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Help needed with extended Euclidean algorithm (Bézout's identity) watch

    • Thread Starter
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    I suppose this is rather a stupid question, but I'm currently looking at the working for a linear diophantine equation, and I can't understand where certain things come from.

    The numbers in question are:

    a = 75
    b = 34
    c = (a,b)

    And the working is as follows:

    1 = 7 - 1 * 6

    = 7 - 1 * ( 34 - 4 * 7 )

    = 5 * 7 - 1 * 34

    = 5 * ( 75 - 2 * 34 ) - 1 * 34

    = 5 * 75 - 11 * 34


    It's probably something really simple that I'm missing, but I don't understand how you go from 7 - 1 * ( 34 - 4 * 7 ) to 5 * 7 - 1 * 34 , and from 5 * ( 75 - 2 * 34 ) - 1 * 34 to 5 * 75 - 11 * 34 .

    I hope someone can help me.

    Thanks
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    (Original post by und)
    = 7 - 1 * ( 34 - 4 * 7 )

    = 5 * 7 - 1 * 34
    It's just a re-arrangement.

    = 7 - 1 * ( 34 - 4 * 7 )

    = 7 - 1* 34 +1* 4 * 7 )

     = 7 -1*34 + 4*7

    =5*7-1*34

    and your second query is similar.
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    (Original post by ghostwalker)
    It's just a re-arrangement.

    = 7 - 1 * ( 34 - 4 * 7 )

    = 7 - 1* 34 +1* 4 * 7 )

     = 7 -1*34 + 4*7

    =5*7-1*34

    and your second query is similar.
    Nevermind. I really am being stupid. I don't know why they didn't just write out the proof in letters. It would have been easier to understand.

    Anyway, thanks for the help.
 
 
 

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