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Math Help - [SOLVED] Inequality proof involving natural numbers.

  1. #1
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    [SOLVED] Inequality proof involving natural numbers.

    Suppose m1 < m2 and n1 < n2. Show that m1n1 < m2n2.

    ANy advice...THanks in advance.
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  2. #2
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    Inequality

    Hello jzellt
    Quote Originally Posted by jzellt View Post
    Suppose m1 < m2 and n1 < n2. Show that m1n1 < m2n2.

    ANy advice...THanks in advance.
    Not true. 3<8 and -2<-1, but -6>-8.

    Perhaps you meant to say that these are natural numbers? In which case, m_1 < m_2 \Rightarrow m_1 +m = m_2, for some m \in \mathbb{N}, and n_1 < n_2 \Rightarrow n_1+n = n_2, for some n \in \mathbb{N}.

    So m_2n_2 = (m_1 +m)(n_1 + n)

    = m_1n_1 + mn_1+nm_1+mn

    > m_1n_1

    Grandad
    Last edited by Grandad; February 13th 2009 at 04:23 AM. Reason: Added proof of suggested amendment
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  3. #3
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    A bit simpler, I think: If m_1< m_2, and n_1> 0 then m_1n_1< m_2n_1. If n_1< n_2 and m_2> 0 then m_2n_1< m_2n_2. Those together ("<" is a transitive relation) give m_1n_1< m_2n_2.

    Notice that does not say anything about m_1. This only requires that 3 of the four numbers be positive.
    Last edited by HallsofIvy; February 13th 2009 at 08:48 AM.
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