Problem 2 - A Book of Abstract Algebra

Question

If \(a b=b a\), then \((a b)^{n}=a^{n} b^{n}\) for every positive integer \(n .\) Prove by induction.

Step-by-Step Solution

Answer

For every positive integer \(n\), \((ab)^n = a^n b^n\) holds true by induction.

Step 1: Base Case

Start with the base case where \(n = 1\). We need to show that \((ab)^1 = a^1 b^1\).\Given \(ab = ba\), we directly have \((ab)^1 = ab\) and the expression \(a^1 b^1 = ab\). Therefore, for \(n = 1\), the expression holds true.

Step 2: Inductive Step Assumption

Assume that for some positive integer \(k\), the expression \((ab)^k = a^k b^k\) holds true. This is our induction hypothesis.

Step 3: Inductive Step Proof

Now we prove that the expression also holds for \(k+1\), i.e., \((ab)^{k+1} = a^{k+1} b^{k+1}\). Start with the left side:\[(ab)^{k+1} = (ab)^k \cdot ab\]By the induction hypothesis, \((ab)^k = a^k b^k\). Substitute this into the equation:\[(ab)^{k+1} = (a^k b^k) \cdot ab\]Since \(ab = ba\), relate \(b^k \cdot a\) and \(a \cdot b^k\) as:\[(a^k b^k) \cdot ab = a^k \cdot (b^k \cdot ab) = a^k \cdot (a \cdot b^k) \cdot b \]\[= a^k \cdot a \cdot b^k \cdot b = a^{k+1} b^{k+1}\]

Step 4: Conclusion

By mathematical induction, since the base case holds and the inductive step is proven, \((ab)^n = a^n b^n\) for every positive integer \(n\).

Key Concepts

Commutative Property

The commutative property is a fundamental principle in mathematics, particularly in algebra. It states that the order in which two elements are added or multiplied does not affect the result. When applied to multiplication, the commutative property can be expressed as:

• For any two numbers, \(a\) and \(b\), \(a \cdot b = b \cdot a\).

This property is crucial in simplifying expressions and solving equations, as it allows for the rearrangement of terms. In the context of the given problem, the commutative property of multiplication is the reason why the expression \((ab)^1 = a^1b^1\) holds true since \(ab = ba\). Understanding this property helps in recognizing why the multiplication of any two numbers or expressions is adaptable to changes in order without impacting the product.

Exponents

Exponents are powerful tools in mathematics that represent repeated multiplication of a number by itself. The number being multiplied is called the base, and the exponent indicates how many times the base is used as a factor. Here's the basic format of exponentiation:

• \(a^n\) means the base \(a\) is multiplied \(n\) times, forming the expression \(a \times a \times \cdots \times a\) (\(n\) times).

In the problem, when we say \((ab)^n\), it means the expression \(ab\) is multiplied by itself \(n\) times. Similarly, \(a^n\) and \(b^n\) mean that \(a\) and \(b\) are individually multiplied by themselves \(n\) times. Recognizing how exponents work enables one to apply them correctly in algebraic manipulations and proofs, such as induction proofs involving exponential terms.

Inductive Hypothesis

The inductive hypothesis is a key component of the method of mathematical induction, a powerful proof technique. In this process, the inductive hypothesis involves assuming that a statement holds true for a given integer \(k\). It's critical because it serves as the foundation for proving that the statement holds for \(k+1\). The general steps involve:

• Assuming a statement is true for \(n = k\).
• Using this assumption to demonstrate that the statement must then be true for \(n = k+1\).

In our exercise, this hypothesis assumes the formula \((ab)^k = a^k b^k\) is true. Using this assumption, you can manipulate the algebraic expressions to show that it also applies to \((ab)^{k+1} = a^{k+1} b^{k+1}\). The default acceptance of this hypothesis, pending proof of the next step, solidifies the continuity of validity across all natural numbers when using induction.

Base Case

In mathematical induction, the base case is the initial step where you prove the statement is true for the initial value, typically \(n = 1\). It is essential because it serves as the starting point for the inductive process. Ensuring the base case is true establishes the baseline of the domino effect that induction causes.

• The statement \((ab)^1 = a^1b^1\) forms the base case for this particular problem.
• Given that \(ab = ba\), it is straightforward to see that the statement holds since both sides evaluate to \(ab\).

By confirming the base case, you ensure the induction process starts on solid ground. If the base case didn't hold, the entire induction proof structure would collapse, similar to how a domino setup fails if the first tile isn't ready to topple.