Help Solve Analysis Problem: Showing α is a Cut & αα=β

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SUMMARY

The discussion centers on a mathematical problem involving cuts in the rational numbers, specifically demonstrating that the set α, defined as α={p in Q: p ≤ 0 or p² in β}, is a cut and that αα=β. The problem leverages the established fact that the squares of rational numbers are dense in the positive rationals, which ensures that for any two rationals r1 and r2, there exists a rational p such that r1 < p² < r2. This foundational concept is crucial for proving the properties of the cut α.

PREREQUISITES
  • Understanding of rational numbers and their properties
  • Familiarity with the concept of cuts in ordered sets
  • Knowledge of mathematical density, particularly in relation to rational numbers
  • Basic algebraic manipulation, including squaring rational numbers
NEXT STEPS
  • Study the properties of cuts in ordered fields
  • Explore the concept of density in rational numbers
  • Learn about the implications of the Archimedean property in real analysis
  • Investigate the relationship between cuts and real numbers in mathematical analysis
USEFUL FOR

Mathematics students, particularly those studying real analysis or number theory, as well as educators seeking to understand cuts and density in rational numbers.

zmac
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Please help!
Here is a homework question that I need help starting!
It is a fact that the squares of the rational numbers are dense in the positive rational numbers, that is, for any r1,r2 in Q with 0< r1< r2, there exists p in Q with r1< p2<r2. Let β be a cut such that 0*< β. Let α be the subset of Q defined by α={p in Q: p< or =0 or p2 in β}. Use the abovementioned density of the squares of rationals to show that α is a cut and that αα=β.

ps How do I put mathematical notation here?

Any foresight you can give would be greatly appreciated!
Thanks
 
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The usual questions: what have you tried, where are you stuck, what are the definitions, etc?
 
I am having a little trouble interpreting the question. By p2, do you mean p squared?

Thanks,

Steve
 

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