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Chapter 3: Problem 3

(a) Write the contrapositive of the following statement: For all positive real numbers \(a\) and \(b\), if \(\sqrt{a b} \neq \frac{a+b}{2},\) then \(a \neq b\). (b) Is this statement true or false? Prove the statement if it is true or provide a counterexample if it is false.

Short Answer

Expert verified
The contrapositive of the given statement is: For all positive real numbers \(a\) and \(b\), if \(a = b\), then \(\sqrt{a b} = \frac{a+b}{2}\). The statement is true, as we can prove it by showing that the contrapositive is true. If \(a = b\), then \(\sqrt{a b} = a\) and \(\frac{a+b}{2} = a\), which makes the contrapositive true, and hence the original statement is also true.

Step by step solution

01

Write the contrapositive

The contrapositive of a statement of the form "if P, then Q" is "if not Q, then not P." To find the contrapositive of the given statement, we must negate both the hypothesis (the "if" part) and the conclusion (the "then" part). The original statement is: For all positive real numbers \(a\) and \(b\), if \(\sqrt{a b} \neq \frac{a+b}{2},\) then \(a \neq b\). The contrapositive statement is: For all positive real numbers \(a\) and \(b\), if \(a = b\), then \(\sqrt{a b} = \frac{a+b}{2}\). 2.
02

Determine if the statement is true or false

Suppose the statement is true. If we let \(a = b\) be any positive real number, then the quantity \(\sqrt{a b}\) is equal to \(\sqrt{a^2} = a\) (since \(a\) is positive) and \(\frac{a+b}{2} = \frac{2a}{2} = a\). Therefore, the contrapositive statement is true, and thus the original statement is also true. 3.
03

Prove the statement

To prove the statement, consider two positive real numbers \(a\) and \(b\). We want to show that if \(\sqrt{a b} \neq \frac{a+b}{2},\) then \(a \neq b\). We will use proof by contrapositive. We have shown that the contrapositive is true. Hence, the original statement is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Mean
The geometric mean is a type of average that indicates the central tendency of a set of numbers by using the product of their values. It is especially useful for sets of numbers that are exponentially related or for growth rates.
The geometric mean of two positive numbers, say \(a\) and \(b\), is given by the formula: \[ \sqrt{a b} \]
  • The geometric mean is always less than or equal to the arithmetic mean, reflecting the balance between the highest and lowest numbers.
  • It can also be interpreted as the side length of a square that has the same area as a rectangle with sides \(a\) and \(b\).
This measure is particularly insightful in understanding the original problem, as it helps connect the relationship between \(a\) and \(b\) with their average values. Its role in mathematical problems often highlights multiplicative relationships.
Arithmetic Mean
The arithmetic mean is what most people refer to simply as the "average." It is used to calculate the central value of a set of numbers by adding them up and dividing by the count of numbers. For two numbers, \(a\) and \(b\), the arithmetic mean is given by: \[ \frac{a + b}{2} \]
  • This is the sum of \(a\) and \(b\) divided equally.
  • The arithmetic mean is a linear measure, used when the numbers in the set are equally important.
In the context of the contrapositive statement, the arithmetic mean provides a benchmark to compare with the geometric mean. Understanding how these means relate helps in analyzing whether certain conditions set in mathematical statements, like equalities or inequalities, hold true.
Proof by Contrapositive
Proof by contrapositive is a powerful mathematical technique for proving mathematical statements. It essentially involves proving that the opposite of a statement leads to the opposite of the conclusion.
When given a statement in the form "if P, then Q," the contrapositive is "if not Q, then not P."
By proving the contrapositive, the original statement is thereby proven to be true.
Here鈥檚 how it works in this problem:
  • The initial statement involves properties of real numbers and mean calculations.
  • We looked at the contrapositive: if \(a = b\), then \(\sqrt{a b} = \frac{a + b}{2}\).
  • Since this contrapositive is verifiable through calculation, it confirms that if the geometric mean isn鈥檛 equal to the arithmetic mean, \(a\) and \(b\) cannot be equal.
Effective use of contrapositive proofs often simplifies proofs where direct methods might be cumbersome.
Real Numbers
Real numbers are the set of numbers that include all the numbers on the number line. This set consists of rational numbers, such as fractions and integers, and irrational numbers, such as \(\sqrt{2}\) and \(\pi\).
  • They are used to measure continuous quantities.
  • Real numbers do not include imaginary numbers or complex numbers.
  • Their real number property allows for a wide range of mathematical operations such as addition, subtraction, multiplication, and division.
In the exercise, real numbers are crucial because the problem deals with positive real number values for \(a\) and \(b\). Understanding this ensures that the operations involving geometric and arithmetic means are valid under the rules of real number arithmetic.

