Question

Let \( V \) be the subset of \( \mathbb{R} \) defined by \[ V = \left\{ \frac{a + b \sqrt{2}}{c + d \sqrt{2}} : a, b, c, d \in \mathbb{Q}, c^2 + d^2 \neq 0 \right\}. \]
Which of the following statements is/are FALSE?

• A. \( V \) is closed under the usual addition in \( \mathbb{R} \).
• B. \( V \) is a subspace of the real vector space \( \mathbb{R} \).
• C. \( V \) is a two-dimensional subspace of the vector space \( \mathbb{R} \) over \( \mathbb{Q} \).
• D. \( V \) is a four-dimensional subspace of the vector space \( \mathbb{R} \) over \( \mathbb{Q} \).

Hint:

When analyzing subspaces, check the dimension and ensure that closure under addition and scalar multiplication holds.

Answer 1

The Correct Option is B, D

Step 1: Analyze option (A).
The set \( V \) is closed under addition because the sum of any two elements of the form \( \frac{a + b \sqrt{2}}{c + d \sqrt{2}} \) is also of the same form, with rational coefficients. Hence, option (A) is true.

Step 2: Analyze option (B).
To be a subspace of \( \mathbb{R} \), the set \( V \) must satisfy the properties of closure under addition and scalar multiplication. Since \( V \) consists of rational multiples of real numbers involving \( \sqrt{2} \), it forms a subspace of \( \mathbb{R} \). Hence, option (B) is true.

Step 3: Analyze option (C).
The set \( V \) is spanned by two elements, \( 1 \) and \( \sqrt{2} \), over \( \mathbb{Q} \), making it a two-dimensional subspace of \( \mathbb{R} \) over \( \mathbb{Q} \). Hence, option (C) is true.

Step 4: Analyze option (D).
Since \( V \) is two-dimensional over \( \mathbb{Q} \), it cannot be a four-dimensional subspace. Therefore, option (D) is false.

Step 5: Conclusion.
The correct answer is (D), as \( V \) is a two-dimensional subspace, not four-dimensional.