| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The Lebesgue measure of the set E can be determined by considering the conditions given for the points (x, y) in E. The set E is defined as the set of points (x, y) in the unit square I = [0,1]×[0,1] such that sin(x) < 1/2 and cos(x + y) is irrational. First, let's consider the condition sin(x) < 1/2. The sine function is less than 1/2 in the intervals [0, π/6] and [5π/6, π]. However, since we are only considering the interval [0,1], we are only interested in the interval [0, π/6]. This means that the set of x values for which sin(x) < 1/2 is [0, π/6] ∩ [0,1] = [0, π/6]. Next, let's consider the condition that cos(x + y) is irrational. The cosine function takes on irrational values almost everywhere, except for a countable set of points. This means that the set of points (x, y) for which cos(x + y) is irrational has full measure in the unit square I. Therefore, the set E can be written as the intersection of the set of points (x, y) for which sin(x) < 1/2 and the set of points (x, y) for which cos(x + y) is irrational. The Lebesgue measure of E is then equal to the Lebesgue measure of the set of points (x, y) for which sin(x) < 1/2, which is the product of the lengths of the intervals [0, π/6] and [0,1]. Thus, the Lebesgue measure of E is: m(E) = m([0, π/6]) * m([0,1]) = (π/6) * 1 = π/6 Therefore, the answer is π/6. However, it's important to note that the actual value of the Lebesgue measure of E might be different due to the complexity of the condition that cos(x + y) is irrational. The above solution assumes that the set of points (x, y) for which cos(x + y) is irrational has full measure in the unit square I. |
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| deepseekr1 | To find the Lebesgue measure of the set \( E = \{(x, y) \in [0,1] \times [0,1] : \sin(x) < \frac{1}{2}, \cos(x+y) \text{ is irrational}\} \), we analyze the two conditions defining \( E \). 1. **Condition on \( x \):** The inequality \( \sin(x) < \frac{1}{2} \) holds for \( x \in [0, \frac{\pi}{6}) \) since \( \sin(\frac{\pi}{6}) = \frac{1}{2} \) and \( \sin(x) \) is increasing on \([0, 1]\). The measure of this interval is \( \frac{\pi}{6} \). 2. **Condition on \( \cos(x+y) \):** The set of \( \theta \) where \( \cos(\theta) \) is rational is countable. This is because for each rational \( q \in [-1, 1] \), the solutions \( \theta \) to \( \cos(\theta) = q \) form a countable set. Hence, the set of \( \theta \) where \( \cos(\theta) \) is rational has measure zero. Therefore, the set where \( \cos(x+y) \) is irrational is the complement of a measure zero set, which has full measure. Combining these conditions, the set \( E \) is the intersection of the vertical strip \( [0, \frac{\pi}{6}) \times [0,1] \) with the set where \( \cos(x+y) \) is irrational. Since the latter set has full measure, the measure of \( E \) is the same as the measure of the vertical strip, which is \( \frac{\pi}{6} \times 1 = \frac{\pi}{6} \). Thus, the Lebesgue measure of \( E \) is \(\boxed{\dfrac{\pi}{6}}\). |
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| gpto1 | To find the Lebesgue measure of the set \( E \subset I = [0,1] \times [0,1] \) defined by \[ E = \{(x, y) \in I : \sin(x) < \tfrac{1}{2}, \ \cos(x + y) \text{ is irrational}\}, \] we proceed step by step by analyzing the conditions and utilizing properties of the Lebesgue measure and irrational numbers. **Step 1: Determine the range of \( x \) satisfying \( \sin(x) < \tfrac{1}{2} \).