I've been using ChatGPT (GPT-4) to help with creating quiz questions, solving math problems, and answering students' queries in my classes. Although it sometimes gets things wrong, with the right prompts, it usually provides remarkably helpful answers.
Our university provides students access to ChatGPT (GPT 4). However, not many are using it. They become frustrated if they don't understand the answers or find errors on their first try.
I'm thinking about a couple of things:
Learning to Use AI: Should teaching students to use AI tools like ChatGPT be part of their education? It's not in our syllabus, but it might be a useful skill.
Increasing Usage: Have any teachers successfully got their students to use AI tools regularly? What strategies worked?
I'd appreciate any advice or examples of what has worked in other math classes.
Just to give you an example, what follows is a actual conversion I had with ChatGPT (GPT 4.0). You can judge if it is good enough to be a tool for students. (It did get the 4th one wrong.)
Also note that if I add "Answer the question one by one, and only continue if I say so.", then ChatGPT would correct answer all five. It has problem with long reasoning, breaking the reasoning into small steps increases accuracy.
Me:
Which of the following five statement in linear algebra are true? Explain your answer to each one briefly.
- If there exists a set $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ that spans $V$, then $\dim V \leq p$.
- If there exists a linearly independent set $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ in $V$, then $\dim V \leq p$.
- If there exists a linearly dependent set $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ in $V$, then $\dim V \geq p$.
- If $\dim V = p > 0$, there may not exist a set of $p+1$ vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_{p+1}\}$ such that $\mathrm{Span}\left\{\mathbf{v}_1, \ldots, \mathbf{v}_{p+1}\right\} = V$.
- If for every set of $p$ vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_{p}\}$ in $V$, $\mathrm{Span}\left\{\mathbf{v}_1, \ldots, \mathbf{v}_{p}\right\} \ne V$, then $\dim V \ge p+2$.
ChatGPT:
Let's evaluate each statement one by one:
If there exists a set $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ that spans $V$, then $\dim V \leq p$.
- True. If a set of vectors spans $V$, it means every vector in $V$ can be expressed as a linear combination of these vectors. If $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ spans $V$, the dimension of $V$ is the maximum number of linearly independent vectors in $V$, which cannot exceed $p$. If there are more than $p$ linearly independent vectors, the set $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ would not be able to generate all vectors in $V$, contradicting the assumption that it spans $V$.
If there exists a linearly independent set $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ in $V$, then $\dim V \leq p$.
- False. The dimension of $V$ is defined by the maximum size of a linearly independent set. A linearly independent set of $p$ vectors does not necessarily span $V$; it could be part of a larger linearly independent set. Thus, $\dim V$ can be greater than $p$ if additional vectors can extend the set $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ while maintaining linear independence.
If there exists a linearly dependent set $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ in $V$, then $\dim V \geq p$.
- False. A linearly dependent set means at least one vector in the set can be written as a linear combination of the others. This does not provide information on the dimension of $V$ relative to $p$ because the redundancy (linear dependency) might imply that the effective number of independent vectors is less than $p$. Therefore, $\dim V$ could be less than $p$.
If $\dim V = p > 0$, there may not exist a set of $p+1$ vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_{p+1}\}$ such that $\mathrm{Span}\left\{\mathbf{v}_1, \ldots, \mathbf{v}_{p+1}\right\} = V$.
- True. If $\dim V = p$, any set of $p+1$ vectors in $V$ must be linearly dependent, since the maximum number of linearly independent vectors in $V$ is $p$. Thus, while a set of $p+1$ vectors can still span $V$, they cannot all be linearly independent, and thus some of the vectors in this set are not necessary to span $V$.
If for every set of $p$ vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_{p}\}$ in $V$, $\mathrm{Span}\left\{\mathbf{v}_1, \ldots, \mathbf{v}_{p}\right\} \ne V$, then $\dim V \ge p+2$.
- False. If no set of $p$ vectors spans $V$, it implies that $\dim V > p$. However, $\dim V$ could be exactly $p+1$, where just one more vector is needed to span $V$. Claiming that $\dim V \geq p+2$ assumes that at least two additional vectors are needed, which is not necessarily the case.
