Problem 1 - Applied Discrete Structures

Question

(a) Use the definition of a field to show that \(\mathbb{Q}(\sqrt{2})\) is a field. (b) Use the definition of vector space to show that \(\mathbb{Q}(\sqrt{2})\) is a vector space over \(\mathbb{Q}\). (c) Prove that \(\\{1, \sqrt{2}\\}\) is a basis for the vector space \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\), and, therefore, the dimension of \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\) is 2 .

Step-by-Step Solution

Answer

\(\mathbb{Q}(\sqrt{2})\) is a field and a vector space over \(\mathbb{Q}\). \(\{1, \sqrt{2}\}\) forms a basis, making the dimension 2 over \(\mathbb{Q}\).

Step 1: Definition of a Field

A field is a set equipped with two operations (addition and multiplication) satisfying the following properties: 1. Closure under addition and multiplication. 2. Associativity of addition and multiplication. 3. Commutativity of addition and multiplication. 4. Additive and multiplicative identities (usually 0 and 1, respectively). 5. Additive inverses (every element has a negative). 6. Multiplicative inverses (every non-zero element has a reciprocal).

Step 2: Demonstrating Field Properties in \(\mathbb{Q}(\sqrt{2})\)

To show \(\mathbb{Q}(\sqrt{2})\) is a field, consider an element \(a + b\sqrt{2}\) where \(a,b \in \mathbb{Q}\):1. **Closure**: Add or multiply any two elements \((a + b\sqrt{2}) + (c + d\sqrt{2}) = (a+c) + (b+d)\sqrt{2}\) and \((a + b\sqrt{2})(c + d\sqrt{2}) = (ac + 2bd) + (ad+bc)\sqrt{2}\), all results are in \(\mathbb{Q}(\sqrt{2})\).2. **Associativity and Commutativity**: Follow directly from \(\mathbb{Q}\) and extend to such expressions.3. **Identities**: \(0 + 0\sqrt{2}\) for addition, and \(1 + 0\sqrt{2}\) for multiplication.4. **Inverses**: The additive inverse is \(-a - b\sqrt{2}\). For multiplicative inverse of \(a+b\sqrt{2}\), if nonzero, use \(\frac{a-b\sqrt{2}}{a^2 - 2b^2}\).

Step 3: Definition of a Vector Space

A vector space over a field \(F\) is a set \(V\) with two operations (vector addition and scalar multiplication) that satisfies the following axioms:1. Closure under addition and scalar multiplication.2. Associativity and commutativity of vector addition.3. Additive identity and inverses.4. Distributive properties connecting scalar multiplication and vector addition.5. Compatibility of scalar multiplication with field multiplication.6. Identity element of scalars acts as unity on vectors.

Step 4: Demonstrating Vector Space Properties in \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\)

For any \(a + b\sqrt{2} \in \mathbb{Q}(\sqrt{2})\) and any scalar (rational number) \(q \in \mathbb{Q}\):1. **Closure under addition and scalar multiplication**: The result of \((a+b\sqrt{2}) + (c+d\sqrt{2})\) and \(q(a+b\sqrt{2}) = qa + qb\sqrt{2}\) are in \(\mathbb{Q}(\sqrt{2})\).2. **Other vector space properties**: Follow the properties of \(\mathbb{Q}\) being associative, commutative, and containing identity and inverse elements for addition, etc.3. **Distributive Laws**: Holds as shown \(q(a+b\sqrt{2}) + (r(a+b\sqrt{2})) = (q+r)(a+b\sqrt{2})\).

Step 5: Showing \(\{1, \sqrt{2}\}\) is a Basis

For a set \(\{1, \sqrt{2}\}\) to be a basis for \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\), it must be linearly independent and span \(\mathbb{Q}(\sqrt{2})\).1. **Span**: Any element \(a + b\sqrt{2}\) in \(\mathbb{Q}(\sqrt{2})\) can be expressed as a linear combination \(a\cdot1 + b\cdot\sqrt{2}\).2. **Linear Independence**: Suppose \(x\cdot1 + y\cdot\sqrt{2} = 0\). Separating gives \(x + y\sqrt{2} = 0\) means both \(x = 0\) and \(y = 0\) as \(\sqrt{2}\) is irrational, proving independence.

