Problem 72 - Algebra and Trigonometry

Question

Suppose that \(a, b, c,\) and \(d\) are positive numbers such that $$ \frac{a}{b}<\frac{c}{d} $$ show that $$ \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d} $$

Step-by-Step Solution

Answer

The inequality \( \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d} \) is true using cross-multiplication.

Step 1: Express Inequality

The given inequality is \( \frac{a}{b} < \frac{c}{d} \). We need to prove that \( \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d} \).

Step 2: Cross-Multiplication for First Inequality

To show \( \frac{a}{b} < \frac{a+c}{b+d} \), cross-multiply to compare: \( a(b+d) < b(a+c) \). Simplifying, you get \( ab + ad < ab + bc \) which implies \( ad < bc \). This is true since \( \frac{a}{b} < \frac{c}{d} \).

Step 3: Cross-Multiplication for Second Inequality

Next, show \( \frac{a+c}{b+d} < \frac{c}{d} \). Cross-multiply to find \( (a+c)d < c(b+d) \). Simplifying gives \( ad + cd < bc + cd \), which simplifies to \( ad < bc \), and we know this is true.

Step 4: Conclude the Compound Inequality

Since both \( \frac{a}{b} < \frac{a+c}{b+d} \) and \( \frac{a+c}{b+d} < \frac{c}{d} \) have been shown to be true under the condition \( \frac{a}{b} < \frac{c}{d} \), the double inequality \( \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d} \) holds.

Key Concepts

Cross-Multiplication

Cross-multiplication is a helpful method to compare two fractions or rational expressions. It involves multiplying the numerator of one fraction by the denominator of the other, and vice versa. By utilizing this approach, you simplify comparisons without needing to compute actual division.

Imagine you have two fractions: \( \frac{a}{b} \) and \( \frac{c}{d} \). To compare these using cross-multiplication, you would compute \( ad \) and \( bc \). If \( ad < bc \), it can be concluded that \( \frac{a}{b} < \frac{c}{d} \).

This technique is especially useful in situations where finding a common denominator is complicated or impractical. By cross-multiplying, you're leveraging arithmetic to maintain the inequality relationship without altering the fractions' values.

Compound Inequality

A compound inequality is an expression with two or more inequalities joined together. It shows that a value must satisfy all inequalities. For example, in the statement \( \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d} \), there are two separate inequalities:

• \( \frac{a}{b} < \frac{a+c}{b+d} \)
• \( \frac{a+c}{b+d} < \frac{c}{d} \)

Both parts must be true for the entire statement to hold.

These types of inequalities help describe a range of values that satisfy a given condition. They are widely used in algebra to better understand the possible values arising from scenarios described by inequalities. Working through them can often involve solving each part separately then finding the intersection of solutions to achieve an overall answer.

Rational Expressions

Rational expressions are fractions in which the numerator and/or the denominator are polynomials. They follow the same rules as regular fractions, but operations might involve additional complexities due to polynomial behavior.

When dealing with rational expressions, it is important to be cautious of undefined values. These occur when the denominator is equal to zero. As a rule, always check for values that make the denominator zero, as these need to be excluded from the solution set.

In many algebraic problems, rational expressions present a higher level of mathematical challenge. Simplifying them often requires factoring, finding least common denominators, or using techniques like cross-multiplication. Despite their complexity, mastering rational expressions enhances problem-solving abilities and understanding of functional relationships within algebra.