Problem 84 - Calculus with Concepts in Calculus

Question

Prove that if \(a

Step-by-Step Solution

Answer

The arithmetic mean \(\frac{a+b}{2}\) lies between \(a\) and \(b\) when \(a

Step 1: Understand the Given Inequality

We need to prove that if \(a

Step 2: Prove the First Part of the Inequality

From the condition \(a

Step 3: Prove the Second Part of the Inequality

Using the original condition \(a

Step 4: Conclusion

Since we have shown that \(a < \frac{a+b}{2}\) and \(\frac{a+b}{2} < b\), it follows that \(a < \frac{a+b}{2} < b\). Therefore, the arithmetic mean \(\frac{a+b}{2}\) is indeed between \(a\) and \(b\) on the real line.

Key Concepts

Understanding Inequalities

Inequalities are expressions used to show the relationship between two values where they are not equal. They are critical in mathematics to help us understand how numbers compare to one another. In the given exercise, we need to show that the arithmetic mean of two numbers is between these numbers by using inequalities effectively.
One important aspect of inequalities is their properties, such as adding or multiplying both sides by the same number. For instance, if we have two real numbers such that \(a < b\), we can add the same number to both sides to maintain the inequality, like in this case: \(a + a < a + b\). This property is central to demonstrating that the arithmetic mean is between the numbers.
By applying these principles, you can deduce that \(a < \frac{a+b}{2} < b\), showing a clear order on the real line and helping illustrate the concept of arithmetic mean.

Role of Real Numbers

Real numbers include all the rational and irrational numbers. They form the complete set of numbers that can be represented on the number line. In this context, they are crucial as they ensure that \(a\) and \(b\) in our exercise can be any numbers on the continuum of real numbers, not just integers or fractions.
Understanding real numbers is fundamental when dealing with arithmetic mean, because the mean, or average, of two numbers needs the sum and division to make sense over the continuous scale of real numbers.
When you compute \(\frac{a+b}{2}\), you're essentially finding a point on this line that is exactly halfway between \(a\) and \(b\), underlining how versatile real numbers are in mathematical expressions and solutions.

Constructing a Proof

Constructing proofs in mathematics is a methodical process that relies on logical reasoning. In proving the inequality \(a < \frac{a+b}{2} < b\), we follow a series of steps to validate our assumptions and reach a concrete conclusion.
Start by understanding the inequality you need to prove: first, we show that \(a < \frac{a+b}{2}\); then, \(\frac{a+b}{2} < b\). Each segment of the proof uses the foundational assumption \(a < b\) to assert both parts of the inequality.
The process includes clearly stating each logical step, using algebraic manipulation like adding the same number to both sides or dividing an inequality to maintain its truth. By systematically tackling each part of the problem, the proof eventually shows that the arithmetic mean lies perfectly between the two real numbers, demonstrating a simple yet powerful callback to the fundamental essence of inequalities on real numbers.

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Problem 83

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Problem 85

Other exercises in this chapter

Problem 81

Use the definition of absolute value to prove the following. a. \(|a b|=|a||b|\) b. \(-|b| \leq b \leq|b|\) c. \(|a-b|=|b-a|\)

View solution

Problem 83

Show that \(|a+b|=|a|+|b|\) if and only if \(a b \geq 0\) (which means that \(a=0, b=0\), or \(a\) and \(b\) have the same sign).

View solution

Problem 85

Prove that if \(00 .\right)\)

View solution

Problem 87

Let \(0

View solution

Key Concepts

Understanding Inequalities

Inequalities are expressions used to show the relationship between two values where they are not equal. They are critical in mathematics to help us understand how numbers compare to one another. In the given exercise, we need to show that the arithmetic mean of two numbers is between these numbers by using inequalities effectively.
One important aspect of inequalities is their properties, such as adding or multiplying both sides by the same number. For instance, if we have two real numbers such that \(a < b\), we can add the same number to both sides to maintain the inequality, like in this case: \(a + a < a + b\). This property is central to demonstrating that the arithmetic mean is between the numbers.
By applying these principles, you can deduce that \(a < \frac{a+b}{2} < b\), showing a clear order on the real line and helping illustrate the concept of arithmetic mean.

Role of Real Numbers

Real numbers include all the rational and irrational numbers. They form the complete set of numbers that can be represented on the number line. In this context, they are crucial as they ensure that \(a\) and \(b\) in our exercise can be any numbers on the continuum of real numbers, not just integers or fractions.
Understanding real numbers is fundamental when dealing with arithmetic mean, because the mean, or average, of two numbers needs the sum and division to make sense over the continuous scale of real numbers.
When you compute \(\frac{a+b}{2}\), you're essentially finding a point on this line that is exactly halfway between \(a\) and \(b\), underlining how versatile real numbers are in mathematical expressions and solutions.

Constructing a Proof

Constructing proofs in mathematics is a methodical process that relies on logical reasoning. In proving the inequality \(a < \frac{a+b}{2} < b\), we follow a series of steps to validate our assumptions and reach a concrete conclusion.
Start by understanding the inequality you need to prove: first, we show that \(a < \frac{a+b}{2}\); then, \(\frac{a+b}{2} < b\). Each segment of the proof uses the foundational assumption \(a < b\) to assert both parts of the inequality.
The process includes clearly stating each logical step, using algebraic manipulation like adding the same number to both sides or dividing an inequality to maintain its truth. By systematically tackling each part of the problem, the proof eventually shows that the arithmetic mean lies perfectly between the two real numbers, demonstrating a simple yet powerful callback to the fundamental essence of inequalities on real numbers.