Step 6: Determine the Dimension of \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\)

The dimension is defined as the number of elements in the basis. Since \(\{1, \sqrt{2}\}\) is a basis, and it contains two elements, the dimension of \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\) is 2.

Key Concepts

Field Theory

Field theory is a branch of abstract algebra that studies fields, a fundamental algebraic structure in mathematics. A field consists of a set equipped with two operations: addition and multiplication. These operations must satisfy particular properties to qualify as a field. Understanding these properties is crucial for mastering concepts in both higher mathematics and applications like coding theory.
Here are the critical properties of a field:

• Closure: The result of adding or multiplying any elements of the field remains in the field.
• Associativity: Addition and multiplication are associative; meaning, for any elements \(a\), \(b\), and \(c\) in the field, \((a+b)+c = a+(b+c)\) and \((a\cdot b)\cdot c = a\cdot(b\cdot c)\).
• Commutativity: Addition and multiplication are commutative; thus, \(a+b = b+a\) and \(a\cdot b = b\cdot a\).
• Identities: There exist distinct elements 0 and 1, acting as the additive and multiplicative identities, respectively.
• Inverses: Every element has an additive inverse, and every non-zero element has a multiplicative inverse.

In our case, \(\mathbb{Q}(\sqrt{2})\) is established as a field since it satisfies all these conditions when working with elements of the form \(a + b\sqrt{2}\).

Vector Spaces

A vector space over a field \(F\) is a set \(V\) that plays a key role in linear algebra. It serves as the arena where vectors live and multiple algebraic operations interact. Understanding vector spaces is essential for progressing in many areas of mathematics and physics, as they provide a framework for studying linear equations and transformations.
A vector space must satisfy certain axioms:

• Closure: Adding vectors, or multiplying them by scalars from the field, results in new vectors that are still within the vector space.
• Associativity and Commutativity: Vector addition is associative and commutative.
• Identity and Inverses: There's a zero vector (the additive identity), and every vector has an additive inverse.
• Scalar Multiplication: Vectors can be scaled by elements of the field \(F\), showcasing distributive properties.

For example, \(\mathbb{Q}(\sqrt{2})\) constitutes a vector space over \(\mathbb{Q}\) because operations like scalar multiplication by a rational number and vector addition behave according to these rules.

Linear Independence

Linear independence is a key concept in the study of vector spaces, revealing how vectors relate to each other. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others. This concept is vital for understanding the structure of vector spaces and for defining unique solutions to linear equations.
To determine if a set of vectors \(\{v_1, v_2, \ldots, v_n\}\) is linearly independent:

• Form a linear combination: Check if \(c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0\) only when all \(c_i = 0\).
• Unique solutions: For linear independence, the only solution to expressing one vector as a combination of others is the trivial one, where all coefficients \(c_i\) are zero.

In the specific context of \(\mathbb{Q}(\sqrt{2})\), the set \(\{1, \sqrt{2}\}\) is linearly independent over \(\mathbb{Q}\) because neither 1 nor \(\sqrt{2}\) can be expressed as a rational multiple of the other. This property is crucial to understanding vector space structures and calculating bases.

Basis and Dimension

The concepts of basis and dimension are cornerstone concepts in linear algebra. A basis of a vector space is a set of vectors that are linearly independent and span the vector space. The dimension of the space is the number of vectors in a basis set, giving it a practical and theoretical measure of the space's "size."

A basis can be thought of as the "building blocks" for creating every vector in the space through linear combinations. For instance, to find a basis:

• Span: The set must span the vector space, meaning any vector in the space can be expressed as a linear combination of the basis vectors.
• Linear Independence: The basis vectors must be linearly independent.

In the case of \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\), the basis \(\{1, \sqrt{2}\}\) is both linearly independent and spans the space since any element can be produced using these two vectors. Accordingly, the dimension of this vector space is 2, indicating the minimal number of vectors needed to generate the entire space.