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Math Help - Show that if a, b∈ R with a < b, then there exists a number x ∈ I with a < x < b

  1. #1
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    Show that if a, b∈ R with a < b, then there exists a number x ∈ I with a < x < b

    Let I be the set of real numbers that are not rational. (We call
    elements of I irrational numbers. Show that if a, b are real
    numbers with a < b, then there exists a number x ∈ I with
    a < x < b.

    Im not sure if this is right:

    Using the density theorem to the real numbers a/√2 and b/√2, we obtain a rational number such that:

    a/√2 < r < b/√2

    where r ≠ 0

    This gives ---> x = r√2, which is an irrational number because it contains √2, which itself is irrational. This satisfies b < x < a

    Is this right? or do i need more in the proof?
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  2. #2
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    No. That is the usual proof. It works.
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