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: Recognise the set.
Every element looks like $\frac{a+b\sqrt2}{c+d\sqrt2}$ with rational $a,b,c,d$. Multiplying top and bottom by $c-d\sqrt2$ turns this into $p+q\sqrt2$ with rational $p,q$. So $V=\mathbb{Q}(\sqrt2)=\{p+q\sqrt2\}$.

Step 2: Closed under addition.
Adding two numbers of the form $p+q\sqrt2$ gives another of the same form. So $V$ is closed under addition. That statement is true, so it is not the false one.

Step 3: A subspace over $\mathbb{Q}$.
With rational scalars, $V$ is spanned by $1$ and $\sqrt2$. So over $\mathbb{Q}$ it is a $2$ dimensional space. That makes the '2 dimensional over $\mathbb{Q}$' statement true as well.

Step 4: Not a real subspace.
As a subspace of $\mathbb{R}$ over the reals, $V$ would need to be closed under any real multiple. But $\pi\cdot 1=\pi$ is not in $V$. So $V$ is not a subspace of the real vector space $\mathbb{R}$. That statement is false.

Step 5: Not four dimensional.
Since $V$ has basis $\{1,\sqrt2\}$ over $\mathbb{Q}$, its dimension is $2$, not $4$. So the '4 dimensional' statement is false too.

Step 6: Conclusion.
The false statements are the real subspace claim and the four dimensional claim.
\[ \boxed{B \text{ and } D \text{ are FALSE}} \]