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    Ok, say I have an inequality...

    x_2n+1 < x_2n+3 < x_2n+4 < x_2n+2

    If I prove that x_2n+1 < x_2n+2 is true, is that the same as proving the entire inequality in four terms is true? Ie if I prove

    x_2n+1 < x_2n+2

    I have also proven

    x_2n+1 < x_2n+3 < x_2n+4 < x_2n+2

    Cheers
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    Wat does x_2n mean?

    coz im pretty sure that ur inequality isnt true... x_2n = k

    k+1<k+3<k+4<k+2 isnt true
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    They are sequences. It is true, but it doesnt matter what the inequality is...

    Say you had..

    a < b < c < d

    If I prove a < d do I consequentially prove a < b < c < d
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    (Original post by Ewan)
    They are sequences. It is true, but it doesnt matter what the inequality is...

    Say you had..

    a < b < c < d

    If I prove a < d do I consequentially prove a < b < c < d

    imo not.
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    (Original post by Willis123)
    imo not.
    EDIT: No I didn't question is right.. anyway...
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    Up we go, someone must know this, its not that hard a question.
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    (Original post by Ewan)
    They are sequences. It is true, but it doesnt matter what the inequality is...

    Say you had..

    a < b < c < d

    If I prove a < d do I consequentially prove a < b < c < d
    No, why would you? If you had a = 4 and d = 6, you could have b = \pi and c = 20349230940234, which obviously wouldn't satisfy the inequality.
 
 
 
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Updated: November 15, 2008
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