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Chapter 1: Problem 9

Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational.

Short Answer

Expert verified
The sum of an irrational number and a rational number is irrational.

Step by step solution

01

State the Claim

We want to prove that the sum of an irrational number and a rational number is irrational.
02

Assume the Opposite

Assume that the sum of an irrational number and a rational number is rational. We will call this sum and aim to find a contradiction.
03

Define Variables

Let the irrational number be denoted as \( x \) and the rational number as \( y \). Assume their sum is rational, say \( z \). Hence, we write: \[ x + y = z \]
04

Rearrange the Equation

Solve for the irrational number \( x \): \[ x = z - y \].
05

Analyze the Result

Since both \( z \) and \( y \) are rational numbers, their difference \( z - y \) must also be a rational number.
06

Reach a Contradiction

Our assumption leads to the conclusion that \( x \) is rational, which contradicts the original premise that \( x \) is irrational.
07

Conclude the Proof

Since assuming that the sum of an irrational number and a rational number is rational leads to a contradiction, it must be that the sum of an irrational number and a rational number is irrational. Therefore, the original claim is proven.

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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 numbers that cannot be expressed as a fraction of two integers. This means that they cannot be written in the form \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b eq 0 \).
They have non-repeating, non-terminating decimals. For example, the number \( \pi \) or the square root of 2 (\( \sqrt{2} \)) are both irrational.
  • They can't be written as simple fractions.
  • The decimal goes on forever without repeating.
To better understand this, you could think of \( \pi = 3.141592653589... \). There鈥檚 no repeating pattern, and it never ends.
rational numbers
Rational numbers are the opposite of irrational numbers. They can be expressed as a fraction of two integers, \( \frac{a}{b} \), where both \( a \) and \( b \) are integers and \( b eq 0 \) . Some examples include \( \frac{1}{2} \), \( 4 \), and \( -7 \), all of which can be written as fractions.
Typical rational numbers include:
  • Whole numbers: 0, 1, 2, 3...
  • Fractions: \( \frac{3}{4} \), \( \frac{2}{5} \)
  • Terminating decimals: 0.5, 3.75
  • Repeating decimals: 0.333鈥�, 1.272727...
A key property of rational numbers is that they are either terminating or repeating decimals. This makes them easier to deal with in many cases because they have a predictable pattern.
contradiction
Contradiction is a method used in proofs to demonstrate that a statement must be true because the assumption that it is false leads to an illogical conclusion. Here's how it works in steps:
  • Assume the opposite of what you want to prove.
  • Show that this assumption leads to a contradiction or an impossibility.
  • Conclude that the original statement must be true since the opposite is false.
For example, in the provided exercise, we assumed that the sum of an irrational number and a rational number was rational. We then showed that this assumption led to a contradiction since it implied that an irrational number was rational. Because this is impossible, we concluded that the sum of an irrational number and a rational number must be irrational.

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Most popular questions from this chapter

A statement is in prenex normal form (PNF) if and only if it is of the form $$ Q_{1} x_{1} Q_{2} x_{2} \cdots Q_{k} x_{k} P\left(x_{1}, x_{2}, \ldots, x_{k}\right) $$ where each \(Q_{i}, i=1,2, \ldots, k,\) is either the existential quantifier or the universal quantifier, and \(P\left(x_{1}, \ldots, x_{k}\right)\) is a predicate involving no quantifiers. For example, \(\exists x \forall y(P(x, y) \wedge Q(y))\) is in prenex normal form, whereas \(\exists x P(x) \vee \forall x Q(x)\) is not (because the quantifiers do not all occur first). Every statement formed from propositional variables, predicates, \(\mathbf{T},\) and \(\mathbf{F}\) using logical connectives and quantifiers is equivalent to a statement in prenex normal form. Exercise 51 asks for a proof of this fact. Show how to transform an arbitrary statement to a statement in prenex normal form that is equivalent to the given statement. (Note: A formal solution of this exercise requires use of structural induction, covered in Section \(5.3 . )\)

Use a proof by cases to show that 10 is not the square of a positive integer. [Hint: Consider two cases: \((i) 1 \leq x \leq 3\) , (ii) \(x \geq 4.1\)

Show that the propositions \(p_{1}, p_{2}, p_{3}, p_{4},\) and \(p_{5}\) can be shown to be equivalent by proving that the conditional statements \(p_{1} \rightarrow p_{4}, p_{3} \rightarrow p_{1}, p_{4} \rightarrow p_{2}, p_{2} \rightarrow p_{5}\) and \(p_{5} \rightarrow p_{3}\) are true.

Construct a truth table for each of these compound propositions. a) \(p \rightarrow(\neg q \vee r)\) b) \(\neg p \rightarrow(q \rightarrow r)\) c) \((p \rightarrow q) \vee(\neg p \rightarrow r)\) d) \((p \rightarrow q) \wedge(\neg p \rightarrow r)\) e) \((p \rightarrow q) \vee(\neg q \rightarrow r)\) f) \((\neg p \leftrightarrow \neg q) \leftrightarrow(q \leftrightarrow r)\)

Suppose that the domain of the propositional function \(P(x)\) consists of \(-5,-3,-1,1,3,\) and \(5 .\) Express these statements without using quantifiers, instead using only negations, disjunctions, and conjunctions. a) \(\exists x P(x)\) b) \(\quad \forall x P(x)\) c) \(\forall x((x \neq 1) \rightarrow P(x))\) d) \(\exists x((x \geq 0) \wedge P(x))\) e) \(\exists x(\neg P(x)) \wedge \forall x((x<0) \rightarrow P(x))\)
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