Proofs & Rational Number Btwn $\sqrt2$ & $\sqrt2 + \frac{1}{1000}$

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SUMMARY

This discussion focuses on proving that there exists a rational number between the square root of 2 and the square root of 2 plus 1/1000. Participants emphasize the importance of understanding the properties of real numbers and rational numbers. Additionally, they explore the positivity and negativity of the function f(x) = (3x - 2)(x - 1)(x + 1) and seek alternative methods for analyzing its intervals. The conversation highlights foundational concepts in real analysis and algebra.

PREREQUISITES
  • Understanding of real numbers and rational numbers
  • Familiarity with the properties of functions and their intervals
  • Knowledge of polynomial functions and their behavior
  • Basic skills in mathematical proofs and inequalities
NEXT STEPS
  • Study the proof of the density of rational numbers in real numbers
  • Learn about the Intermediate Value Theorem in calculus
  • Explore methods for finding intervals of positivity and negativity in polynomial functions
  • Investigate the use of derivatives to determine critical points and behavior of functions
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Students of mathematics, particularly those studying real analysis and algebra, as well as educators seeking to enhance their understanding of rational numbers and polynomial functions.

KelvinMa
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1. Show that between any two real numbers there is a rational number.
You may assume that for each a>0 and b>0 in R there is a positive integer satisfying b<ma.

2.Find the intervals on which f(x) = (3x -2)(x 1)(x + 1) is positive, and
the intervals on which it is negative.

3.Find a rational number p/q between root of 2 and root of 2 + (1/1000)
 
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Welcome to PF!

Hi KelvinMa! Welcome to PF! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
thx tiny-tim : D
i really don't have any ideas about the first question and the third question
i ve looked up on my notes and i couldn't find any clues
the second question i ll find the max and min points but is there any other way to do it?
 

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