and q is an irrational number, we reach a contradiction based on the defined properties of these numbers. This kind of logical reasoning is not only crucial for mathematics but also enhances critical thinking skills by challenging students to consider all possible outcomes and examine the consistencies and inconsistencies within arguments.

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Most popular questions from this chapter

(a) Prove that for each real number \(x,(x+\sqrt{2})\) is irrational or \((-x+\sqrt{2})\) is irrational. (b) Generalize the proposition in Part (a) for any irrational number (instead of just \(\sqrt{2}\) ) and then prove the new proposition.

Consider the following proposition: Proposition. For all integers \(m\) and \(n,\) if \(n\) is odd, then the equation $$ x^{2}+2 m x+2 n=0 $$ has no integer solution for \(x\). (a) What are the solutions of the equation when \(m=1\) and \(n=-1 ?\) That is, what are the solutions of the equation \(x^{2}+2 x-2=0 ?\) (b) What are the solutions of the equation when \(m=2\) and \(n=3\) ? That is, what are the solutions of the equation \(x^{2}+4 x+6=0 ?\) (c) Solve the resulting quadratic equation for at least two more examples using values of \(m\) and \(n\) that satisfy the hypothesis of the proposition. (d) For this proposition, why does it seem reasonable to try a proof by contradiction? (e) For this proposition, state clearly the assumptions that need to be made at the beginning of a proof by contradiction. (f) Use a proof by contradiction to prove this proposition.

Prove the following proposition: Let \(a\) and \(b\) be integers with \(a \neq 0\). If \(a\) does not divide \(b\), then the equation \(a x^{3}+b x+(b+a)=0\) does not have a solution that is a natural number. Hint: It may be necessary to factor a sum of cubes. Recall that $$ u^{3}+v^{3}=(u+v)\left(u^{2}-u v+v^{2}\right) $$

(Exercise (17), Section 3.2) Let \(a\) and \(b\) be natural numbers such that \(a^{2}=\) \(b^{3} .\) Prove each of the propositions in Parts (6a) through (6d). (The results of Exercise (1) and Theorem 3.10 from Section 3.2 may be helpful.) (a) If \(a\) is even, then 4 divides \(a\). (b) If 4 divides \(a,\) then 4 divides \(b\). (c) If 4 divides \(b,\) then 8 divides \(a\). (d) If \(a\) is even, then 8 divides \(a\). (e) Give an example of natural numbers \(a\) and \(b\) such that \(a\) is even and \(a^{2}=b^{3},\) but \(b\) is not divisible by 8.

Recall that a Pythagorean triple consists of three natural numbers \(a, b,\) and \(c\) such that \(a

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