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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

Suppose that the domain of \(Q(x, y, z)\) consists of triples \(x, y, z,\) where \(x=0,1,\) or \(2, y=0\) or \(1,\) and \(z=0\) or \(1 .\) Write out these propositions using disjunctions and conjunctions. $$ \begin{array}{ll}{\text { a) } \forall y Q(0, y, 0)} & {\text { b) } \exists x Q(x, 1,1)} \\ {\text { c) } \exists z \neg Q(0,0, z)} & {\text { d) } \exists x \neg Q(x, 0,1)}\end{array} $$

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