** First, consider the function \( \sin(x) \) on the interval \( [0,1] \). Since \( \sin(x) \) is an increasing function from \( x = 0 \) to \( x = \frac{\pi}{2} \approx 1.5708 \), within \( [0,1] \), \( \sin(x) \) increases from \( \sin(0) = 0 \) to \( \sin(1) \approx 0.8415 \). We find the value of \( x \) where \( \sin(x) = \tfrac{1}{2} \): \[ x_0 = \arcsin\left(\tfrac{1}{2}\right) = \frac{\pi}{6} \approx 0.5236. \] Thus, \( \sin(x) < \tfrac{1}{2} \) when \( x \in [0, \frac{\pi}{6}) \). **Step 2: Analyze the condition \( \cos(x + y) \) being irrational.** For any real number \( \theta \), \( \cos(\theta) \) is rational only at countably many points. Specifically, the set of \( \theta \) such that \( \cos(\theta) \) is rational is countable. This implies that for almost every \( \theta \in \mathbb{R} \), \( \cos(\theta) \) is irrational. Given \( x \in [0, \tfrac{\pi}{6}) \), for almost every \( y \in [0,1] \), \( \cos(x + y) \) is irrational. The set of \( y \) such that \( \cos(x + y) \) is rational has Lebesgue measure zero. **Step 3: Compute the Lebesgue measure of \( E \).** Since for each \( x \in [0, \tfrac{\pi}{6}) \), \( \cos(x + y) \) is irrational for almost every \( y \in [0,1] \), the measure of the set of such \( y \) is 1. Therefore, the Lebesgue measure \( m(E) \) is: \[ m(E) = \int_{0}^{\frac{\pi}{6}} \left( \int_{0}^{1} \chi_{\{\cos(x + y) \text{ is irrational}\}}(y) \, dy \right) dx. \] Since the inner integral equals 1 for each \( x \in [0, \frac{\pi}{6}) \): \[ m(E) = \int_{0}^{\frac{\pi}{6}} 1 \, dx = \frac{\pi}{6}. \] **Conclusion:** The Lebesgue measure of \( E \) is \( \frac{\pi}{6} \). **Answer:** The Lebesgue measure of \( E \) is \( \dfrac{\pi}{6} \). |
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| GPT-4o | To determine the Lebesgue measure of the set \( E = \{(x, y) \in I : \sin(x) < \frac{1}{2}, \cos(x+y) \text{ is irrational}\} \), we will proceed step by step. 1. **Condition on \(\sin(x)\)**: \[\sin(x) < \frac{1}{2}\] The sine function \(\sin(x)\) ranges from \(-1\) to \(1\). To find the values of \(x\) for which \(\sin(x) < \frac{1}{2}\), we solve the inequality: \[ \sin(x) < \frac{1}{2} \] This inequality holds when \(x\) is in the intervals: \[ x \in \left[0, \arcsin\left(\frac{1}{2}\right)\right) \cup \left(\pi - \arcsin\left(\frac{1}{2}\right), 2\pi\right] \] Since we are restricted to the interval \([0, 1]\), we need to find the intersection of \([0, 1]\) with the intervals where the sine function is less than \(\frac{1}{2}\). \[ \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \approx 0.5236 \] Therefore, within \([0, 1]\): \[ x \in [0, 0.5236) \cup (1 - 0.5236, 1] \approx [0, 0.5236) \cup (0.4764, 1] \] The total measure of the \(x\)-intervals where \(\sin(x) < \frac{1}{2}\) is: \[ 0.5236 + (1 - 0.4764) = 0.5236 + 0.5236 = 1.0472 \] 2. **Condition on \(\cos(x+y)\)**: \(\cos(x+y)\) must be irrational. The cosine function can take any value in \([-1, 1]\) over its domain. The set of rational numbers in any interval is countable, while the set of irrational numbers is uncountable. For fixed \(x\), the function \(y \mapsto \cos(x+y)\) is continuous and takes on a dense set of values in \([-1,1]\). The irrational numbers in \([-1,1]\) form an uncountable set, and the measure of the set of \(y\) values for which \(\cos(x+y)\) is irrational is the same as the measure of the entire interval since the rationals form a set of measure zero. Therefore, for each fixed \(x\) in the intervals \([0, 0.5236)\) or \((0.4764, 1]\), almost every \(y \in [0, 1]\) will make \(\cos(x+y)\) irrational. 3. **Measure of \(E\)**: The measure of \(E\) is the product of the measures of the \(x\)-intervals and the \(y\)-intervals satisfying the conditions. Since the irrational condition holds for almost every \(y\) in \([0, 1]\) for each \(x\): \[ \text{Measure of } E = \text{Measure of } \{x : \sin(x) < \frac{1}{2}\} \times \text{Measure of } [0, 1] = 1.0472 \times 1 = 1.0472 \] Therefore, the Lebesgue measure of \(E\) is \(1.0472\). |
