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Use forward reasoning to show that if \(x\) is a nonzero real number, then \(x^{2}+1 / x^{2} \geq 2 .[\text { Hint: Start with the in- }\) equality \((x-1 / x)^{2} \geq 0,\) which holds for all nonzero real numbers \(x . ]\)

Short Answer

Expert verified
\(x^2 + \frac{1}{x^2} \geq 2\)

Step by step solution

01

- Start with the Given Hint

Begin with the inequality \[ (x - \frac{1}{x})^2 \geq 0 \] which is true for all nonzero real numbers \(x\) because the square of any real number is always non-negative.
02

- Expand the Square

Expand the left side of the inequality: \[ (x - \frac{1}{x})^2 = x^2 - 2 \cdot x \cdot \frac{1}{x} + (\frac{1}{x})^2 \] Simplify the terms to get: \[ x^2 - 2 + \frac{1}{x^2} \geq 0 \]
03

- Isolate the Desired Expression

Add 2 to both sides of the inequality: \[ x^2 - 2 + \frac{1}{x^2} + 2 \geq 0 + 2 \] Simplify to obtain: \[ x^2 + \frac{1}{x^2} \geq 2 \] This is the required inequality.

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Key Concepts

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

Nonzero Real Numbers
In this exercise, we deal with nonzero real numbers. A real number is any value that can represent a distance along a number line, including both positive and negative numbers, as well as zero. However, when we specify 'nonzero', we exclude zero from our range. That’s because we can’t divide by zero—such an operation is undefined in mathematics.
In our problem, we are asked to start with an inequality involving a nonzero real number, specifically: \((x - \frac{1}{x})^2 \geq 0\). To understand why this is always true for nonzero \(x\), recall that the square of any real number (positive or negative) is always non-negative. Therefore, this inequality will hold for any nonzero value of \(x\).
Algebraic Manipulations
Algebraic manipulation is the process of rearranging and simplifying expressions to get a desired form or solution. Here's how we apply that to our given problem:
  • Starting with the inequality: \((x - \frac{1}{x})^2 \geq 0\)
  • We expand the left side: \((x - \frac{1}{x})^2 = x^2 - 2 \cdot x \cdot \frac{1}{x} + (\frac{1}{x})^2 \)
After simplifying, we get: \x^2 - 2 + \frac{1}{x^2} \geq 0\. Now, to isolate the desired expression, we add 2 to both sides of the inequality:
\ x^2 - 2 + \frac{1}{x^2} + 2 \geq 0 + 2 \ which simplifies to: \ x^2 + \frac{1}{x^2} \geq 2 \.
This step-by-step algebraic manipulation helps us derive the required inequality from the original one.
Inequations
Inequations, often referred to as inequalities, are mathematical statements that are used to compare the sizes of two values or expressions. Unlike equations, which state that two expressions are equal, inequations indicate that quantities are not necessarily equal but have a specific order relation (greater than, less than, etc.).
In our example, the original inequation \((x - \frac{1}{x})^2 \geq 0\) uses the 'greater than or equal to' relation to demonstrate that the square of any real number is non-negative. This is fundamental because it allows us to safely perform algebraic manipulations to derive the conclusion \x^2 + \frac{1}{x^2} \geq 2\. By understanding and applying properties of real numbers and inequalities, we can demonstrate this important mathematical truth effectively.

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Most popular questions from this chapter

Use a proof by exhaustion to show that a tiling using dominoes of a \(4 \times 4\) checkerboard with opposite corners removed does not exist. [Hint: First show that you can assume that the squares in the upper left and lower right corners are removed. Number the squares of the original checkerboard from 1 to \(16,\) starting in the first row, moving right in this row, then starting in the leftmost square in the second row and moving right, and so on. Remove squares 1 and \(16 .\) To begin the proof, note that square 2 is covered either by a domino laid horizontally, which covers squares 2 and \(3,\) or vertically, which covers squares 2 and \(6 .\) Consider each of these cases separately, and work through all the subcases that arise. \(]\)

Prove that given a real number \(x\) there exist unique numbers \(n\) and \(\epsilon\) such that \(x=n-\epsilon, n\) is an integer, and \(0 \leq \epsilon<1 .\)

Formulate a conjecture about the final two decimal digits of the square of an integer. Prove your conjecture using a proof by cases.

Exercises \(48-51\) establish rules for null quantification that we can use when a quantified variable does not appear in part of a statement. Establish these logical equivalences, where \(x\) does not occur as a free variable in \(A\) . Assume that the domain is nonempty. $$ \begin{array}{l}{\text { a) } \forall x(A \rightarrow P(x)) \equiv A \rightarrow \forall x P(x)} \\ {\text { b) } \exists x(A \rightarrow P(x)) \equiv A \rightarrow \exists x P(x)}\end{array} $$

Exercises \(40-44\) deal with the translation between system specification and logical expressions involving quantifiers. Translate these system specifications into English, where the predicate \(S(x, y)\) is \(^{\prime} x\) is in state \(y\) " and where the domain for \(x\) and \(y\) consists of all systems and all possible states, respectively. $$ \begin{array}{l}{\text { a) } \exists x S(x, \text { open })} \\ {\text { b) } \forall x(S(x, \text { malfunctioning }) \vee S(x, \text { diagnostic) }} \\\ {\text { c) } \exists x S(x, \text { open }) \vee \exists x S(x, \text { diagnostic) }}\end{array} $$ $$ \begin{array}{l}{\text { d) } \exists x \rightarrow S(x, \text { available })} \\\ {\text { e) } \forall x \neg S(x, \text { working })}\end{array} $$

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