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Math Help - direct and indirect proof

  1. #1
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    direct and indirect proof

    hello all, someone posted a similar question to this but he wrote the question wrong so the subsequent answers were wrong.

    I need to prove directly and by contradiction that

    if x is a positive real number and x < 1, then x^2 < 1

    all help will be greatly appreciated
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  2. #2
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    Quote Originally Posted by murielx View Post
    hello all, someone posted a similar question to this but he wrote the question wrong so the subsequent answers were wrong.

    I need to prove directly and by contradiction that

    if x is a positive real number and x < 1, then x^2 < 1

    all help will be greatly appreciated
    If 0 < x < 1, then we can write x = \frac{1}{y}, where y > 1.


    So x^2 = \left(\frac{1}{y}\right)^2 = \frac{1}{y^2}.

    Since y > 1, y^2 > 1.

    Therefore \frac{1}{y^2} < 1.

    Therefore x^2 < 1.
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    Quote Originally Posted by murielx View Post
    hello all, someone posted a similar question to this but he wrote the question wrong so the subsequent answers were wrong.

    I need to prove directly and by contradiction that

    if x is a positive real number and x < 1, then x^2 < 1

    all help will be greatly appreciated
    0 < x < 1.

    Assume x^2 \geq 1

    Then \sqrt{x^2} \geq \sqrt{1}

    |x| \geq 1

    x \leq -1 or x \geq 1.


    This contradicts what we said originally. So x^2 < 1.
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