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| llama3.1-70B-Instruct | To find the Lebesgue measure of the set $E$, we need to break down the problem into manageable parts. ## Step 1: Understand the Conditions for E The set $E$ is defined by two conditions within the unit square $I = [0,1] \times [0,1]$: $sin(x) < \frac{1}{2}$ and $cos(x+y)$ is irrational. ## Step 2: Analyze the First Condition The condition $sin(x) < \frac{1}{2}$ defines a range for $x$. Since $sin(x)$ is periodic with period $2\pi$, and within the interval $[0, 2\pi]$, $sin(x) < \frac{1}{2}$ for $x \in [0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, 2\pi]$, we consider the portion of this range within $[0,1]$. This implies $x \in [0, \frac{\pi}{6})$ since $\frac{5\pi}{6} > 1$. ## Step 3: Analyze the Second Condition The condition that $cos(x+y)$ is irrational is more complex. The cosine function is irrational for most of its domain, except for specific values where it is rational (like $cos(0) = 1$, $cos(\frac{\pi}{2}) = 0$, etc.). However, because $x+y$ varies continuously and the set of rational numbers is countable (and thus has Lebesgue measure 0), the set of $x+y$ values for which $cos(x+y)$ is rational has Lebesgue measure 0. ## Step 4: Combine the Conditions Since the set of $x+y$ values for which $cos(x+y)$ is rational has Lebesgue measure 0, and the condition $sin(x) < \frac{1}{2}$ defines a set of positive measure, the Lebesgue measure of $E$ is essentially determined by the measure of the set defined by $sin(x) < \frac{1}{2}$ within the unit square. ## Step 5: Calculate the Lebesgue Measure The Lebesgue measure of the set defined by $sin(x) < \frac{1}{2}$ within $[0,1]$ is the length of the interval $[0, \frac{\pi}{6})$, which is $\frac{\pi}{6}$. However, since we are considering the unit square $I = [0,1] \times [0,1]$, and the condition on $y$ is essentially that $cos(x+y)$ is irrational (which does not restrict the measure in a significant way due to the nature of irrational numbers), the measure of $E$ is the area of the rectangle defined by $x \in [0, \frac{\pi}{6})$ and $y \in [0,1]$, which is $\frac{\pi}{6} \times 1 = \frac{\pi}{6}$. The final answer is: $\boxed{\frac{\pi}{6}}$ |
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| llama3.1-405B-Instruct | A nice problem in real analysis! To find the Lebesgue measure of $E$, we can use the fact that the Lebesgue measure is countably additive and that the set $E$ can be written as a countable union of disjoint sets. First, let's analyze the condition $\sin(x) < \frac{1}{2}$. This inequality is satisfied when $x \in \left( \frac{\pi}{6}, \frac{5\pi}{6} \right)$. Since $x \in [0,1]$, we have $x \in \left( \frac{\pi}{6}, 1 \right)$. Now, let's consider the condition $\cos(x+y)$ is irrational. This is equivalent to saying that $x+y$ is not a rational multiple of $\pi$. Since $x \in \left( \frac{\pi}{6}, 1 \right)$ and $y \in [0,1]$, we have $x+y \in \left( \frac{\pi}{6}, 2 \right)$. The set of rational multiples of $\pi$ in $\left( \frac{\pi}{6}, 2 \right)$ is countable, so we can write it as $\{q_1, q_2, \ldots\}$. For each $q_i$, the set $\{(x,y) \in I : x+y = q_i\}$ is a line segment in $I$ with Lebesgue measure 0. Now, we can write $E$ as: $$E = \bigcup_{i=1}^\infty \left\{(x,y) \in I : x \in \left( \frac{\pi}{6}, 1 \right), x+y \neq q_i \right\}$$ Each set in the union has Lebesgue measure equal to the length of the interval $\left( \frac{\pi}{6}, 1 \right)$, which is $\frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}$. Since the sets are disjoint, the Lebesgue measure of $E$ is the sum of the measures of each set, which is: $$\mu(E) = \sum_{i=1}^\infty \frac{2\pi}{3} = \infty$$ However, this is not the final answer. Since $E \subseteq I$, we have $\mu(E) \leq \mu(I) = 1$. Therefore, the Lebesgue measure of $E$ is: $$\mu(E) = 1$$ So, the final answer is 1. |
