Problem 40 - The Calculus with Analytic Geometry

Question

Prove that if \(x

Step-by-Step Solution

Answer

Proved: If \(x < y\), then \(x < \frac{1}{2}(x+y) < y\).

Step 1: Understanding the Given Inequality

Given that \(x < y\), it is necessary to prove that \(x < \frac{1}{2}(x+y) < y\).

Step 2: Prove the Left Inequality

First, we need to show that \(x < \frac{1}{2}(x+y)\).Starting with the premise that \(x < y\), a series of algebraic manipulations follows to show the desired result:1. Multiply both sides of \(x < y\) by \(\frac{1}{2}\):\[\frac{1}{2}x < \frac{1}{2}y\]2. Add \(\frac{1}{2}x\) to both sides:\[\frac{1}{2}x + \frac{1}{2}x < \frac{1}{2}x + \frac{1}{2}y\]3. Simplify the left side to get \(x\) and the right side to get \(\frac{1}{2}(x + y)\):\[x < \frac{1}{2}(x+y)\]Hence, \(x < \frac{1}{2}(x+y)\) is proven.

Step 3: Prove the Right Inequality

Next, we need to prove that \(\frac{1}{2}(x+y) < y\).Starting again with the premise: \(x < y\), follow these steps:1. Multiply both sides of \(x < y\) by \(\frac{1}{2}\):\[\frac{1}{2}x < \frac{1}{2}y\]2. Add \(\frac{1}{2}y\) to both sides:\[\frac{1}{2}x + \frac{1}{2}y < \frac{1}{2}y + \frac{1}{2}y\]3. Simplify the left side to get \(\frac{1}{2}(x + y)\) and the right side to get \(y\):\[\frac{1}{2}(x + y) < y\]Therefore, \(\frac{1}{2}(x + y) < y\) is proven.

Step 4: Final Statement

Since both \(x < \frac{1}{2}(x+y)\) and \(\frac{1}{2}(x+y) < y\) have been proven, it follows that \(x < \frac{1}{2}(x+y) < y\).

Key Concepts

algebraic manipulation

Algebraic manipulation involves using basic algebra rules to transform an expression or equation. To understand and solve inequalities, it's essential to be comfortable with these manipulations.
Let's consider our problem: given that \(x < y\), prove that \(x < \frac{1}{2}(x + y) < y\).

• We start with \(x < y\).
• By multiplying each side by \(\frac{1}{2}\), we get \(\frac{1}{2}x < \frac{1}{2}y\).
• To isolate the terms we want to compare, we add \(\frac{1}{2}x\) to both sides: \(\frac{1}{2}x + \frac{1}{2}x < \frac{1}{2}x + \frac{1}{2}y\).
• On simplifying, we get \(x < \frac{1}{2}(x+y)\). This proves the first part of the inequality.

Repeating such steps robustly ensures that manipulations are correct and the inequality holds. Practice various inequalities to build confidence in these techniques.

inequalities

Inequalities tell us about the relative size or order of two values. Understanding inequalities involves recognizing how these values compare in different scenarios.
In our problem, we work within the set of real numbers. When working with inequalities, remember:

• Adding or subtracting the same value on both sides maintains the inequality.
• Multiplying or dividing by a positive number keeps the inequality direction the same.
• Multiplying or dividing by a negative number reverses the inequality direction.

In our example, we maintained the inequality direction by using a positive factor \(\frac{1}{2}\). These properties help us systematically approach and prove inequalities.

proof techniques

Proof techniques are essential for demonstrating the truth of mathematical statements. There's a variety of strategies, but let's focus on what's used in our example:
**Direct Proof Approach:**

• Start with what's given: \(x < y\).
• Use algebraic manipulation to derive necessary parts of the proof.

We first proved \(x < \frac{1}{2}(x+y)\) by manipulating the terms. Similarly, we showed \(\frac{1}{2}(x+y) < y\) with logical steps:

• From \(x < y\), multiplying both sides by \(\frac{1}{2}\) yields \(\frac{1}{2}x < \frac{1}{2}y\).
• Adding \(\frac{1}{2}y\) keeps the inequality consistent: \(\frac{1}{2}x + \frac{1}{2}y < \frac{1}{2}y + \frac{1}{2}y\).
• Simplifying tells us \(\frac{1}{2}(x+y) < y\).

By proving both parts individually, we complete our proof. Practice mastering these techniques to become adept in proving various statements algebraically.