Problem 1 - A Book of Abstract Algebra

Question

Prove that \(a^{m} a^{n}=a^{m+n}\) in the following cases: (i) \(m=0\) (ii) \(m<0\) and \(n>0\) (iii) \(m<0\) and \(n<0\)

Step-by-Step Solution

Answer

Proven for all cases: (i) \(a_0a^n = a^n\), (ii) \(m < 0, n > 0\), (iii) \(m < 0, n < 0\) using definitions and exponent properties.

Step 1: Understanding the Problem

We need to prove the properties of exponents, specifically that \(a^m \times a^n = a^{m+n}\) for different cases involving the exponents \(m\) and \(n\). We'll examine each scenario separately: \(m = 0\), \(m < 0, n > 0\), and \(m < 0, n < 0\).

Step 1: Case m = 0

We want to prove \(a^0 \times a^n = a^{0+n} = a^n\). By definition, \(a^0 = 1\) for any non-zero \(a\). Therefore, \(a^0 \times a^n = 1 \times a^n = a^n\), which is equal to \(a^{0+n}\).

Step 2: Case m 0

For this case, let's set \(m = -p\) where \(p > 0\) and \(n = q\) where \(q > 0\). Then, \(a^m = a^{-p} = \frac{1}{a^p}\) and \(a^n = a^q\).\[ a^m \times a^n = \frac{1}{a^p} \times a^q = \frac{a^q}{a^p} = a^{q-p} \]\( a^{m+n} = a^{q-p} \), so the expression holds.

Step 3: Case m < 0 and n < 0

Let \(m = -p\) and \(n = -q\) where \(p, q > 0\). Then \(a^m = a^{-p} = \frac{1}{a^p}\) and \(a^n = a^{-q} = \frac{1}{a^q}\).\[ a^m \times a^n = \frac{1}{a^p} \times \frac{1}{a^q} = \frac{1}{a^{p+q}} = a^{-(p+q)} \]This simplifies to \(a^{m+n}\) because \(m+n = -p - q\).

Key Concepts

Exponentiation

Exponentiation is a mathematical operation that involves raising a number, known as the base, to the power of an exponent. The exponent indicates how many times the base is multiplied by itself. For example, in the expression \(a^n\), "\(a\)" is the base, and "\(n\)" is the exponent.

• If \(n = 2\), then \(a^2 = a \times a\).
• When \(n = 3\), it becomes \(a^3 = a \times a \times a\).

Exponentiation simplifies repetitive multiplication. It is fundamental in algebra, providing a compact way to express repeated multiplication. Understanding how exponents work is essential to solving algebraic expressions, especially when dealing with exponential functions and equations.

Negative Exponents

Negative exponents represent the reciprocal of the base raised to the corresponding positive power. That is, if you have \(a^{-n}\), it translates to \(\frac{1}{a^n}\). This concept can be a bit tricky initially but is crucial for simplifying expressions.

• For example, \(a^{-1} = \frac{1}{a}\).
• Similarly, \(b^{-2} = \frac{1}{b^2}\).

This property is particularly useful when dealing with fractions and when simplifying complex algebraic expressions, as seen in the exercise where you multiply different powers. Understanding negative exponents helps in proving relationships involving exponential expressions, as it allows simplifying equations by converting them to more manageable forms.

Zero Exponent

The zero exponent rule states that any non-zero number raised to the power of zero is equal to one. Thus, for any base \(a\), \(a^0 = 1\), provided \(a\) is not zero. This rule might seem counterintuitive, but it fits naturally with the rules of exponents for maintaining consistency.

• If \(a = 0\), \(a^0\) is usually undefined.
• The concept holds true regardless of whether \(a\) is a real or complex number.

For example, when multiplying powers like in the given problem, \(a^0 \times a^n = 1 \times a^n = a^n\), maintaining this rule ensures expressions are simplified correctly. Recognizing the zero exponent is key to grasping more difficult algebraic concepts.

Algebraic Proofs

Algebraic proofs involve demonstrating the truth of mathematical statements using logical reasoning and algebraic techniques. The exercise explores proofs about the laws of exponents. Specifically, the associative property: \(a^m \times a^n = a^{m+n}\). Proving such properties requires understanding and applying the rules of exponentiation.

• Case analysis for various possibilities, like zero or negative exponents, ensures comprehensive proofs.
• Breaking down equations into simpler forms often involves substituting expressions like \(a^{-p} = \frac{1}{a^p}\).

Clear algebraic proofs help solidify foundational concepts in mathematics, serving as a basis for more advanced topics. Engaging with these proofs encourages logical thinking and problem-solving skills, essential for tackling complex algebraic challenges.