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| DeepSeek-Coder-V2 | To determine the Lebesgue measure of the set \( E = \{(x, y) \in I: \sin(x) < \frac{1}{2}, \cos(x+y) \text{ is irrational}\} \), we will follow a step-by-step approach. 1. **Identify the region where \(\sin(x) < \frac{1}{2}\)**: - The inequality \(\sin(x) < \frac{1}{2}\) holds for \(x \in [0, 1]\) when \(x \in [0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, 1]\). - Therefore, the set \(E\) is restricted to these intervals in \(x\). 2. **Determine the measure of the region where \(\sin(x) < \frac{1}{2}\)**: - The length of the interval \([0, \frac{\pi}{6})\) is \(\frac{\pi}{6}\). - The length of the interval \((\frac{5\pi}{6}, 1]\) is \(1 - \frac{5\pi}{6}\). - Thus, the total length of the intervals where \(\sin(x) < \frac{1}{2}\) is: \[ \frac{\pi}{6} + (1 - \frac{5\pi}{6}) = 1 - \frac{2\pi}{3} \] 3. **Consider the condition \(\cos(x+y)\) is irrational**: - The function \(\cos(x+y)\) takes irrational values for almost all \(y \in [0, 1]\) when \(x+y\) is not an integer multiple of \(\pi\). - Since \(x \in [0, 1]\), \(x+y\) will be irrational for almost all \(y \in [0, 1]\) because the sum of a rational number (x) and an irrational number (y) is irrational. 4. **Combine the conditions**: - The set \(E\) is the intersection of the region where \(\sin(x) < \frac{1}{2}\) and the region where \(\cos(x+y)\) is irrational. - Since \(\cos(x+y)\) is irrational for almost all \(y \in [0, 1]\), the measure of the set where \(\cos(x+y)\) is irrational is 1 for each fixed \(x\). 5. **Calculate the Lebesgue measure of \(E\)**: - The Lebesgue measure of \(E\) is the product of the measure of the interval where \(\sin(x) < \frac{1}{2}\) and the measure of the interval where \(\cos(x+y)\) is irrational. - Therefore, the Lebesgue measure of \(E\) is: \[ (1 - \frac{2\pi}{3}) \times 1 = 1 - \frac{2\pi}{3} \] Thus, the Lebesgue measure of \(E\) is: \[ \boxed{1 - \frac{2\pi}{3}} \] |
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| Qwen2-72B-Instruct | The Lebesgue measure of the set E can be determined by considering the conditions given for the points (x, y) in E. The set E is defined as the set of points (x, y) in the unit square I = [0,1]×[0,1] such that sin(x) < 1/2 and cos(x + y) is irrational. First, let's consider the condition sin(x) < 1/2. The sine function is less than 1/2 in the intervals [0, π/6] and [5π/6, π]. However, since we are only considering the interval [0,1], we are only interested in the interval [0, π/6]. This means that the set of x values for which sin(x) < 1/2 is [0, π/6] ∩ [0,1] = [0, π/6]. Next, let's consider the condition that cos(x + y) is irrational. The cosine function takes on irrational values almost everywhere, except for a countable set of points. This means that the set of points (x, y) for which cos(x + y) is irrational has full measure in the unit square I. Therefore, the set E can be written as the intersection of the set of points (x, y) for which sin(x) < 1/2 and the set of points (x, y) for which cos(x + y) is irrational. The Lebesgue measure of E is then equal to the Lebesgue measure of the set of points (x, y) for which sin(x) < 1/2, which is the product of the lengths of the intervals [0, π/6] and [0,1]. Thus, the Lebesgue measure of E is: m(E) = m([0, π/6]) * m([0,1]) = (π/6) * 1 = π/6 Therefore, the answer is π/6. However, it's important to note that the actual value of the Lebesgue measure of E might be different due to the complexity of the condition that cos(x + y) is irrational. The above solution assumes that the set of points (x, y) for which cos(x + y) is irrational has full measure in the unit square I. |
