Explore BrainMass

Explore BrainMass

    Real Analysis of Irrational Numbers

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Proof:
    Given any two real numbers a<b ,there exists an irrational number t satisfying a<t<b

    © BrainMass Inc. brainmass.com March 4, 2021, 6:02 pm ad1c9bdddf
    https://brainmass.com/math/real-analysis/real-analysis-irrational-numbers-25355

    Solution Preview

    Let decimal expansions of two real numbers a and b (a<b) first differ in the nth digit.
    First, if the nth digit is not the last digit of b, consider the number bn obtained from b by cutting all the digits of b after the nth, then: a<bn<b.
    there is two cases here:
    (1)b has non zero digits after the nth, Append to bn one zero digit, give the result number to t;
    (2)b has zero digits right after the nth. Let bm be ...

    Solution Summary

    The expert provides a proof for given any two real numbers a<b ,there exists an irrational number t satisfying a<t<b

    $2.19

    ADVERTISEMENT