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Chapter 1: Problem 3

Use a proof by cases to show that 100 is not the cube of a positive integer. [Hint: Consider two cases: (i) \(1 \leq x \leq 4\) , (ii) \(x \geq 5 . ]\)

Short Answer

Expert verified
100 is not the cube of any positive integer, as shown by examining integers 1-4 and integers 鈮�5.

Step by step solution

01

Identify Case 1

For the first case, consider positive integers from 1 through 4. The goal is to determine if any of these numbers cubed equals 100. Compute: \( 1^3 = 1 \), \( 2^3 = 8 \), \( 3^3 = 27 \), and \( 4^3 = 64 \). None of these values is equal to 100, so we can confidently say that 100 is not the cube of any number between 1 and 4.
02

Identify Case 2

For the second case, consider positive integers greater than or equal to 5. Calculate the cubes for these numbers and note that \( 5^3 = 125 \). Since 125 already exceeds 100, any number greater than 5 will have a cube much larger than 100.
03

Conclusion

Each positive integer was tested: for \( 1 \leq x \leq 4 \), cubes are less than 100; for \( x \geq 5 \), cubes exceed 100. Therefore, 100 is not the cube of any positive integer.

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Key Concepts

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

cube of a number
The cube of a number is calculated by multiplying the number by itself two more times. In mathematical terms, if you have a number \( x \), its cube is \( x^3 \) which means \( x \times x \times x \).

Cubing a number makes it grow very quickly. For example, the cube of 1 is 1, but the cube of 2 is 8, and the cube of 3 is 27.

These values show how rapidly the cubes of small numbers can increase. This is important to remember because it helps us understand why 100 is not a cube of any positive integer. The cubes of all numbers either fall significantly below or go far above 100.
positive integers
Positive integers are the set of all positive whole numbers starting from 1. They do not include zero or any negative numbers.

They are often represented as \{1, 2, 3, 4, 5, ...\}. When solving problems involving cubes, considering only positive integers helps to simplify the calculations, since we're dealing with straightforward, increasing values.

In our solution to the given exercise, we specifically looked at the positive integers \( 1 \) through \( 4 \) and again numbers starting at \( 5 \). Knowing their properties allowed us to efficiently exclude ranges of values that couldn't produce the cube of 100.
mathematical proof
Mathematical proofs are logical arguments that demonstrate the truth of a statement. Proof by cases is a fundamental method where different scenarios (or cases) are considered separately to make a comprehensive argument.

In our exercise, we used proof by cases to show that 100 cannot be the cube of any positive integer. By dividing positive integers into two cases, we checked all possible values systematically:
  • Case 1: Considering integers from \(1\) through \(4\), none of their cubes equal 100.
  • Case 2: Considering integers 5 and above, their cubes are all greater than 100.
Combining these cases allows us to conclude that no positive integer's cube equals 100.

Proof by cases is effective because it leaves no room for doubt. By addressing every possible case, you ensure your proof covers all possible scenarios, making the argument solid and complete.

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Most popular questions from this chapter

Prove that there are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers.

Prove or disprove that if you have an 8 -gallon jug of water and two empty jugs with capacities of 5 gallons and 3 gallons, respectively, then you can measure 4 gallons by successively pouring some of or all of the water in a jug into another jug.

Express each of these statements using quantifiers. Then form the negation of the statement so that no negation is to the left of a quantifier. Next, express the negation in simple English. (Do not simply use the phrase "It is not the case that.") a) No one has lost more than one thousand dollars playing the lottery. b) There is a student in this class who has chatted with exactly one other student. c) No student in this class has sent e-mail to exactly two other students in this class. d) Some student has solved every exercise in this book. e) No student has solved at least one exercise in every section of this book.

Suppose the domain of the propositional function \(P(x, y)\) consists of pairs \(x\) and \(y,\) where \(x\) is \(1,2,\) or 3 and \(y\) is \(1,2,\) or \(3 .\) Write out these propositions using disjunctions and conjunctions. $$ \begin{array}{ll}{\text { a) } \forall x \forall y P(x, y)} & {\text { b) } \exists x \exists y P(x, y)} \\ {\text { c) } \exists x \forall y P(x, y)} & {\text { d) } \forall y \exists x P(x, y)}\end{array} $$

Let \(P(x)\) be the statement "The word \(x\) contains the letter \(a\) ." What are these truth values? \(\begin{array}{ll}{\text { a) } P(\text { orange })} & {\text { b) } P(\text { lemon })} \\ {\text { c) } P(\text { true })} & {\text { d) } P(\text { false })}\end{array}\)
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