Problem 12 - Analysis 1: Ein Lehr- und Arbeitsbuch für Studienanfänger

Question

Seien a,b \(\in \mathbb{R}\) mit \(a

Step-by-Step Solution

Answer

The inequality holds because \(a < \frac{a+b}{2} < b\) is equivalent to \(a < b\), which is given as true.

Step 1: Understanding the Problem

We need to prove the inequality \(a < \frac{a+b}{2} < b\) given that \(a < b\) and that both \(a\) and \(b\) are real numbers. This involves showing the middle term is greater than \(a\) and less than \(b\).

Step 2: Simplifying Inequality 1

First, let's show that \(a < \frac{a+b}{2}\). Start by multiplying both sides of \(a < \frac{a+b}{2}\) by 2: \(2a < a + b\). This simplifies to \(a < b\), which we know to be true, so the inequality holds.

Step 3: Simplifying Inequality 2

Next, show that \(\frac{a+b}{2} < b\). Similarly, multiply both sides by 2 to get \(a + b < 2b\). This simplifies to \(a < b\), which is known to be true, confirming the inequality.

Key Concepts

Real Numbers

In the realm of mathematics, when we speak of real numbers, we are referring to the vast set of numbers that include all the rational numbers, such as integers and fractions, as well as the irrational numbers. This collection is crucial because it provides us with a continuous number line where numbers can stretch infinitely in both positive and negative directions.

Real numbers are essential in proofs and inequalities because they follow well-defined properties such as order, arithmetic, and operations. These properties allow mathematicians and students alike to evaluate and compare values with straightforward rules, like additive and multiplicative functions.

• Rational numbers: These are numbers which can be expressed as a quotient or fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q eq 0 \).
• Irrational numbers: These numbers cannot be written as a simple fraction, examples include \( \sqrt{2} \) or \( \pi \).

Given that \( a \) and \( b \) are real numbers, the inequality \( a < b \) places them on a straight line where \( a \) is assuredly to the left of \( b \). This forms the basis of the proof, as we know that any point between \( a \) and \( b \) has certain predictable properties.

Mathematical Proofs

Mathematical proofs are foundational structures in mathematics, constructed to confirm the truth of a given statement using logical steps. Proving an inequality often involves manipulating the expressions involved in a way that transforms a complex relationship into simpler, already established truths.

In the case of proving \( a < \frac{a+b}{2} < b \), the objective is to demonstrate that the average of \( a \) and \( b \) lies strictly between them. This can be achieved using strategies like direct proof or contradiction.

• Direct Proof: Involves logically progressing from the initial assumption to the conclusion using known facts and assumptions.
• Proof by Contradiction: Assumes the negation of the statement to derive a contradiction with known facts, thus proving the statement must be true.

Through direct manipulation of the initial inequality \( a < b \), both sub-inequalities are simplified to confirm that the average indeed lies between \( a \) and \( b \).

Inequality Simplification

Simplifying inequalities involves breaking down complex expressions into easier ones which can be more directly compared. This relies on valid mathematical operations which do not alter the essence of the inequality, such as adding, subtracting, multiplying, or dividing both sides by the same positive number.

The simplification of \( a < \frac{a+b}{2} \) and \( \frac{a+b}{2} < b \) are key steps in the solution, illustrating how manipulation can lead directly to a proof. In both scenarios:

• The process starts by removing fractions by multiplying through by 2, thus eliminating denominators.
• The resulting inequalities \( 2a < a + b \) and \( a + b < 2b \) align with the premise \( a < b \).

This straightforward approach helps to confirm that \( \frac{a+b}{2} \) is indeed confined between \( a \) and \( b \). Thus, through simplification, we reaffirm the logical ordering of the three values.