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Thread: 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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    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 $\displaystyle 0 < x < 1$, then we can write $\displaystyle x = \frac{1}{y}$, where $\displaystyle y > 1$.


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

    Since $\displaystyle y > 1, y^2 > 1$.

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

    Therefore $\displaystyle x^2 < 1$.
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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
    $\displaystyle 0 < x < 1$.

    Assume $\displaystyle x^2 \geq 1$

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

    $\displaystyle |x| \geq 1$

    $\displaystyle x \leq -1$ or $\displaystyle x \geq 1$.


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