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| baichuan4 | To find the Lebesgue measure of set $E$, we need to consider the conditions that define $E$. Set $E$ consists of points $(x, y) \in I$ such that $\sin(x) < \frac{1}{2}$ and $\cos(x+y)$ is irrational. We will analyze these conditions separately and then combine them to find the measure of $E$. 1. Consider the condition $\sin(x) < \frac{1}{2}$. The function $\sin(x)$ is periodic with period $2\pi$. In the interval $[0, 1]$, the inequality $\sin(x) < \frac{1}{2}$ is satisfied in the following subintervals: $$[0, \frac{\pi}{6}], [\frac{5\pi}{6}, \frac{7\pi}{6}], [\frac{11\pi}{6}, 1]$$ Note that these intervals are within one period of the sine function. The total length of these intervals is: $$\frac{\pi}{6} + (\frac{7\pi}{6} - \frac{5\pi}{6}) + (1 - \frac{11\pi}{6}) = \frac{\pi}{6} + \frac{2\pi}{6} - \frac{5\pi}{6} = 1 - \frac{2\pi}{3}$$ Since the sine function is symmetric about the $y$-axis, the measure of the set of $x$ values in $[0, 1]$ for which $\sin(x) < \frac{1}{2}$ is $2(1 - \frac{2\pi}{3})$. 2. Now consider the condition $\cos(x+y)$ is irrational. The function $\cos(x+y)$ is continuous and takes all real values as $x$ and $y$ vary in $[0, 1]$. Since the set of irrational numbers is dense in $\mathbb{R}$, the set of $(x, y)$ pairs for which $\cos(x+y)$ is irrational has full measure in $I$. In other words, the measure of this set is $1$. Combining the two conditions, we see that the set $E$ is the intersection of two sets: one defined by the inequality $\sin(x) < \frac{1}{2}$ and the other by the irrationality of $\cos(x+y)$. Since the measure of the set defined by the irrationality of $\cos(x+y)$ is full, the measure of $E$ is determined by the first condition. Thus, the Lebesgue measure of $E$ is $2(1 - \frac{2\pi}{3})$. |
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| doubao | For the condition $\sin(x) \lt \frac{1}{2}$ in the interval $[0,1]$, this is satisfied when $0 \leq x \lt \frac{\pi}{6}$. For the condition $\cos(x+y)$ to be irrational, this is a more complex condition and it's difficult to precisely determine the measure of the set where this holds for all $(x,y) \in I$. However, we know that the set where $0 \leq x \lt \frac{\pi}{6}$ has measure $\frac{\pi}{6}$. And since the second condition is more difficult to analyze precisely in terms of measure, we can't determine the exact Lebesgue measure of $E$ beyond saying that it is at most $\frac{\pi}{6}$. So, in conclusion, we can't determine the exact Lebesgue measure of $E$ but we know it is no more than $\frac{\pi}{6}$. |
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| minimax | To find the Lebesgue measure of the set \( E \), we need to consider the two conditions given for the elements \( (x, y) \) in \( E \): 1. \( \sin(x) < \frac{1}{2} \) 2. \( \cos(x + y) \) is irrational Let's analyze each condition separately: 1. \( \sin(x) < \frac{1}{2} \) The sine function is periodic with period \( 2\pi \), and within the interval \([0, 1]\), the values of \( \sin(x) \) range from \( \sin(0) = 0 \) to \( \sin(1) \approx 0.84147 \). The inequality \( \sin(x) < \frac{1}{2} \) is satisfied for \( x \) in the interval \( I_1 = [0, \arcsin(\frac{1}{2})] \cup [\pi - \arcsin(\frac{1}{2}), \pi] \). Since \( \arcsin(\frac{1}{2}) = \frac{\pi}{6} \), the interval \( I_1 \) becomes \( [0, \frac{\pi}{6}] \cup [\frac{5\pi}{6}, \pi] \). However, since we are only considering the interval \([0, 1]\), we only take the part of \( I_1 \) that falls within this interval, which is \( [0, \frac{\pi}{6}] \). 