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Chapter 3: Problem 21

For all real numbers \(x\), if \(x>1\) then \(x^{2}>x\).

Short Answer

Expert verified
To prove that for all real numbers x > 1, x^2 > x, we start by assuming x > 1 and then subtract 1 from both sides, obtaining x - 1 > 0. Next, we multiply both sides by (x + 1), which is also positive, giving (x - 1)(x + 1) > 0. Expanding this, we find x^2 - 1 > 0, or x^2 > 1. Since x > 1, it follows that x^2 > 1 > x, proving that x^2 > x for all real numbers x > 1.

Step by step solution

01

Recall the properties of inequalities

We should first recall that inequalities, like equations, have their own set of rules or properties. Some of these properties are: 1. If a > b, then a+c > b+c for all real numbers a, b, and c. 2. If a > b and c > 0, then a * c > b * c for all real numbers a, b, and c.
02

Assume x is a real number greater than 1

Let x be any real number such that x > 1. We want to show that x^2 > x.
03

Subtract 1 from both sides of the inequality

If x > 1, we can subtract 1 from both sides of the inequality without changing its direction: x - 1 > 0
04

Multiply both sides by (x + 1)

We know that x > 1, so x + 1 > 2. Therefore, x + 1 is greater than 0. Now, we can multiply both sides of the inequality (x - 1 > 0) by (x + 1): (x - 1)(x + 1) > 0
05

Expand (x - 1)(x + 1)

Let's expand the left side of the inequality (x - 1)(x + 1): x^2 - 1 > 0
06

Add 1 to both sides of the inequality

Add 1 to both sides of the inequality to isolate x^2: x^2 > 1
07

Observe and conclude

We started with the assumption that x > 1. After performing algebraic operations, we reached the conclusion that x^2 > 1. Since x > 1, it follows that x^2 > 1 > x. Thus, x^2 > x, which is what we needed to prove for all real numbers x > 1.

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Key Concepts

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

Real Numbers
Real numbers are a fundamental part of mathematics and include all the numbers on the number line. They encompass all the rational numbers like fractions and integers, as well as the irrational numbers.
A crucial aspect of real numbers is their ability to be ordered, meaning we can compare them to determine which is greater or smaller. This property is particularly useful in solving inequalities.
Knowing that a number is real also assures us it behaves consistently under operations like addition, subtraction, multiplication, and division (except by zero).
  • Rational numbers such as 2, 7.5, and -3/4.
  • Irrational numbers like \(\sqrt{2}\) and \(\pi\).
  • Real numbers can be compared: either \(a > b\), \(a < b\), or \(a = b\).
Understanding these concepts helps us explore further mathematical theories and applications.
Properties of Inequalities
Inequalities have rules that are similar but distinct from equations. These properties allow us to manipulate inequalities and make valid conclusions. When dealing with inequalities in real numbers, you can perform certain operations without disrupting the inequality.
A few key properties of inequalities include:
  • If \(a > b\), then adding or subtracting the same number \(c\) keeps the sense of the inequality: \(a+c > b+c\).
  • If \(a > b\) and \(c > 0\), then multiplying both sides by \(c\) keeps the inequality direction: \(a \cdot c > b \cdot c\).
  • If \(a > b\) and \(c < 0\), multiplying by \(c\) reverses the inequality: \(a \cdot c < b \cdot c\).
These properties enable consistent and predictable results while simplifying more complex mathematical expressions. They help in solving real-world problems where such inequalities frequently arise.
Algebraic Manipulation
Algebraic manipulation refers to the process of transforming equations or inequalities to achieve a desired form, allowing easier analysis or solution.
In the context of inequalities, algebraic manipulation often involves adding, subtracting, multiplying, or dividing both sides by the same value, just like in equations. However, one must always respect the properties of inequalities to maintain the correct direction of the inequality.
  • We can add or subtract: Transforming \(x > 1\) to \(x - 1 > 0\) without altering inequality direction.
  • Multiplying factors: By multiplying both sides of \(x - 1 > 0\) by a positive \(x + 1\), we get \((x - 1)(x + 1) > 0\).
  • Expansion: Distributing terms helps express inequalities in simpler forms, like expanding \((x - 1)(x + 1)\) to \(x^2 - 1\).
These steps illustrate how to navigate through inequalities carefully, ensuring each move adheres to the mathematical rules, leading to a valid conclusion.

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Most popular questions from this chapter

Definition: The least common multiple of two nonzero integers \(a\) and \(b\), denoted \(\operatorname{lcm}(a, b)\), is the positive integer \(c\) such that a. \(a \mid c\) and \(b \mid c\) b. for all integers \(m\), if \(a \mid m\) and \(b \mid m\), then \(c \mid m\). Prove that for all integers \(a\) and \(b, \operatorname{gcd}(a, b) \mid \operatorname{lcm}(a, b)\).

(Two integers are consecutive if, and only if, one is one more than the other.) Any product of four consecutive integers is one less than a perfect square.

Use proof by contradiction to show that for all integers \(m\), \(7 m+4\) is not divisible by 7 .

If \(r\) is any rational number and \(s\) is any irrational number, then \(r / s\) is irrational.

Let \(N=2 \cdot 3 \cdot 5 \cdot 7+1\). What remainder is obtained when \(N\) is divided by 2 ? 3 ? 5 ? 7 ? Is \(N\) prime? Justify your answer.
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