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Most popular questions from this chapter

Determine if each of the following statements is true or false. Provide a counterexample for statements that are false and provide a complete proof for those that are true. (a) For all real numbers \(x\) and \(y, \sqrt{x y} \leq \frac{x+y}{2}\). (b) For all real numbers \(x\) and \(y, x y \leq\left(\frac{x+y}{2}\right)^{2}\). (c) For all nonnegative real numbers \(x\) and \(y, \sqrt{x y} \leq \frac{x+y}{2}\).

Let \(n\) be a natural number greater than 4 and let \(a\) be an integer that has a remainder of \(n-2\) when it is divided by \(n\). Make whatever conclusions you can about the remainder of \(a^{2}\) when it is divided by \(n\). Justify all conclusions.

Prove that for each natural number \(n, \sqrt{3 n+2}\) is not a natural number.

We have already seen examples of Pythagorean triples, which are natural numbers \(a, b,\) and \(c\) where \(a^{2}+\) \(b^{2}=c^{2}\). For example, \(3,4,\) and 5 form a Pythagorean triple as do \(5,\) \(12,\) and \(13 .\) One of the famous mathematicians of the 17 th century was Pierre de Fermat \((1601-1665)\). Fermat made an assertion that for each natural number \(n\) with \(n \geq 3,\) there are no positive integers \(a, b,\) and \(c\) for which \(a^{n}+b^{n}=c^{n}\). This assertion was discovered in a margin of one of Fermat's books after his death, but Fermat provided no proof. He did, however, state that he had discovered a truly remarkable proof but the margin did not contain enough room for the proof. This assertion became known as Fermat's Last Theorem but it more properly should have been called Fermat's Last Conjecture. Despite the efforts of mathematicians, this "theorem" remained unproved until Andrew Wiles, a British mathematician, first announced a proof in June of \(1993 .\) However, it was soon recognized that this proof had a serious gap, but a widely accepted version of the proof was published by Wiles in \(1995 .\) Wiles' proof uses many concepts and techniques that were unknown at the time of Fermat. We cannot discuss the proof here, but we will explore and prove the following proposition, which is a (very) special case of Fermat's Last Theorem. Proposition. There do not exist prime numbers \(a, b,\) and \(c\) such that \(a^{3}+b^{3}=c^{3}\) Although Fermat's Last Theorem implies this proposition is true, we will use a proof by contradiction to prove this proposition. For a proof by contradiction, we assume that there exist prime numbers \(a, b,\) and \(c\) such that \(a^{3}+b^{3}=c^{3}\). Since 2 is the only even prime number, we will use the following cases: (1) \(a=b=2 ;\) (2) \(a\) and \(b\) are both odd; and (3) one of \(a\) and \(b\) is odd and the other one is 2 . (a) Show that the case where \(a=b=2\) leads to a contradiction and hence, this case is not possible. (b) Show that the case where \(a\) and \(b\) are both odd leads to a contradiction and hence, this case is not possible. (c) We now know that one of \(a\) or \(b\) must be equal to 2 . So we assume that \(b=2\) and that \(a\) is an odd prime. Substitute \(b=2\) into the equation \(b^{3}=c^{3}-a^{3}\) and then factor the expression \(c^{3}-a^{3}\). Use this to obtain a contradiction. (d) Write a complete proof of the proposition.

Is the following statement true or false? For all positive real numbers \(x\) and \(y, \sqrt{x+y} \leq \sqrt{x}+\sqrt{y}\).
See all solutions

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