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Prove each of the following: (a) For each nonzero real number \(x,\left|x^{-1}\right|=\frac{1}{|x|}\). (b) For all real numbers \(x\) and \(y,|x-y| \geq|x|-|y|\) Hint: An idea that is often used by mathematicians is to add 0 to an expression "intelligently". In this case, we know that \((-y)+y=0\). Start by adding this "version" of 0 inside the absolute value sign of \(|x|\). (c) For all real numbers \(x\) and \(y, \| x|-| y|| \leq|x-y|\).

Short Answer

Expert verified
In summary, we have proved the following three statements: (a) Let x be a nonzero real number, then |x^(-1)| = 1/|x|. (b) For all real numbers x and y, |x-y| ≥ |x|-|y|. (c) For all real numbers x and y, ||x|-|y|| ≤ |x-y|.

Step by step solution

01

Let x be a nonzero real number. Then, x^(-1) is the multiplicative inverse of x, meaning it satisfies the equation x(x^(-1)) = 1. #Step A2: Take the absolute value of both sides#

We know that |ab| = |a||b|. Therefore, applying this to the equation x(x^(-1)) = 1, we have |x||x^(-1)| = |1|. #Step A3: Simplify the equation and solve for |x^(-1)|#
02

Since |1| = 1, we have |x||x^(-1)| = 1. Now, divide both sides by |x| to get |x^(-1)| = 1/|x|. This completes our proof for statement (a). #Step B1: Start with the original inequality, and use the hint to add 0 intelligently#

For statement (b), we have |x - y|. Following the hint provided, we can write: |x - y| = |x + (-y + y) - y|. #Step B2: Use the triangle inequality and properties of absolute values#
03

The triangle inequality states that |a+b| ≤ |a|+|b|. Applying this to our re-arranged expression from step B1, we have: |x - y| = |(x-y) + (y)| ≥ |x-y| - |y|. #Step B3: Simplify further and complete the proof for statement (b)#

Since we have |x-y| ≥ |x-y| - |y| and it's clear that |x-y| + |y| = |x|, the inequality simplifies to |x-y| ≥ |x| - |y|, completing the proof for statement (b). #Step C1: Begin with the original inequality for statement (c)#
04

For statement (c), we want to prove that ||x| - |y|| ≤ |x-y|. #Step C2: Apply triangle inequality to |x - y| and simplify#

By applying triangle inequality to |x - y|, we have |x - y| ≥ ||x| - |y||, since |a+b| ≥ |a|-|b| is also a valid form of the triangle inequality. This completes the proof for statement (c), ||x|-|y|| ≤ | borderSideym|. Now, we have proved all three statements, (a), (b), and (c), as required.

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Key Concepts

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

Triangle Inequality
The triangle inequality is a foundational principle in mathematics, particularly in the study of metric spaces like the set of real numbers. It states that for any two real numbers, say, a and b, the absolute value of their sum is less than or equal to the sum of their absolute values, formally written as
\[ |a + b| \: \: \leq |a| + |b| \].
This principle gets its name from the geometric fact that in any triangle, the length of one side must be less than the sum of the lengths of the other two sides.

Understanding with an example

Consider the numbers 3 and -2. According to the triangle inequality: \[ |3 + (-2)| \: \: \leq |3| + |-2| \] which simplifies to \[ |1| \: \: \leq 3 + 2 \]. As 1 is indeed less than or equal to 5, the inequality holds. The triangle inequality is essential for proving various mathematical concepts, including those relevant to absolute value properties and the relationship between distances in geometry.
Absolute Value Properties
The absolute value measures the distance of a number from zero on the real number line, irrespective of direction. It's denoted by two vertical bars surrounding the number or expression, such as \( |x| \). Some key properties of absolute values involve their interaction with other numbers:
  • Non-negativity: \( |x| \: \: \geq 0 \) for any real number \( x \).
  • Identity: \( |x| = x \) if \( x \: \: \geq 0 \) and \( |x| = -x \) if \( x < 0 \).
  • Product: \( |ab| = |a||b| \), meaning the absolute value of a product is the product of the absolute values.
  • Quotient: \( |\frac{a}{b}| = \frac{|a|}{|b|} \), as long as \( b \) is not zero.

These properties are vital when we manipulate equations and inequalities in algebra. They underpin proofs, like the one in the exercise which demonstrates the relationship between the absolute value of a reciprocal and the reciprocal of an absolute value.
Real Numbers
Real numbers are the collection of all rational and irrational numbers, encompassing a wide array of value types including integers, fractions, decimals, and numbers like \( \pi \) and \( \sqrt{2} \). They can be plotted on a one-dimensional line, known as the real number line.
  • Rational Numbers: Can be expressed as the quotient of two integers, like \( \frac{3}{4} \) or \( -\frac{5}{2} \).
  • Irrational Numbers: Cannot be written as exact fractions, such as \( \pi \) or \( \sqrt{3} \), and have non-repeating, non-terminating decimal expansions.

