Problem 1 - A Book of Abstract Algebra

Question

Prove each of the following, using the principle of mathematical induction $$ 1+3+5+\cdots+(2 n-1)=n^{2} $$ (The sum of the first \(\mathrm{n}\) odd integers is \(\mathrm{n}^{2}\).)

Step-by-Step Solution

Answer

The formula \( 1+3+5+\cdots+(2n-1)=n^2 \) is true for all positive integers \( n \) by induction.

Step 1: Base Case

To prove by induction, we first verify the base case. When \( n = 1 \), the left side is \( 1 \) and the right side is \( 1^2 = 1 \). Therefore, the equation \( 1 = 1 \) holds true for the base case \( n = 1 \).

Step 2: Inductive Hypothesis

Assume that the formula holds true for some arbitrary positive integer \( k \). That is, assume \( 1+3+5+\cdots+(2k-1) = k^2 \). This assumption is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove that the formula holds for \( n = k + 1 \), that is, \( 1+3+5+\cdots+(2k-1) + (2(k+1)-1) = (k+1)^2 \).

Step 4: Expand and Simplify

Consider the left side from the inductive hypothesis: \( k^2 + (2(k+1)-1) \). Simplifying \( 2(k+1)-1 \) gives \( 2k + 2 - 1 = 2k + 1 \). Thus, the left side becomes \( k^2 + 2k + 1 \).

Step 5: Verify Inductive Step

Simplify the right side of the equation: \( (k+1)^2 = k^2 + 2k + 1 \). Notice that this matches the simplified left side: \( k^2 + 2k + 1 = k^2 + 2k + 1 \). Therefore, the statement holds for \( n = k + 1 \) too.

Step 6: Conclusion

By the principle of mathematical induction, since the base case holds and the inductive step is proven, the equation \( 1+3+5+\cdots+(2n-1)=n^2 \) is true for all positive integers \( n \).

Key Concepts

Understanding Odd Integers

Odd integers are numbers that cannot be evenly divided by 2. They always have a remainder of 1 when divided by 2. For example, 1, 3, 5, and 7 are odd integers. When looking at a series such as 1, 3, 5, ... , these numbers are generated by a simple formula: the $n$-th odd integer is expressed as $2n-1$. This formula helps us find odd integers in a sequence quickly without manual counting. Here's how it works:

• For $n=1$, $2(1)-1 = 1$
• For $n=2$, $2(2)-1 = 3$
• For $n=3$, $2(3)-1 = 5$

Knowing this formula is essential when exploring series that involve odd integers, such as in our problem where we evaluate the sum of the first $n$ odd integers.

Inductive Hypothesis Explained

The inductive hypothesis is a key component of mathematical induction. It involves assuming that a statement is true for a certain case, typically denoted as \(k\), in order to prove it for the next case, \(k+1\).In the context of our exercise, the inductive hypothesis is:

• Assume \(1 + 3 + 5 + \cdots + (2k-1) = k^2\) is true for some positive integer \(k\).

This assumption sets the stage to prove the statement for the next integer, \(k+1\). The power of this hypothesis lies in how it connects known cases to establish the truth for the subsequent one. It's like setting up dominoes; by toppling one case, we can trigger a chain reaction proving many others.

The Inductive Step

The inductive step is where we put the inductive hypothesis to work. We use it to show that if a statement holds for $n=k$, it must also hold for $n=k+1$. For our problem:

• Start with the left side from the inductive hypothesis: $k^2 + (2(k+1)-1)$
• Simplify the new term: $2(k+1)-1 = 2k + 1$
• Plug it in to get $k^2 + 2k + 1$

This simplifies perfectly to match the right side: $(k+1)^2 = k^2 + 2k + 1$. By demonstrating that the expression is true for $k+1$, the inductive step solidifies the connection between the two successive cases of the hypothesis.

Base Case in Induction

The base case is where induction begins. It's the simplest instance of a problem that we verify to be true. In our exercise, the base case considers $n=1$:

• The left side: $1$
• The right side: $1^2 = 1$

Both sides are equal, confirming that the formula works when $n=1$. Establishing this initial truth is critical because it serves as the foundational domino in our metaphorical row of dominoes. Without a true base case, the entire chain of inductive reasoning wouldn't work. It's the solid step we need to proceed with confidence to solve for larger $n$. Wouldn't it be interesting how such a small step leads to comprehensive proof extending to all positive integers?