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Chapter 1: Problem 20

(a) If \(0n^{2}\) and \(1 / m>1 / n\)

Short Answer

Expert verified
For the given inequality \(m<n\), if both m and n are positive, squaring both sides maintains the inequality direction, and taking their reciprocals reverses the direction. On the other hand, if both are negative, squaring reverses the inequality direction and taking the reciprocals also reverses the direction.

Step by step solution

01

Analyzing the first part of the exercise with positive integers

The first part of the exercise states that if \(0<m<n\), then \(m^{2}<n^{2}\) and \(1 / n<1 / m\). If we multiply both sides of the inequality \(m<n\) with m or n, since both m and n are positive, the results will still hold true. Similarly, since they are in the denominator, if m is less than n, then \(1 / m\) is greater than \(1 / n\).
02

Proving the first part of the exercise

Given that \(0<m<n\), if we square both sides \(m^{2}<n^{2}\). Similarly, if we take reciprocal of both sides, since they are positive, \(1 / n<1 / m\).
03

Analyzing the second part of the exercise with negative integers

The second part of the exercise states that if \(mn^{2}\) and \(1 / m>1 / n\). Here, multiplying the inequality by itself will reverse the sign of the inequality because m and n are negative. Also, since they are in the denominator, m is more negative (or less) than n, then \(1 / m\) is greater than \(1 / n\).
04

Proving the second part of the exercise

Given that \(mn^{2}\), because if we square a more negative number, it becomes more positive. Similarly, if we take reciprocal of both sides, since they are negative, \(1 / m>1 / n\).

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Key Concepts

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

Positive and Negative Integers
Understanding the properties of positive and negative integers is crucial when dealing with inequalities.
The set of integers is composed of zero, positive numbers (1, 2, 3, ...), and negative numbers (-1, -2, -3, ...). Any number without a decimal or fractional part is an integer.
Positive integers are greater than zero and increase in value as they move away from zero (+1, +2, +3, ...). They are used to count or measure things that can only exist in whole amounts.
Negative integers, on the other hand, are less than zero and decrease in value as they move further below zero (-1, -2, -3, ...). They often represent debts or deficits.
  • A positive multiplied by a positive, or a negative multiplied by a negative, results in a positive.
  • A positive multiplied by a negative results in a negative.
  • The product of any number and zero is zero.
Inequalities involving integers rely on these fundamental rules, and as showcased in the provided exercise, manipulating these integers by various operations (like squaring or taking reciprocals) produces different outcomes depending on their sign.
Inequalities
Inequalities represent the relation between two values showing that one is less than, greater than, equal to, or not equal to another value.
In mathematics, common inequality symbols are <, >, \(鈮) (less than or equal to), and \(鈮) (greater than or equal to). They are crucial in the study of order and structure within the set of real numbers and have profound implications in various fields such as economics, physics, and statistics.
Handling inequalities requires attention to operations performed on them. For example:
  • If we add or subtract the same number from both sides of an inequality, the relation does not change.
  • When we multiply or divide both sides of an inequality by a positive number, the inequality remains the same.
  • However, multiplying or dividing both sides of an inequality by a negative number will switch the direction of the inequality (e.g., from < to >).
With these rules, the given exercise is solved by careful square multiplication of the integer values and taking their reciprocals, keeping in mind the behavior of inequalities under these operations.
Reciprocal Relations
In mathematics, the reciprocal of a number is simply one divided by that number. For instance, the reciprocal of 2 is \(\frac{1}{2}\), and the reciprocal of \(\frac{1}{2}\) is 2.
Reciprocals are important when dealing with division and fractions. One of the essential properties of reciprocals is the fact that a number times its reciprocal always equals 1 (\(x * \frac{1}{x} = 1\)).
Reciprocal relations influence inequalities in an interesting way.
  • If we have two positive numbers where one is greater than the other (e.g., \(a > b > 0\)), then their reciprocals will flip the inequality: \(\frac{1}{b} > \frac{1}{a}\).
  • For negative numbers, since they are less than zero, the relationships switch. So, for \(a < b < 0\), \(\frac{1}{a} > \frac{1}{b}\) holds true.
This concept is used in the exercise's second scenario, revealing the effects of taking reciprocals of negative integers within an inequality.

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Most popular questions from this chapter

Let \(\left\\{a_{n}\right.\) ' be any sequence of numbers converging to 0 , and let \(\sigma_{n}\) be the sequence of arithmetic means (averages). $$ \sigma_{n}=\begin{gathered} a_{1}+a_{2}+a_{3}+\cdots+a_{n} \\ n \end{gathered} $$ Prove that \(\lim _{n \rightarrow \infty} \sigma_{n}=0\).

Show that \(\left|p_{1}+p_{2}+p_{3}+\cdots+p_{n}\right| \leq\left|p_{1}\right|+\left|p_{2}\right|+\cdots+\left|p_{n}\right|\)

In the triangle \(A B C\). join \(A\) to a point 4 of the way from \(B\) toward \(C\), join \(B\) to a point \(\frac{1}{3}\) of the way from \(C\) toward \(A\), and join \(C\) to a point \(f\) of the way from \(A\) toward \(B\). Express the vertices of the smaller triangle thus formed in terms of \(A, B\), and \(C\).

What are the cluster points for the set $$ S=\left\\{\text { all }\left(\begin{array}{ll} 1 & 1 \\ n & m \end{array}\right) \text { with } n=1,2, \ldots, m=1,2, \ldots\right. $$

Prove that \(\lim _{n \rightarrow x} c^{1 / n}=1\) for any \(c>1\) by setting \(a_{n}=c^{1 / n}-1\), and then deriving the estimate \(0 \leq a_{n} \leq(c-1) / n\)
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