Is Q a Vector Space Over R: Exploring the Axioms and Properties

The set of rational numbers, denoted as ( mathbb{Q} ), is often explored in the context of vector spaces, but specifically over the field of real numbers, denoted as ( mathbb{R} ). This article delves into whether ( mathbb{Q} ) satisfies the axioms required to be a vector space over ( mathbb{R} ), and explores related concepts such as uncountability and cardinality.

Vector Space Axioms

To be a vector space, a set ( V ) must satisfy several axioms. These include closure under vector addition and scalar multiplication, the existence of an additive identity and additive inverses, and the fulfillment of vector space properties such as associativity, commutativity, and distributivity.

For the set of rational numbers ( mathbb{Q} ) to be a vector space over the real numbers ( mathbb{R} ), it must:

Be closed under vector addition: For any two rational numbers ( u, v in mathbb{Q} ), their sum ( u v ) must also be a rational number. Be closed under scalar multiplication: For any rational number ( v in mathbb{Q} ) and any real number ( c in mathbb{R} ), the product ( c cdot v ) must also be a rational number. Contain an additive identity element, typically denoted as 0. Contain additive inverses for every element in the set. Satisfy the standard vector space properties regarding vector addition and scalar multiplication.

Verification of Vector Space Properties

The set of rational numbers ( mathbb{Q} ) does indeed satisfy these axioms when considered over the field of real numbers ( mathbb{R} ). The operations of addition and scalar multiplication in ( mathbb{R} ) are well-defined for the elements of ( mathbb{Q} ), and ( mathbb{Q} ) is closed under these operations. Therefore, ( mathbb{Q} ) is a vector space over ( mathbb{R} ).

Counter Arguments and Further Discussions

Is Q a Vector Space Over R - No?

One common argument is that ( mathbb{Q} ) cannot be a vector space over ( mathbb{R} ) because, for any choice of scalar multiplication ( c cdot x_1 ), ( x_1 ) must be a rational number for all real numbers ( x ). However, this argument overlooks the requirement that the scalar must come from ( mathbb{R} ), not just ( mathbb{Q} ).

Consider the function ( f(x) x^{1} ), which is an injective function. This implies that ( mathbb{Q} ) must contain a subset of the same cardinality as ( mathbb{R} ). Since the real numbers ( mathbb{R} ) are uncountable, and the rational numbers ( mathbb{Q} ) are countable, this is a contradiction. Therefore, ( mathbb{Q} ) cannot be a vector space over ( mathbb{R} ).

Is R a Vector Space Over Q? Yes.

On the other hand, the set of real numbers ( mathbb{R} ) can indeed be considered a vector space over the set of rational numbers ( mathbb{Q} ). This is because every field can be regarded as a vector space over itself or a sub-field of itself. In this case, ( mathbb{R} ) is an infinite-dimensional vector space over ( mathbb{Q} ) and has the cardinality of the continuum.

Final Consideration: Multiply 1 by (sqrt{2})

Another argument involves the specific case of multiplying 1 (a rational number) by ( sqrt{2} ), which is a real number but not a rational number. This is a counterexample that directly shows ( mathbb{Q} ) is not closed under scalar multiplication over ( mathbb{R} ).

Thus, the set of rational numbers ( mathbb{Q} ) is not a vector space over the field of real numbers ( mathbb{R} ) due to the lack of closure under scalar multiplication.

By understanding these axioms, properties, and counterexamples, one can better grasp the nuanced nature of vector spaces and mathematical structures.