2. \( \cos(x + y) \) is irrational The cosine function is also periodic with period \( 2\pi \), and it takes on both rational and irrational values. However, the set of points \( (x, y) \) for which \( \cos(x + y) \) is irrational is dense in the plane and has full measure. This is because the set of rational numbers is countable and has measure zero, while the set of irrational numbers is uncountable and has full measure in any interval. Therefore, almost every point in the plane will have an irrational cosine value when plugged into \( x + y \). Combining these two conditions, we see that the set \( E \) is the intersection of the set where \( \sin(x) < \frac{1}{2} \) and the set where \( \cos(x + y) \) is irrational. Since the second condition holds almost everywhere, the measure of \( E \) will be determined by the first condition. The set \( [0, \frac{\pi}{6}] \) has a Lebesgue measure of \( \frac{\pi}{6} \), and since \( y \) can take any value in \([0, 1]\), the measure of \( E \) in the \( y \)-direction is 1. Therefore, the Lebesgue measure of \( E \) is the product of the measures in the \( x \)- and \( y \)-directions: \[ \text{Lebesgue measure of } E = \frac{\pi}{6} \times 1 = \frac{\pi}{6} \] So the Lebesgue measure of \( E \) is \( \frac{\pi}{6} \). |
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| kimi | To find the Lebesgue measure of the set \( E \), we need to consider the conditions given for \( E \). The set \( E \) is defined as: \[ E = \{(x, y) \in I: \sin(x) < \frac{1}{2}, \cos(x + y) \text{ is irrational} \} \] Let's break down the problem into parts to find the measure of \( E \). 1. **The condition \( \sin(x) < \frac{1}{2} \)**: This condition defines a subset of the interval \([0, 1]\) for \( x \). The sine function is less than \( \frac{1}{2} \) in the interval \([0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \pi]\). However, since \( x \) is in \([0, 1]\), we only consider the interval \([0, \frac{\pi}{6})\). The length of this interval is \(\frac{\pi}{6}\). Since \( x \) is in \([0, 1]\), the measure of this condition is \(\frac{\pi}{6}\). 2. **The condition \( \cos(x + y) \) is irrational**: The cosine function is irrational for almost all real numbers (in the sense of Lebesgue measure). This is because the set of rational numbers is countable and has measure zero, while the set of irrational numbers is uncountable and has full measure. Therefore, for almost all \( y \) in \([0, 1]\), \( \cos(x + y) \) is irrational. Now, we need to combine these two conditions. Since \( \cos(x + y) \) is irrational for almost all \( y \), and we are considering \( \sin(x) < \frac{1}{2} \), we can say that for almost all \( y \) in \([0, 1]\), the pair \((x, y)\) will satisfy both conditions. The product measure of the two sets is the measure of \( E \). Since the measure of \( \sin(x) < \frac{1}{2} \) is \(\frac{\pi}{6}\) and the measure of \( \cos(x + y) \) being irrational is almost 1 for each \( x \), the Lebesgue measure of \( E \) is approximately: \[ \text{Measure of } E \approx \frac{\pi}{6} \times 1 = \frac{\pi}{6} \] So, the Lebesgue measure of \( E \) is \(\frac{\pi}{6}\). |
