text: Hw 1
  
  \begin{document}
  
  \maketitle
  
  \section{Problem 5}
  
  Theorem: There exist irrational numbers a,b such that $a^b$ is rational.
  
  Proof:
  We know $\sqrt{2}$ is irrational if $a=\sqrt{2}$ and $b=\sqrt{2}$ then $a^b=(\sqrt{2})^\sqrt{2}$, then there are two cases, $a^b$ is either rational or irrational.
  
  If $(\sqrt{2})^\sqrt{2}$ is rational then there exist irrational numbers a,b such that $a^b$ is rational.
  
  If $(\sqrt{2})^\sqrt{2}$ is irrational then let $a=(\sqrt{2})^\sqrt{2}$ and $b=\sqrt{2}$ then $a^b=((\sqrt{2})^\sqrt{2})^\sqrt{2}$ $=(\sqrt{2})^2=2$ and $a^b$. Since we know that 2 is rational, this shows that there exist irrational numbers a,b such that $a^b$ is rational.
  
  \end{document}
  
  page revision: 1, last edited: 25 Jan 2018 23:08
explanation: The document is a mathematical proof that demonstrates the existence of irrational numbers \(a\) and \(b\) such that \(a^b\) is rational. It does not contain a question but rather a theorem and its proof.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Hw 1
  
  \begin{document}
  
  \maketitle
  
  \section{Problem 5}
  
  Theorem: There exist irrational numbers a,b such that $a^b$ is rational.
  
  Proof:
  We know $\sqrt{2}$ is irrational if $a=\sqrt{2}$ and $b=\sqrt{2}$ then $a^b=(\sqrt{2})^\sqrt{2}$, then there are two cases, $a^b$ is either rational or irrational.
  
  If $(\sqrt{2})^\sqrt{2}$ is rational then there exist irrational numbers a,b such that $a^b$ is rational.
  
  If $(\sqrt{2})^\sqrt{2}$ is irrational then let $a=(\sqrt{2})^\sqrt{2}$ and $b=\sqrt{2}$ then $a^b=((\sqrt{2})^\sqrt{2})^\sqrt{2}$ $=(\sqrt{2})^2=2$ and $a^b$. Since we know that 2 is rational, this shows that there exist irrational numbers a,b such that $a^b$ is rational.
  
  \end{document}
  
  page revision: 1, last edited: 25 Jan 2018 23:08
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a mathematical proof that demonstrates the existence of irrational numbers \(a\) and \(b\) such that \(a^b\) is rational. It does not contain a question but rather a theorem and its proof.
  
  Label: no-question