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Thread: Absolute value proofs

  1. #1
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    Absolute value proofs

    I need to prove the implication,

    $\displaystyle |a| < |b| \Rightarrow a^2 < b^2$ I know that $\displaystyle x^2 \mbox{is equal to} |x|^2 $ but how can I use that in a formal proof?

    Also I need to show that $\displaystyle |a|<|b| \Leftrightarrow a^2 < b^2$

    Once again all I can think of is the thing I wrote above. Thank you!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by meg0529 View Post
    I need to prove the implication,

    $\displaystyle |a| < |b| \Rightarrow a^2 < b^2$ I know that $\displaystyle x^2 \mbox{is equal to} |x|^2 $ but how can I use that in a formal proof?
    note that an alternate definition for $\displaystyle |x|$ is $\displaystyle \sqrt{x^2}$. also note that the square root function is a strictly increasing function, that is, for positive $\displaystyle x$ and $\displaystyle y$, $\displaystyle x < y \implies \sqrt{x} < \sqrt{y}$.

    Now, to prove our implication, we can use the contrapositive: assume $\displaystyle a^2 \ge b^2$, then we have $\displaystyle \sqrt{a^2} \ge \sqrt{b^2}$. But that means $\displaystyle |a| \ge |b|$.

    Also I need to show that $\displaystyle |a|<|b| \Leftrightarrow a^2 < b^2$

    Once again all I can think of is the thing I wrote above. Thank you!
    in light of what i did before, try this one
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    note that an alternate definition for $\displaystyle |x|$ is $\displaystyle \sqrt{x^2}$. also note that the square root function is a strictly increasing function, that is, for positive $\displaystyle x$ and $\displaystyle y$, $\displaystyle x < y \implies \sqrt{x} < \sqrt{y}$.

    Now, to prove our implication, we can use the contrapositive: assume $\displaystyle a^2 \ge b^2$, then we have $\displaystyle \sqrt{a^2} \ge \sqrt{b^2}$. But that means $\displaystyle |a| \ge |b|$.

    in light of what i did before, try this one

    This might be really stupid, but by proving using contrapositive A -> B are we not proving B->A using inverse?
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  4. #4
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    Why not just apply simple rules?
    $\displaystyle \begin{gathered}
    \left| a \right| < \left| b \right|\, \Rightarrow \,\sqrt {a^2 } < \sqrt {b^2 } \, \Rightarrow \,a^2 < b^2 \hfill \\
    \hfill \\
    a^2 < b^2 \, \Rightarrow \,\left| a \right|^2 < \left| b \right|^2 \, \Rightarrow \,\left| a \right| < \left| b \right| \hfill \\
    \end{gathered} $
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