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| wenxinyiyan | 为了解决这个问题，我们首先需要理解集合E的构成。集合E由满足两个条件的点(x, y)组成： 1. $\sin(x) < \frac{1}{2}$ 2. $\cos(x+y)$ 是无理数 我们可以将这个问题分解为两个部分来处理。 ### 第一部分：处理 $\sin(x) < \frac{1}{2}$ * $\sin(x) < \frac{1}{2}$ 在区间 $[0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, 1]$ 内成立。 * 因此，对于x的这部分，我们可以限制x在这些区间内。 ### 第二部分：处理 $\cos(x+y)$ 是无理数 * $\cos(x+y)$ 是无理数的情况稍微复杂一些，因为对于任意的x，我们需要找到使 $\cos(x+y)$ 成为无理数的y值。 * 幸运的是，无理数的集合在实数中是稠密的，这意味着在每个开区间内，我们都可以找到无理数。 * 因此，对于每个满足 $\sin(x) < \frac{1}{2}$ 的x，我们都可以找到一个y，使得 $\cos(x+y)$ 是无理数。 ### 结合两部分 * 对于每个满足 $\sin(x) < \frac{1}{2}$ 的x，我们都可以找到一个y，使得 $\cos(x+y)$ 是无理数。这意味着，对于每个这样的x，存在一个y使得点(x, y)属于集合E。 * 因此，集合E的Lebesgue测度与满足 $\sin(x) < \frac{1}{2}$ 的x的集合的Lebesgue测度相同。 * 由于 $\sin(x) < \frac{1}{2}$ 在区间 $[0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, 1]$ 内成立，这些区间的总长度是 $1 - \frac{\pi}{3}$。 因此，集合E的Lebesgue测度是 $1 - \frac{\pi}{3}$。 |
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| zhipu | To find the Lebesgue measure of the set \( E = \{(x, y) \in I : \sin(x) < \frac{1}{2}, \cos(x+y) \text{ is irrational}\} \), we need to break down the problem into more manageable parts. ### Step 1: Understanding the Set \( E \) The set \( E \) is defined by two conditions: 1. \( \sin(x) < \frac{1}{2} \) 2. \( \cos(x+y) \) is irrational ### Step 2: Analyzing the First Condition \( \sin(x) < \frac{1}{2} \) The condition \( \sin(x) < \frac{1}{2} \) can be analyzed as follows: - \( \sin(x) = \frac{1}{2} \) at \( x = \frac{\pi}{6} + 2k\pi \) and \( x = \frac{5\pi}{6} + 2k\pi \) for \( k \in \mathbb{Z} \). - Therefore, \( \sin(x) < \frac{1}{2} \) for \( x \in \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 1\right) \). Since \( x \in [0, 1] \), we need to consider the intervals within \( [0, 1] \): - \( \left(0, \frac{\pi}{6}\right) \approx \left(0, 0.5236\right) \) - \( \left(\frac{5\pi}{6}, 1\right) \approx \left(0.8727, 1\right) \) ### Step 3: Analyzing the Second Condition \( \cos(x+y) \text{ is irrational} \) The cosine function \( \cos(x+y) \) takes values in \([-1, 1]\). The set of irrational numbers in \([-1, 1]\) has a Lebesgue measure of 1 because the irrational numbers are dense in \([-1, 1]\) and the Lebesgue measure of the rational numbers is zero. ### Step 4: Combining the Conditions We need to find the measure of the set \( E \) where both conditions hold: - For \( x \in \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 1\right) \), the \( y \)-values are not restricted by the first condition. - For each \( x \) in these intervals, \( \cos(x+y) \) being irrational happens almost everywhere (with measure 1) in \( [0, 1] \). ### Step 5: Calculating the Lebesgue Measure The Lebesgue measure of \( E \) can be calculated by considering the measure of the \( x \)-intervals and the \( y \)-interval: - The measure of \( x \in \left(0, \frac{\pi}{6}\right) \) is \( \frac{\pi}{6} \approx 0.5236 \). - The measure of \( x \in \left(\frac{5\pi}{6}, 1\right) \) is \( 1 - \frac{5\pi}{6} \approx 1 - 0.8727 = 0.1273 \). The total measure of the \( x \)-intervals is: \[ \frac{\pi}{6} + \left(1 - \frac{5\pi}{6}\right) = 1 - \frac{4\pi}{6} = 1 - \frac{2\pi}{3} \] Since for each \( x \) in these intervals, the \( y \)-values satisfying \( \cos(x+y) \) being irrational have a measure of 1, the Lebesgue measure of \( E \) is: \[ \left(1 - \frac{2\pi}{3}\right) \times 1 = 1 - \frac{2\pi}{3} \] ### Final Answer The Lebesgue measure of \( E \) is: \[ 1 - \frac{2\pi}{3} \] This result accounts for the \( x \)-intervals where \( \sin(x) < \frac{1}{2} \) and the fact that \( \cos(x+y) \) being irrational occurs almost everywhere in the \( y \)-interval. |
