| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B. 2 | B |
| deepseekr1 | The degree of a field extension \( \mathbb{Q}(\sqrt{2}) \) over \( \mathbb{Q} \) is determined by the dimension of \( \mathbb{Q}(\sqrt{2}) \) as a vector space over \( \mathbb{Q} \). Since \( \sqrt{2} \) is a root of the irreducible polynomial \( x^2 - 2 \) over \( \mathbb{Q} \), its minimal polynomial has degree 2. This implies \( \mathbb{Q}(\sqrt{2}) \) has a basis \( \{1, \sqrt{2}\} \), giving the extension degree 2. Answer: B |
B |
| gpto1 | B | B |
| GPT-4o | B. 2 | B |
| llama3.1-70B-Instruct | B |
B |
| llama3.1-405B-Instruct | B | B |
| DeepSeek-Coder-V2 | B | B |
| Qwen2-72B-Instruct | B. 2 | B |
| baichuan4 | B. 2 | B |
| doubao | B. 2. The degree of a field extension is determined by the degree of the minimal polynomial. The minimal polynomial of √2 over Q is x^2 - 2, which has degree 2. | B |
| minimax | B. 2 | B |
| kimi | B. 2 | B |
| wenxinyiyan | B. 2 Explanation: The degree of a field extension K over F, denoted [K:F], is the dimension of K as a vector space over F. In this case, we have the field extension Q(sqrt(2)) over Q. We can view Q(sqrt(2)) as a vector space over Q with basis {1, sqrt(2)}. This means that every element of Q(sqrt(2)) can be written as a linear combination of 1 and sqrt(2) with coefficients in Q. For example, an element of Q(sqrt(2)) such as 3 + 4*sqrt(2) can be expressed as a linear combination of 1 and sqrt(2) as follows: 3 + 4*sqrt(2) = 3*1 + 4*sqrt(2) Since the basis has 2 elements, the dimension of Q(sqrt(2)) as a vector space over Q is 2. Therefore, the degree of the field extension Q(sqrt(2)) over Q is 2. |
B |
| zhipu | B The degree of the field extension \(\mathbb{Q}(\sqrt{2})\) over \(\mathbb{Q}\) is 2. This is because \(\sqrt{2}\) is a root of the polynomial \(x^2 - 2\), which is irreducible over \(\mathbb{Q}\). Therefore, \(\mathbb{Q}(\sqrt{2})\) is a degree 2 extension of \(\mathbb{Q}\). |
B |
| gpto1 | B | B |
| deepseekv3 | The degree of a field extension \( \mathbb{Q}(\sqrt{2}) \) over \( \mathbb{Q} \) is the dimension of \( \mathbb{Q}(\sqrt{2}) \) as a vector space over \( \mathbb{Q} \). The minimal polynomial of \( \sqrt{2} \) over \( \mathbb{Q} \) is \( x^2 - 2 \), which has degree 2. Therefore, the degree of the extension \( \mathbb{Q}(\sqrt{2}) \) over \( \mathbb{Q} \) is 2. Thus, the correct answer is: **B. 2** |
B |