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Chapter 5: Problem 68

Prove each for \(n \geq 0\) $$A(1, n)=n+2$$

Short Answer

Expert verified
We prove the statement A(1, n) = n+2 for every non-negative integer n by induction. For the base case, when n=0, we plug into the formula and get A(1, 0) = 0 + 2 = 2, which is true. Next, we assume the inductive hypothesis, A(1, k) = k+2 for an arbitrary non-negative integer k. We then show that if the formula holds for n=k, it also holds for n=k+1. Evaluating A(1, k+1), we get A(1, k+1) = (k+1) + 2 = k+3. Since the formula is true for the base case and n=k+1, we conclude by induction that A(1, n)=n+2 holds for all non-negative integers n.

Step by step solution

01

Base case n=0

We need to show that the formula A(1, n)=n+2 is valid for n=0. Plug in n=0 in A(1, n): \(A(1, 0) = 0 + 2 = 2\) Since the formula holds for n=0, it is valid for the base case.
02

Inductive Step (Inductive Hypothesis)

We assume that the formula A(1, n)=n+2 is true for n=k, where k is an arbitrary non-negative integer. This is our inductive hypothesis. That is, we assume: \(A(1, k) = k + 2\)
03

Inductive Step (Show for n=k+1)

Now, we need to show that if the formula holds for n=k, then it also holds for n=k+1. To do this, we evaluate A(1, k+1): \(A(1, k+1) = (k+1) + 2\)
04

Use the Inductive Hypothesis

To prove that the formula also holds for n=k+1, we will use our inductive hypothesis (A(1, k) = k + 2) and manipulate the expression to yield the desired result. We have: \(A(1, k+1) = (k+1) + 2 = k + 3\) Since A(1, k+1)=k+3, the formula holds for n=k+1.
05

Conclude by Induction

Since we have proven that the formula A(1, n)=n+2 holds for the base case (n=0), and also shown that if the formula is true for n=k, then it is true for n=k+1, we can conclude by the principle of induction that the formula is valid for all non-negative integers n.

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Key Concepts

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

Inductive Hypothesis
In mathematical induction, the inductive hypothesis is a pivotal concept. It is essentially an assumption that a statement is true for some arbitrary case, say for a general integer \(k\). This forms the crux of the inductive method.
In our problem, we assume \(A(1, k) = k + 2\) holds true. This assumption is not baseless; instead, it enables us to build a logical bridge to the following step.
The importance of the inductive hypothesis lies in its role as a stepping stone. Without assuming some truth for \(k\), it would be challenging to demonstrate its truth for \(k + 1\). Therefore, this assumption is not just a guess but a strategic part of the proof process.
  • Assumption makes the problem approachable.
  • Serves as a link between known and unknown values.
Base Case
The base case in mathematical induction is where everything begins. It is the simplest instance where you prove your statement holds true.
For our problem, the base case is \(n = 0\). We substitute this value into the equation \(A(1, n) = n + 2\) and find that it holds true because \(A(1, 0) = 0 + 2 = 2\).
The base case is crucial because it confirms the validity of the statement for at least one value. Without it, the induction would not even get off the ground.
  • Foundation of the induction process.
  • Ensures the formula is valid at least once.
By establishing the base case, you lay the groundwork upon which the rest of the proof is built.
Inductive Step
The inductive step is where the magic of induction really happens. Once you assume the statement is true for \(n=k\) (using the inductive hypothesis), the next task is to show it holds true for \(n = k + 1\).
Continuing our example, knowing \(A(1, k) = k + 2\), we must show \(A(1, k+1) = (k+1) + 2 = k + 3\). It's crucial to use the hypothesized truth for \(k\) to demonstrate the same for \(k + 1\).
This step is the key link in induction that allows us to conclude the pattern holds for all subsequent integers beyond the base case.
  • Shows the logical progression from \(k\) to \(k+1\).
  • Allows generalization to all numbers in the domain.
Thus, with a valid inductive step, the proof of our initial formula can extend infinitely.

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