Algorithms can't work over $\mathbb{R}^n$, as you can't represent real numbers in finite space. Also, you can't pick a number uniformly at random from $\mathbb{R}$. Instead, usually we work over a finite field.

Then we can't do any better. Suppose we are working in the finite field with two elements, $GF(2)$. Suppose that

$$AB - C = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}.$$

Then it is easy to verify that Frievald's algorithm is wrong with probability $1/2$, as $(AB-C)v = 0$ holds with probability $1/2$ when we select $v$ uniformly at random. You can generalize this to a $n\times n$ matrix that is all zeros except for a single entry, and then the probability of false positive is $1/2$.

Algorithms can't work over $\mathbb{R}^n$, as you can't represent real numbers in finite space. Also, you can't pick a number uniformly at random from $\mathbb{R}$. Instead, usually we work over a finite field.

Then we can't do any better. Suppose we are working in the finite field with two elements, $GF(2)$. Suppose that

$$AB - C = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}.$$

Then it is easy to verify that Frievald's algorithm is wrong with probability $1/2$, as $(AB-C)v = 0$ holds with probability $1/2$ when we select $v$ uniformly at random. You can generalize this to a $n\times n$ matrix that is all zeros except for a single entry, and then the probability of false positive is $1/2$.

If you are working over $\mathbb{Q}$, then the same matrix $AB-C$ also provides a similar counterexample. As Greg Martin explains, Freivald's algorithm by definition chooses vectors uniformly at random from $\{0,1\}^n$, and then when $v$ is selected from this distribution, $(AB-C)v=0$ holds with probability $1/2$. (And if you are wondering whether it is possible to do better by choosing $v$ differently, there is no way to choose a number uniformly at random from $\mathbb{Q}$, so it's not clear what distribution you would use.)