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| gpto1 | To find the Lebesgue measure of the set \( E \subset I = [0,1] \times [0,1] \) defined by \[ E = \{(x, y) \in I : \sin(x) < \tfrac{1}{2}, \ \cos(x + y) \text{ is irrational}\}, \] we proceed step by step by analyzing the conditions and utilizing properties of the Lebesgue measure and irrational numbers. **Step 1: Determine the range of \( x \) satisfying \( \sin(x) < \tfrac{1}{2} \).** First, consider the function \( \sin(x) \) on the interval \( [0,1] \). Since \( \sin(x) \) is an increasing function from \( x = 0 \) to \( x = \frac{\pi}{2} \approx 1.5708 \), within \( [0,1] \), \( \sin(x) \) increases from \( \sin(0) = 0 \) to \( \sin(1) \approx 0.8415 \). We find the value of \( x \) where \( \sin(x) = \tfrac{1}{2} \): \[ x_0 = \arcsin\left(\tfrac{1}{2}\right) = \frac{\pi}{6} \approx 0.5236. \] Thus, \( \sin(x) < \tfrac{1}{2} \) when \( x \in [0, \frac{\pi}{6}) \). **Step 2: Analyze the condition \( \cos(x + y) \) being irrational.** For any real number \( \theta \), \( \cos(\theta) \) is rational only at countably many points. Specifically, the set of \( \theta \) such that \( \cos(\theta) \) is rational is countable. This implies that for almost every \( \theta \in \mathbb{R} \), \( \cos(\theta) \) is irrational. Given \( x \in [0, \tfrac{\pi}{6}) \), for almost every \( y \in [0,1] \), \( \cos(x + y) \) is irrational. The set of \( y \) such that \( \cos(x + y) \) is rational has Lebesgue measure zero. **Step 3: Compute the Lebesgue measure of \( E \).** Since for each \( x \in [0, \tfrac{\pi}{6}) \), \( \cos(x + y) \) is irrational for almost every \( y \in [0,1] \), the measure of the set of such \( y \) is 1. Therefore, the Lebesgue measure \( m(E) \) is: \[ m(E) = \int_{0}^{\frac{\pi}{6}} \left( \int_{0}^{1} \chi_{\{\cos(x + y) \text{ is irrational}\}}(y) \, dy \right) dx. \] Since the inner integral equals 1 for each \( x \in [0, \frac{\pi}{6}) \): \[ m(E) = \int_{0}^{\frac{\pi}{6}} 1 \, dx = \frac{\pi}{6}. \] **Conclusion:** The Lebesgue measure of \( E \) is \( \frac{\pi}{6} \). **Answer:** The Lebesgue measure of \( E \) is \( \dfrac{\pi}{6} \). |
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| deepseekv3 | To find the Lebesgue measure of the set \( E = \{(x, y) \in I: \sin(x) < \frac{1}{2}, \cos(x+y) \text{ is irrational}\} \), we proceed as follows: 1. **Analyze the condition \(\sin(x) < \frac{1}{2}\):** - The equation \(\sin(x) = \frac{1}{2}\) holds when \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\) in the interval \([0, \pi]\). - Since \(\sin(x)\) is increasing on \([0, \frac{\pi}{2}]\) and decreasing on \([\frac{\pi}{2}, \pi]\), the inequality \(\sin(x) < \frac{1}{2}\) holds for \(x \in [0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, 1]\). - The length of this set in \([0, 1]\) is \(\frac{\pi}{6} + \left(1 - \frac{5\pi}{6}\right) = 1 - \frac{2\pi}{3}\). 2. **Analyze the condition \(\cos(x+y)\) is irrational:** - The function \(\cos(x+y)\) is continuous and non-constant. The set of irrational numbers in \([-1, 1]\) has Lebesgue measure 2 (since the rationals have measure 0). - For fixed \(x\), the set of \(y\) such that \(\cos(x+y)\) is irrational has Lebesgue measure 1 in \([0, 1]\). - Therefore, for each \(x\) satisfying \(\sin(x) < \frac{1}{2}\), the set of \(y\) such that \(\cos(x+y)\) is irrational has full measure. 3. **Compute the measure of \(E\):** - The measure of \(E\) is the product of the measure of the set of \(x\) satisfying \(\sin(x) < \frac{1}{2}\) and the measure of the set of \(y\) satisfying \(\cos(x+y)\) is irrational. - Thus, the measure of \(E\) is \(\left(1 - \frac{2\pi}{3}\right) \times 1 = 1 - \frac{2\pi}{3}\). However, we must ensure that \(x\) and \(y\) are within \([0, 1]\). Since \(\frac{5\pi}{6} \approx 2.61799 > 1\), the upper bound \(1\) is used instead of \(\frac{5\pi}{6}\). Therefore, the measure of the set of \(x\) satisfying \(\sin(x) < \frac{1}{2}\) in \([0, 1]\) is \(\frac{\pi}{6}\). Thus, the Lebesgue measure of \(E\) is: \[ \boxed{\frac{\pi}{6}} \] |
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