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Chapter 1: Problem 73

Using properties of inequalities, prove each of the statements. a) If \(x0,\) then \(\frac{1}{y}<\frac{1}{x}\) b) If \(x<0\frac{1}{x}\)

Short Answer

Expert verified
For (a), take reciprocals to get \(\frac{1}{x} > \frac{1}{y}\). For (b), since \(\frac{1}{x}\) is negative and \(\frac{1}{y}\) is positive, \(\frac{1}{y} > \frac{1}{x}\).

Step by step solution

01

Analyze the inequality for Part (a)

To prove \(x0\), notice that \(0<x<y\). Since \(x\) and \(y\) are positive, we can take reciprocals without changing the direction of the inequality.
02

Apply the property of reciprocals for Part (a)

Taking the reciprocal of the inequality \(x\frac{1}{y}\). This is the desired conclusion.
03

Analyze the inequality for Part (b)

For \(x<00\). Particularly, the sign change for \(x\) when considered in the denominator (as \(\frac{1}{x}\)) is crucial.
04

Determine the inequalities for Part (b)

Since \(x<0\) and \(y>0\), the reciprocal \(\frac{1}{x}\) is negative. Meanwhile, \(\frac{1}{y}\) is positive because \(y>0\). Therefore, \(\frac{1}{y}>\frac{1}{x}\). This is the desired conclusion.

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Key Concepts

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

Reciprocal Inequalities
When dealing with reciprocal inequalities, we're essentially talking about the relationship between the reciprocals of numbers and how this affects inequalities. If you know that two numbers, say `x` and `y`, satisfy an inequality like \( x < y \) and they are both positive, you can find an interesting property with their reciprocals. Here鈥檚 where it gets intriguing!
  • For positive numbers, if \( x < y \), flipping to reciprocals reverses the inequality, so \( \frac{1}{x} > \frac{1}{y} \).
  • This happens because smaller positive numbers have larger reciprocals. Let鈥檚 think about numbers like 1 and 2. Since 1 is less than 2, it follows that \( \frac{1}{1} = 1 \) is greater than \( \frac{1}{2} = 0.5 \).
This property is particularly useful in mathematics, especially when solving inequalities involving fractions and rational expressions.
Inequality Reversal
The trick with inequality reversal lies in understanding when and why we switch the direction of an inequality. This concept pops up notably with two types of operations: multiplication or division by a negative number and taking reciprocals of positive values.
  • When multiplying or dividing both sides of an inequality by a negative number, the inequality sign flips to maintain the truth of the statement.
  • For instance, if \( -x < -y \) with positive \( x \) and \( y \), dividing by -1 gives us \( x > y \).
When working with reciprocals of positive numbers, like in the problem鈥檚 case, if \( x < y \) and both are greater than zero, swapping to \( \frac{1}{x} \) and \( \frac{1}{y} \) means the inequality becomes \( \frac{1}{x} > \frac{1}{y} \). The reversal here happens precisely because of the positive nature of our numbers.
Positive and Negative Fractions
Understanding how positive and negative numbers behave as fractions is crucial for mastering inequalities. Let's decode this in simpler terms.
  • Negative fractions result from a negative numerator or denominator, impacting how they compare to positive fractions.
  • For any positive number \( y \), \( \frac{1}{y} \) is definitely positive, since dividing by anything positive remains positive.
  • Conversely, with a negative number \( x \), \( \frac{1}{x} \) plunges into negative territory.
In situations like the given exercise, understanding that any comparison between a negative (like \( \frac{1}{x} \)) and a positive (like \( \frac{1}{y} \)) fraction naturally leads to the positive being greater forms the backbone of solving these inequalities efficiently.

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Most popular questions from this chapter

Rationalize the denominator of each fractional expression. $$\frac{1}{\sqrt{x+h}+\sqrt{x}}$$

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