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  • How can I show that gcd(a,b) =gcd(b,r) ? Euclidean algorithm? given a=bq+r?

    Anonymous Christian
    Sep 5, 2014
    MATH
    Comments Off on How can I show that gcd(a,b) =gcd(b,r) ? Euclidean algorithm? given a=bq+r?

    Answer

    Let gcd(a,b) = t. … (1)
    So in a = bq + r, let a = ct and b=dt, with gcd(c,d) = 1. so
    ct = dtq + r.
    Thus t divides r. Let r = te. We get,
    c = dq + e. …(2)

    Now let gcd ( d,e) = h. ( h divides d & e) in (2) ,would imply that ( h divides c )! But since ( h divides c ) & ( h divides d ) means that ( h divides c & d ). But gcd ( c,d) = 1. Hence h divides 1, i.e. h = 1. If h is not equal to ‘1’, we have the contradiction a positive integer dividing ‘1’ and not equal to one. No such integer exists.

    In other words, gcd ( d,e ) = 1. Thus,
    gcd ( b , r ) = gcd ( dt, te ) = t * gcd ( d,e) = t*1 = t
    That is gcd ( b,r ) = t …(3)

    “Compare” (1) and (3) to see plainly that
    gcd ( a,b) = gcd ( b,r) because both equals the value ‘t’ !

    QED.

anonymous-christian

Jonathan Ramachandran is founder at AnonymousChristian.org which serves as a Christian Blog discussing the Greater Hope of Biblical Christian in Christ Alone. He is also an IMO (International Math Coach) and a featuring Guitarist for many bands.

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