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Chapter 2: Problem 15

(a) Prove that if \(x=p+\sqrt{q}\) where \(p\) and \(q\) are rational, and \(m\) is a natural number, then \(x^{m}=a+b \sqrt{q}\) for some rational \(a\) and \(b\) (b) Prove also that \((p-\sqrt{q})^{m}=a-b \sqrt{q}\)

Short Answer

Expert verified
(a) Proven by induction that \( x^m = a + b \sqrt{q} \). (b) Proven by induction that \( (p - \sqrt{q})^m = a - b \sqrt{q} \).

Step by step solution

01

Understanding the Base Case

Consider the base case when m = 1: \[ x = p + \sqrt{q} \] Clearly, this is of the form \( a + b \sqrt{q} \) where \( a = p \) and \( b = 1 \).
02

Induction Hypothesis

Assume that for some natural number k, we have \[ x^k = a_k + b_k \sqrt{q}, \] where \( a_k \) and \( b_k \) are rational numbers.
03

Induction Step

To prove the case for \( m = k + 1 \), consider \[ x^{k+1} = x^k \cdot x \] Using the induction hypothesis, we get \[ (a_k + b_k \sqrt{q})(p + \sqrt{q}) = a_k p + a_k \sqrt{q} + b_k p \sqrt{q} + b_k q \] Simplify this to \[ a_{k+1} + b_{k+1} \sqrt{q}, \] where \[ a_{k+1} = a_k p + b_k q \] and \[ b_{k+1} = a_k + b_k p \] are both rational. Therefore, the form holds for \( m = k + 1 \).
04

Conclusion of Part (a)

By induction, it follows that for any natural number \( m \), \[ x^m = a_m + b_m \sqrt{q}, \] where \( a_m \) and \( b_m \) are rational.
05

Base Case for Part (b)

Consider the base case when \( m = 1 \): \[ p - \sqrt{q} \] This is of the form \( a - b \sqrt{q} \) where \( a = p \) and \( b = 1 \).
06

Induction Hypothesis for Part (b)

Assume that for some natural number \( k \), we have \[ (p - \sqrt{q})^k = a_k - b_k \sqrt{q}, \] where \( a_k \) and \( b_k \) are rational.
07

Induction Step for Part (b)

To prove the case for \( m = k + 1 \), consider \[ (p - \sqrt{q})^{k+1} = (p - \sqrt{q})^k (p - \sqrt{q}). \] Using the induction hypothesis, we get \[ (a_k - b_k \sqrt{q})(p - \sqrt{q}) = a_k p - a_k \sqrt{q} - b_k p \sqrt{q} + b_k q \] Simplify this to \[ a_{k+1} - b_{k+1} \sqrt{q}, \] where \[ a_{k+1} = a_k p + b_k q \] and \[ b_{k+1} = a_k + b_k p \]. Both expressions are rational, thus the form holds for \( m = k + 1 \).
08

Conclusion of Part (b)

By induction, it follows that for any natural number \( m \), \[ (p - \sqrt{q})^m = a_m - b_m \sqrt{q}, \] where \( a_m \) and \( b_m \) are rational.

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Key Concepts

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

Rational Numbers
Rational numbers are numbers that can be written as the ratio of two integers. The general form is \(\frac{p}{q}\) where both p and q are integers and q is not zero. The decimal expansion of a rational number can be either terminating or repeating. Examples of rational numbers include 1/2, -3/4, and 5. Understanding rational numbers is crucial in algebra, as they form the basis for many mathematical operations and functions. In the given problem, both p and q are rational numbers, which means they can be expressed as fractions of integers.
Inductive Proof
Inductive proof is a technique in mathematics used to prove statements that are believed to be true for all natural numbers. It has two main steps:
  • Base Case: Verify the statement is true for the initial value, often 1.
  • Induction Step: Assume it's true for an arbitrary value k and then show it must be true for k+1.

The goal is to demonstrate that if the statement holds for one value, it must hold for the next. This process creates a domino effect, proving the statement for all natural numbers. In our exercise, we used induction to prove that the given expression holds for all natural powers m.
Natural Numbers
Natural numbers are the set of positive integers starting from 1. They are usually represented by \(\text{N}\). Natural numbers are foundational in mathematics and are used for counting and ordering. In problem-solving, natural numbers are often used in conjunction with proofs and algorithms. In our case, the natural number m represents the exponent in the expression, and we showed through induction that the given form holds for all such m.
Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and operations such as addition, subtraction, multiplication, and division. They represent a broad range of mathematical statements. An expression can be as simple as \(\text{p + q}\) or more complex like the expressions in our problem. Algebraic expressions are essential for modeling real-world situations and solving equations. In this exercise, we manipulated algebraic expressions to show that \((p + \sqrt{q})^m\) and \((p - \sqrt{q})^m\) fit a certain form for any natural number m.

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