Problem 57

Question

Show that if $a$ and $b$ are real numbers and $a < b$, then \(a<(a+b) / 2

Step-by-Step Solution
Verified

Answer

It has been proven that if $a$ and $b$ are real numbers and $a < b$, then $a < (a+b)/2 < b$. This proves that the arithmetic mean of any two real numbers always lies between those two numbers.

1Step 1: Identify the given

The given is that $a$ and $b$ are real numbers and $a < b$. We have to prove that $a < (a+b)/2 < b$.

2Step 2: Calculate the arithmetic mean

Since $a < b$, we can add $a$ to both sides and divide by 2: $a < (a+b)/2$. This does not change the inequality because 2 is a positive number and increasing any expression by a positive number will always increase the expression. So the calculated arithmetic mean is greater than $a$.

3Step 3: Prove the second part

Similarly, since $a < b$, we can add $b$ to both sides and divide by 2: $(a+b)/2 < b$. This does not change the inequality because 2 is a positive number and increasing any expression by a positive number will always increase the expression. So the calculated arithmetic mean is less than $b$.

Key Concepts

InequalitiesReal NumbersProof Techniques

Inequalities

Inequalities are mathematical expressions that involve a degree of comparison between two values or expressions. They are represented using symbols such as <, >, ≤, and ≥. When we say that one quantity is less than another, it is an inequality.
In the context of this exercise, we are given that $a < b$ for real numbers $a$ and $b$. This is a simple inequality highlighting that $a$ is strictly less than $b$. When dealing with inequalities, performing the same operation on both sides of the inequality is permissible as long as it doesn't violate the rule that governs them.

**Preservation of Inequality:** If you add or subtract the same number to both sides of an inequality, the inequality sign remains the same. For example, adding a constant to both $a$ and $b$ in $a < b$ maintains the inequality.

**Multiplication and Division:** Multiplying or dividing both sides by a positive number also keeps the inequality intact. However, doing so with a negative number reverses the inequality sign.
Understanding how inequalities work is critical in solving and proving statements involving them.

Real Numbers

Real numbers encompass all the possible numbers which are used to measure or count, including rational and irrational numbers. They form a continuous line on the number line, allowing for a meaningful comparison of magnitudes and operations like addition and division.
In the proof provided, $a$ and $b$ are real numbers, allowing us to perform operations like finding the arithmetic mean. Because real numbers are ordered, we can say with confidence that one is larger than the other. Their inclusion property ensures that the arithmetic mean of two distinct real numbers also lies between them.

**Rational Numbers:** These numbers can be expressed as fractions $\frac{p}{q}$ where both $p$ and $q$ are integers and $q eq 0$.

**Irrational Numbers:** These numbers can't be expressed as a fraction and include numbers such as $\pi$ and $\sqrt{2}$.
Knowing that $a$ and $b$ are real ensures there is always a value exactly halfway between them, which is the arithmetic mean.

Proof Techniques

Proving mathematical statements involves a sequence of logical steps that lead from known assumptions to a desired conclusion. Different proof techniques help in logically establishing truth.
For the exercise, proving $a < \frac{a+b}{2} < b$ involves a combination of direct computations with inequalities.

**Direct Proof:** This involves using known facts and logical deductions to prove the given statement directly.

**Algebraic Manipulation:** Using transformations like addition, subtraction, multiplication, and division to reshape expressions while preserving inequalities.
In the solution provided, we directly utilize the properties of inequalities, adding $a$ to both sides and dividing by 2, to establish $a < \frac{a+b}{2}$. Similarly, adding $b$ and dividing helps establish $\frac{a+b}{2} < b$. Each step is a necessary logical leap grounded in prior mathematical properties.

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Other exercises in this chapter

Problem 57
Knowing that $|a+b| \leq|a|+|b| \quad$ for all real $a . b$ show that $$|a|-|b| \leq|a-b| \quad \text { for all real } a . b$$
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Problem 57
Find the real roots of the equation. $x^{2}-x-2=0$.
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Problem 57
Give the domain and range of the function. $$f(x)=1+\tan ^{2} x$$.
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Problem 57
Show that the distance from the origin to the line $A x+B y+$ $C=0$ is given by the formula $$d(0, l)=\frac{|C|}{\sqrt{A^{2}+B^{2}}}$$
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