Understanding real numbers is crucial for grasping concepts in algebra, such as solving equations and inequalities, and in calculus, such as finding limits and derivatives. The set of real numbers is fundamental to all of mathematical analysis.
Inequalities
Inequalities express the relationship between two values when they are not equal. An inequality indicates whether one number is greater than, less than, greater than or equal to, or less than or equal to another number, using symbols like \( > \), \( < \), \( \: \: \geq \), and \( \: \: \leq \).
When working with inequalities, it's important to remember:
  • Adding or subtracting the same number from both sides of an inequality does not change the direction of the inequality. For example, if \( a > b \), then \( a + c > b + c \) for any real number \( c \).
  • Multiplying or dividing both sides by a positive number preserves the inequality direction, while doing the same with a negative number reverses it. For instance, if \( a \) is less than \( b \), and \( c < 0 \), then \( ac \) is greater than \( bc \).

Understanding how inequalities work with the absolute value is particularly important, as it helps us handle absolute differences and relationships, as shown in parts (b) and (c) of the exercise.

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Most popular questions from this chapter

(a) Use cases based on congruence modulo 3 and properties of congruence to prove that for each integer \(n, n^{3}=n(\mathrm{mod} 3)\). (b) Explain why the result in Part (a) proves that for each integer \(n, 3\) divides \(\left(n^{3}-n\right) .\) Compare this to the proof of the same result in Proposition 3.27

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) For all nonzero integers \(a\) and \(b,\) if \(a+2 b \neq 3\) and \(9 a+2 b \neq 1,\) then the equation \(a x^{3}+2 b x=3\) does not have a solution that is a natural number. \mathrm{P} \text { Proof } . We will prove the contrapositive, which is For all nonzero integers \(a\) and \(b\), if the equation \(a x^{3}+2 b x=3\) has a solution that is a natural number, then \(a+2 b=3\) or \(9 a+2 b=1\). So we let \(a\) and \(b\) be nonzero integers and assume that the natural number \(n\) is a solution of the equation \(a x^{3}+2 b x=3\). So we have $$ \begin{aligned} a n^{3}+2 b n &=3 \quad \text { or } \\ n\left(a n^{2}+2 b\right) &=3 \end{aligned} $$ So we can conclude that \(n=3\) and \(a n^{2}+2 b=1\). Since we now have the value of \(n\), we can substitute it in the equation \(a n^{3}+2 b n=3\) and obtain \(27 a+6 b=3\). Dividing both sides of this equation by 3 shows that \(9 a+2 b=1\). So there is no need for us to go any further, and this concludes the proof of the contrapositive of the proposition. (b) For all nonzero integers \(a\) and \(b,\) if \(a+2 b \neq 3\) and \(9 a+2 b \neq 1,\) then the equation \(a x^{3}+2 b x=3\) does not have a solution that is a natural number. We will use a proof by contradiction. Let us assume that there exist nonzero integers \(a\) and \(b\) such that \(a+2 b=3\) and \(9 a+2 b=1\) and \(a n^{3}+2 b n=3,\) where \(n\) is a natural number. First, we will solve one equation for \(2 b\); doing this, we obtain $$ \begin{aligned} a+2 b &=3 \\ 2 b &=3-a . \end{aligned} $$ We can now substitute for \(2 b\) in \(a n^{3}+2 b n=3\). This gives $$ \begin{array}{l} a n^{3}+(3-a) n=3 \\ a n^{3}+3 n-a n=3 \\ n\left(a n^{2}+3-a\right)=3 . \end{array} $$ By the closure properties of the integers, \(\left(a n^{2}+3-a\right)\) is an integer and, hence, equation (2) implies that \(n\) divides \(3 .\) So \(n=1\) or \(n=3\). When we substitute \(n=1\) into the equation \(a n^{3}+2 b n=3,\) we obtain \(a+2 b=3\). This is a contradiction since we are told in the proposition that \(a+2 b \neq 3\). This proves that the negation of the proposition is false and, hence, the proposition is true.

(a) Is the base 2 logarithm of \(32, \log _{2}(32),\) a rational number or an irrational number? Justify your conclusion. (b) Is the base 2 logarithm of \(3, \log _{2}(3),\) a rational number or an irrational number? Justify your conclusion.

(a) Use the result in Proposition 3.33 to help prove that the integer \(m=\) 5,344,580,232,468,953,153 is not a perfect square. Recall that an integer \(n\) is a perfect square provided that there exists an integer \(k\) such that \(n=k^{2} .\) Hint: Use a proof by contradiction. (b) Is the integer \(n=782,456,231,189,002,288,438\) a perfect square? Justify your conclusion.

Is the following statement true or false? For all positive real numbers \(x\) and \(y, \sqrt{x+y} \leq \sqrt{x}+\sqrt{y}\).

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