Problem 73 - Calculus Early Transcendentals: Pearson New International Edition

Question

The number \(\sqrt{a b}\) is called the geometric mean of two positive numbers \(a\) and \(b\). Prove that $$ 0

Step-by-Step Solution

Answer

Proven: \(a < \sqrt{ab} < b\) holds for \(0 < a < b\).

Step 1: Understand the Inequality

The problem states that for two positive numbers with the property \(0 < a < b\), we need to prove the inequality \(a < \sqrt{ab} < b\). This means \(\sqrt{ab}\), the geometric mean of \(a\) and \(b\), should lie between \(a\) and \(b\).

Step 2: Prove \(a < \sqrt{ab}\)

To prove \(a < \sqrt{ab}\), first square both sides to eliminate the square root, resulting in \(a^2 < ab\). This simplifies to \(a < b\) because both sides are positive and \(0 < a < b\). Thus, the inequality \(a < \sqrt{ab}\) holds true.

Step 3: Prove \(\sqrt{ab} < b\)

Similarly, for \(\sqrt{ab} < b\), squaring both sides gives \(ab < b^2\). Dividing both sides by \(b\) (which is greater than \(a\) and positive) yields \(a < b\), a true statement by the given condition that \(0 < a < b\). Thus, \(\sqrt{ab} < b\) holds true as well.

Step 4: Conclude the Proof

Since both inequalities \(a < \sqrt{ab}\) and \(\sqrt{ab} < b\) have been proven, we can conclude that \(a < \sqrt{ab} < b\). Therefore, the geometric mean \(\sqrt{ab}\) lies between \(a\) and \(b\) when \(0 < a < b\).

Key Concepts

Understanding Inequalities

Inequalities are expressions that show the relation of quantity and value between numbers. They explain how one value is greater than, less than, or equal to another value. In this problem, we deal with an inequality involving the geometric mean.
The inequality we are tasked to prove is that the geometric mean, \(\sqrt{ab}\), lies between two positive numbers \(a\) and \(b\), provided that \(0 < a < b\). This means that \(a < \sqrt{ab} < b\), showing that the mean is not just an average of \(a\) and \(b\) but also exhibits a unique position between them.
Understanding this helps us see the balance the geometric mean represents when dealing with positive numbers.

Square Roots and Their Properties

The square root is a fundamental concept in mathematics, symbolized by \(\sqrt{\cdot}\). Taking the square root of a number means finding a value that, when multiplied by itself, gives the original number. For example, with \(\sqrt{ab}\), it means finding a number \(x\) such that \(x \cdot x = ab\).
Square roots are important in understanding the geometric mean because they help derive a middle ground between two numbers, \(a\) and \(b\). This is why removing the square root by squaring both sides of an inequality becomes useful in proofs. In our solution, to show \(a < \sqrt{ab}\), we square both sides to obtain \(a^2 < ab\), simplifying the problem into a more digestible form.
This operation is legal under certain conditions, especially when both sides of the inequality involve positive numbers. This ensures that the direction of the inequality remains unchanged.

Proof by Mathematical Reasoning

In mathematics, a proof is a logical argument that establishes the truth of a statement. Here, we use direct proof to establish that the geometric mean \(\sqrt{ab}\) lies between \(a\) and \(b\).
- First, we prove that \(a < \sqrt{ab}\). We do this by squaring both sides to simplify the inequality to \(a^2 < ab\). Since \(0 < a < b\), multiplying \(a\) by \(b\) indeed results in a larger product, confirming this part of the inequality.- Secondly, we need to show that \(\sqrt{ab} < b\). Again, by squaring both sides, we get \(ab < b^2\). Since \(b\) is greater than \(a\) and positive, dividing through by \(b\) doesn't change the inequality, confirming that \(a < b\).
Once these steps are completed, we have logically shown and concluded both parts of our intended proof, effectively proving \(a < \sqrt{ab} < b\). This methodical approach helps ensure nothing is left to assumption, primarily relying on well-established mathematical reasoning and operations.