Chapter 2: Problem 6
For any field elements \(a, b\) and natural numbers \(m, n,\) prove that (i) \(a^{m} a^{n}=a^{m+n}\) (ii) \(\quad\left(a^{m}\right)^{n}=a^{m n} ;\) (iii) \(\quad(a b)^{n}=a^{n} b^{n} ;\) (iv) \(\quad(m+n) a=m a+n a ;\) (v) \(n(m a)=(n m) \cdot a ;\) (vi) \(\quad n(a+b)=n a+n b\).
Short Answer
Expert verified
Properties can be proven using the laws of exponents and distributive properties in a field.
Step by step solution
01
Prove Property (i)
To prove \(a^m a^n = a^{m+n}\), consider the definition of exponentiation. When multiplying two powers of the same base, you add the exponents: \[a^m a^n = a^{m+n}\]This is because \(a^m\) is the product of \(a\) with itself \(m\) times and \(a^n\) is the product of \(a\) with itself \(n\) times, thus totaling \(m + n\) factors of \(a\). The property is therefore proven.
02
Prove Property (ii)
To prove \((a^m)^n = a^{m n}\), use the rule of exponents: Raising a power to another power means multiplying the exponents. Consider \[(a^m)^n = a^{m imes n}\]This holds because \((a^m)^n\) means multiplying \(a^m\) by itself \(n\) times, leading to a total of \(m \times n\) occurrences of \(a\). This confirms the property.
03
Prove Property (iii)
To prove \((a b)^n = a^n b^n\), expand the expression using the definition of an exponent. \[(a b)^n = (a b)(a b)...(a b) ext{ (n times)}\]Each pair \((a b)\) contributes one \(a\) and one \(b\) in each factor, resulting in \(a^n b^n\) when collected across all \(n\) instances. The distribution of exponents thus holds.
04
Prove Property (iv)
To prove \((m+n)a = ma + na\), use the distributive property of scalar multiplication over addition for a field. For any field element \(a\), \[(m+n)a = ma + na\]The left side \((m+n)a\) corresponds to multiplying \(a\) by the sum of \(m\) and \(n\), producing the equivalent result found by summing \(ma\) and \(na\) separately.
05
Prove Property (v)
To show that \(n(ma) = (nm) \cdot a\), apply associative property of multiplication in a field. Here, \[n(ma) = (nm) imes a\]It indicates performing the multiplication \((m \times n)\) first, then using it to scale \(a\). This is identical to multiplying \(a\) by \(m\) followed by \(n\), confirming the property.
06
Prove Property (vi)
To establish \(n(a+b) = na+ nb\), utilize the distributive property of multiplication over addition in a field: \[n(a+b) = na + nb\]This is accomplished by distributing the multiplication over each element \(a\) and \(b\) in the sum, which matches separately scaled versions, aligning with the rule that multiplication distributes across addition.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Field Elements
In mathematics, a field is a set equipped with two operations, typically called addition and multiplication, and a set of rules (axioms) they must follow. Field elements are the individual members of such a set. Fields are important in both pure and applied mathematics, particularly in linear algebra and number theory.
Fields provide the foundation for many algebraic properties and operations. They support addition, subtraction, multiplication, and division (excluding division by zero). In fields, you can solve equations such as linear equations and polynomial equations. For example, real numbers (\mathbb{R}) and complex numbers (\mathbb{C}) are examples of fields.
Fields provide the foundation for many algebraic properties and operations. They support addition, subtraction, multiplication, and division (excluding division by zero). In fields, you can solve equations such as linear equations and polynomial equations. For example, real numbers (\mathbb{R}) and complex numbers (\mathbb{C}) are examples of fields.
- Additive identity exists: There is an element in the field, typically denoted as 0, such that for any element \(a\), \(a + 0 = a\).
- Multiplicative identity exists: There is an element 1 such that for any element \(a\), \(a \times 1 = a\).
- Every element has an additive inverse, meaning for any element \(a\), there exists an element \(-a\) such that \(a + (-a) = 0\).
- Every non-zero element has a multiplicative inverse, so for any element \(aeq0\), there exists an element \(a^{-1}\) such that \(a \times a^{-1} = 1\).
Exponentiation Rules
Exponentiation is a mathematical operation involving two numbers, the base and the exponent. When dealing with exponentiation in the context of fields, the properties of exponents are governed by specific rules that simplify various operations. The rules are particularly useful when working either with large computations or algebraic simplifications. This section explains some common exponentiation rules.
- The Product of Powers Rule: For any base \(a\) raised to exponents \(m\) and \(n\), this rule states \(a^m \times a^n = a^{m+n}\). Here, you multiply the same base and simply add the exponents because you're essentially working with repeated multiplication.
- The Power of a Power Rule: According to this rule, \((a^m)^n = a^{m \times n}\). This rule applies when a power itself is raised to another power, indicating multiplication of the exponents.
- The Power of a Product Rule: Shows that \((ab)^n = a^n b^n\). When a product of two bases \((a \cdot b)\) is raised to an exponent \(n\), distribute the exponent to each element of the product.
Distributive Property
The distributive property is a fundamental principle that facilitates the simplification of expressions and solving of equations. It states how multiplication is distributed over addition or subtraction, ensuring that each term is multiplied independently by the common factor.
For any numbers or elements \(a, b,\) and \(c\), the distributive property can be expressed as \(a(b + c) = ab + ac\). This property is particularly useful in algebra when you deal with polynomials and factorization.
For any numbers or elements \(a, b,\) and \(c\), the distributive property can be expressed as \(a(b + c) = ab + ac\). This property is particularly useful in algebra when you deal with polynomials and factorization.
- Distributing the multiplication: By distributing a factor across a sum or difference, you break down complex expressions into simpler components, facilitating easier calculation.
- Working in reverse: Factoring out a common factor from a sum, using distributive property in reverse, simplifies expressions and solves equations efficiently.
Scalar Multiplication
Scalar multiplication is a fundamental operation in linear algebra, transforming vectors by stretching or compressing them. When applied to field elements, scalar multiplication involves multiplying a scalar—a single number—by each entry of a vector or matrix, or in simpler terms, multiplying a field element by a constant.
A scalar multiplication can be represented by \(k(a + b) = ka + kb\). This property shows how multiplying a sum by a scalar distributes the operation over each component. It aligns with the distributive property and is used in many mathematical applications.
A scalar multiplication can be represented by \(k(a + b) = ka + kb\). This property shows how multiplying a sum by a scalar distributes the operation over each component. It aligns with the distributive property and is used in many mathematical applications.
- Scalar multiplication with vectors: Each component of the vector is multiplied by the scalar, thereby resizing the vector proportionally.
- Scaling field elements: Multiplying field elements by a scalar alters their magnitude but maintains proportionality and direction.
- Applications in physics and engineering: Commonly used in contexts like force calculations, where you might multiply a set of force vectors by a scalar.