Problem 87 - Calculus with Concepts in Calculus

Question

Let \(0

Step-by-Step Solution

Answer

The inequality is proved as \(\sqrt{a} < \sqrt{b}\), given \(0 < a < b\).

Step 1: Square Both Sides of the Inequality

To eliminate the square roots, let's first start by squaring both sides of the inequality \(\sqrt{b} - \sqrt{a} < \sqrt{b-a}\). This yields:\((\sqrt{b} - \sqrt{a})^2 < (\sqrt{b-a})^2\).On the right side, the square root and the square cancel out, giving us \(b-a\).

Step 2: Expand the Left Side

Expand the left side of the inequality \((\sqrt{b} - \sqrt{a})^2\) using the binomial formula:\((\sqrt{b} - \sqrt{a})^2 = b - 2\sqrt{b}\sqrt{a} + a\).

Step 3: Simplify the Inequality

Substituting the expansion into the inequality gives:\(b - 2\sqrt{b}\sqrt{a} + a < b - a\).Now, subtract \(b\) from both sides:\(-2\sqrt{b}\sqrt{a} + a < -a\).

Step 4: Rearrange and Simplify

Now, rearrange the terms:\(a - 2\sqrt{b}\sqrt{a} < -a\).Add \(a\) to both sides of the inequality:\(a - 2\sqrt{b}\sqrt{a} + a < 0\), which simplifies to:\(2a < 2\sqrt{b}\sqrt{a}\) or \(a < \sqrt{b}\sqrt{a}\).

Step 5: Further Simplification

Divide both sides of the inequality \(a < \sqrt{b}\sqrt{a}\) by \(\sqrt{a}\) (since \(a > 0\)):\(\sqrt{a} < \sqrt{b}\).This is true because we are given that \(0 < a < b\).

Step 6: Conclude the Proof

The inequality \(\sqrt{a} < \sqrt{b}\) is satisfied under the given condition \(0 < a < b\). Therefore, \(\sqrt{b} - \sqrt{a} < \sqrt{b-a}\) is proved.

Key Concepts

Square Roots and Inequalities

Understanding how to manipulate inequalities involving square roots is crucial in calculus. This often involves removing the square roots by squaring both sides of an inequality. Doing so simplifies the problem and makes it easier to work with. However, be cautious as squaring can sometimes introduce extraneous solutions, so it's important to ensure the variable conditions justify the step. In our problem, we started with the inequality \( \sqrt{b} - \sqrt{a} < \sqrt{b-a} \). By squaring both sides, we transitioned to a format devoid of radicals: \((\sqrt{b} - \sqrt{a})^2 < b-a \). This step is valid under the knowledge that squaring is a monotonic function when dealing with positive values.

Binomial Expansion

The binomial expansion is a key algebraic tool that helps in simplifying expressions, particularly when they involve sums or differences. In this context, the expression \((\sqrt{b} - \sqrt{a})^2\) was expanded using the binomial formula:\
\[\text{(first term)}^2 - 2 \times \text{(first term)} \times \text{(second term)} + \text{(second term)}^2\].\
Applying this, we arrived at \(b - 2\sqrt{b}\sqrt{a} + a\). The binomial formula helps break down complex expressions into manageable components, often leading to simplifications in problems involving inequalities.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations or inequalities to find solutions. In this problem, we used such manipulations to refine the inequality\
\( b - 2\sqrt{b}\sqrt{a} + a < b-a ;\)
which was simplified to \(-2\sqrt{b}\sqrt{a} + a < -a\). By isolating terms and systematically simplifying, \( 2a < 2\sqrt{b}\sqrt{a} \) was obtained. Simple algebraic manipulations, like subtracting or adding terms on both sides, helped to hone in on the true relationship between \(a\) and \(b\). These steps are vital when dealing with proofs involving complex operations.

Proof Techniques

Proving mathematical expressions, especially inequalities, usually requires a structured approach. The first step often involves rewriting or transforming the problem into a more tractable form, as done by squaring the inequality. Logical reasoning is used when justifying each algebraic manipulation step to ensure the initial relationships hold. In this exercise, starting from significant information such as \(0 < a < b\), allowed us to conclude \( \sqrt{a} < \sqrt{b} \) which confirmed the original inequality. Important proof techniques include assumption testing and logical deduction to verify conclusions are consistent with original conditions.