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Chapter 2: Problem 46

Prove that for all \(n \in \mathbf{Z}, n\) is even if and only if \(n+2\) is even.

Short Answer

Expert verified
The claim is proven to be true. With the two steps, we formally prove that a number \(n\) is even if and only if \(n+2\) is also even.

Step by step solution

01

Prove the First Condition

To solve the first condition, we start from the premise that \(n\) is even. An even number can be written in the form \(2k\), where \(k\) is an integer. Thus, if \(n\) is even, we can write it as \(n = 2k\). By adding 2 to both sides, we get \(n + 2 = 2k + 2 = 2(k + 1)\). Here, we see that \(n+2\) is also in the form of \(2m\) where \(m = k + 1\). Since \(k\) and \(1\) are both integers, \(m = k + 1\) is also an integer. Hence, \(n+2\) is even.
02

Prove the Second Condition

The second condition to solve is that if \(n+2\) is even, then \(n\) is also even. Considering \(n+2 = 2k\), where \(k\) is an integer, shows that \(n+2\) is even. By subtracting 2 from both sides, we result in \(n = 2k - 2 = 2(k - 1)\). Again, this is in the form of \(2m\) where \(m = k - 1\). Since \(k\) is an integer, then \(k - 1\) is also an integer, which means \(n\) is even.

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Key Concepts

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

Parity of integers
Understanding parity is all about recognizing whether a number is even or odd. In mathematics, an integer can be either even or odd. An even integer can be written as \(2k\), where \(k\) is any integer. Conversely, an odd integer can be expressed as \(2k+1\). The concept of parity helps us determine how integers interact during operations and transformations.

In the original problem, recognizing even numbers is key. If \(n\) is even, then \(n\) can be expressed as \(2k\). By adding 2, which itself is even, the resulting value \(n+2\) retains its even property.
This principle serves as the foundation for proving the problem's statement.
  • Even + Even = Even
  • Odd + Odd = Even
Mathematical proofs
Mathematical proofs provide a way to demonstrate that certain statements or assertions hold true. A proof uses logical reasoning, starting from established premises and following through to a logical conclusion. This structured approach ensures that all parts of a statement have been verified.

In this particular exercise, the goal was to use proofs to establish the equivalency of statements involving the parity of integers. The first part of the proof showed that if \(n\) is even, then \(n+2\) is automatically even.
The second part of the proof reversed this logic, showing that if \(n+2\) is even, \(n\) must also be even.
  • Start with assumptions (even or odd nature of integer)
  • Apply arithmetic operations
  • Conclude based on observations
These steps ensure that both directions of implication are seamlessly verified.
Integer operations
Integer operations are basic arithmetic manipulations involving integers. They include addition, subtraction, multiplication, and division. Understanding how these operations affect the parity of numbers is crucial to solving parity problems.

In the exercise, we used addition and subtraction to manipulate the numbers. When adding 2 (an even number) to \(n\), the parity of \(n\) is preserved. Similarly, subtracting 2 from \(n + 2\) also maintains uniform parity.
This arises from the rule that the sum or difference of even numbers is always even.
  • Even + Even = Even
  • Subtracting an even number from another even number results in an even outcome
Thus, understanding the mechanics of these integer operations helps in verifying the parity conditions in mathematical proofs.

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Most popular questions from this chapter

What is wrong with the following argument, which allegedly shows that any two positive integers are equal? We use induction on \(n\) to "prove" that if \(a\) and \(b\) are positive integers and \(n=\max \\{a, b\\},\) then \(a=b\). If \(a\) and \(b\) are positive integers and \(1=\max \\{a, b\\},\) we must have \(a=b=1\) Assume that if \(a^{\prime}\) and \(b^{\prime}\) are positive integers and \(n=\max \left\\{a^{\prime}, b^{\prime}\right\\},\) then \(a^{\prime}=b^{\prime} .\) Suppose that \(a\) and \(b\) are positive integers and that \(n+1=\max \\{a, b\\} .\) Now \(n=\) \(\max \\{a-1, b-1\\} .\) By the inductive hypothesis, \(a-1=b-1\) Therefore, \(a=b\). Since we have verified the Basis Step and the Inductive Step, by the Principle of Mathematical Induction, any two positive integers are equal!

Use induction to prove the statement. \(6 \cdot 7^{n}-2 \cdot 3^{n}\) is divisible by \(4,\) for all \(n \geq 1\)

Prove that if \(a\) and \(b\) are real numbers with \(a

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1 \quad \text { and } \quad \frac{p}{q}<\frac{1}{n-1} .$$ (c) Show that if \(p / q=1 / n,\) the proof is complete. (d) Assume that \(1 / n
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