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Magazine Chapter 5: Problem 31
Use mathematical induction in Exercises \(31-37\) to prove divisibility facts. Prove that 2 divides \(n^{2}+n\) whenever \(n\) is a positive integer.
Short Answer
Expert verified
2 divides \(n^2 + n\) for all positive integers \(n\) by mathematical induction.
Step by step solution
01
- Understanding the Problem
Given the statement to prove: 2 divides \(n^2 + n\) for any positive integer \(n\). This means that \(n^2 + n\) must be an even number for all positive integers \(n\).
02
- Base Case
First, verify the base case for \(n = 1\). Calculate \(1^2 + 1 = 2\). Since 2 is divisible by 2, the base case holds.
03
- Inductive Hypothesis
Assume the statement is true for some positive integer \(k\). That is, assume that 2 divides \(k^2 + k\). This means \(k^2 + k = 2m\) for some integer \(m\).
04
- Inductive Step
Show that if the statement is true for \(k\), then it is also true for \(k + 1\). Calculate \((k + 1)^2 + (k + 1)\). Expand and simplify the expression: \[(k + 1)^2 + (k + 1) = k^2 + 2k + 1 + k + 1 = k^2 + 3k + 2\]Using the inductive hypothesis, replace \(k^2 + k\) with \(2m\): \[k^2 + 3k + 2 = (k^2 + k) + 2k + 2 = 2m + 2k + 2\]Factor out the 2: \[2m + 2k + 2 = 2(m + k + 1)\]Since \(2(m + k + 1)\) is divisible by 2, the statement holds for \(k + 1\).
05
- Conclusion
By mathematical induction, the statement is true for all positive integers \(n\). That is, 2 divides \(n^2 + n\) whenever \(n\) is a positive integer.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Base Case
To begin a mathematical induction proof, you first test a specific instance of the statement. This is called the base case. It sets the foundation for the induction. For example, if we need to show that 2 divides \(n^2 + n\) for any positive integer \(n\), we start with \(n = 1\). We calculate:
- \( 1^2 + 1 = 2 \)
This clearly shows that 2 divides 2, so the base case holds true.
Explaining the base case in more detail:
- By verifying this initial step, we confirm that there is at least one value of \(n\) for which the statement holds.
- This provides the anchor for the next part of our proof involving an inductive hypothesis.
Inductive Hypothesis
The next step in mathematical induction is to assume the statement is true for some arbitrary integer \(k\). This assumption is known as the inductive hypothesis. In our example:
This crucial step builds a bridge to connect the base case with the goal of proving the statement for all integers. Important points to remember about the inductive hypothesis:
- We assume 2 divides \(k^2 + k\)
This means that there exists an integer \(m\) such that:
\( k^2 + k = 2m \).
This crucial step builds a bridge to connect the base case with the goal of proving the statement for all integers. Important points to remember about the inductive hypothesis:
- It's essentially taking the leap of faith, assuming the statement's truth for an arbitrary case.
- It simplifies our work for the next part of the proof, as we can use this assumption to tackle the problem for \(k + 1\).
Inductive Step
Once we have our inductive hypothesis, we need to prove that the statement holds for \(k+1\). This is called the inductive step. For our example, our aim is to show that: \( (k + 1)^2 + (k + 1) \) is divisible by 2
We start by expanding the expression:
We start by expanding the expression:
- \( (k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = k^2 + 3k + 2 \) Using our inductive hypothesis, we substitute \(k^2 + k\) with 2m:
- \( k^2 + 3k + 2 = (k^2 + k) + 2k + 2 = 2m + 2k + 2 \) Notice that we can factor out a 2:
- \( 2m + 2k + 2 = 2(m + k + 1) \) This final expression shows that it is divisible by 2, thus proving that if the statement is true for some integer \(k\), it is also true for \(k+1\).
- Showcase how the statement for \(k\) (inductive hypothesis) helps in proving it for \(k + 1\).
- This step solidifies the induction process by proving the continuity of the statement from one integer to the next.
Important takeaways from the inductive step:
Divisibility
In the context of our example, we are working with the concept of divisibility. Specifically, we want to show that the expression \(n^2 + n\) is divisible by 2 for any positive integer \(n\). In simpler terms:
- The expression must be an even number, as even numbers are divisible by 2. To understand why \(n^2 + n\) is always even:
- When \(n\) is even, \(n^2\) and \(n\) are both even, making their sum even.
- When \(n\) is odd, \(n^2\) is odd, and adding one (an odd number) results in an even sum. Important concepts about divisibility:
- An expression is divisible by a number if, after division, the result is an integer with no remainder.
- In our proof, we make use of algebraic manipulation and factorization to demonstrate divisibility. The concept of divisibility is fundamental in many math problems, not just in the context of mathematical induction. It helps determine the relationship between numbers and their factors. In this proof, it reassures that \(n^2 + n\) being even confirms its divisibility by 2.
