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Math Help - PROVE THE FOLLOWING has 3 parts

  1. #1
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    PROVE THE FOLLOWING has 3 parts

    1.
    i) if p is prime, a,b are integers, and p divdes ab, then p divides a or p divides b[Hint use result when proving the theorem there exists integers m, n such that g = m*a=n*b]

    ii) If r is a rational number then there exist integers m,n such that n>0, r = m/n and m, n have no common factors.. [USE WELL ORDERING PRINCIPLE]


    iii) PROVE SQRT(21) is irrational [Hint use part 1 and part 2]
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  2. #2
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    Quote Originally Posted by ruprotein View Post
    1.
    i) if p is prime, a,b are integers, and p divdes ab, then p divides a or p divides b[Hint use result when proving the theorem there exists integers m, n such that g = m*a=n*b]
    Let a,b>0.

    If p|(ab) then \gcd(a,p)=1,p (definition of prime: no non-trivial proper divisors). If \gcd(a,p)=p then p|a and proof is complete. It \gcd(a,p)=1 then p|b (by Euclid's Lemma)*

    *)Let a,b,c>0, and if c|ab and \gcd(a,c)=1 then a|b.
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    Quote Originally Posted by ruprotein View Post
    ii) If r is a rational number then there exist integers m,n such that n>0, r = m/n and m, n have no common factors.. [USE WELL ORDERING PRINCIPLE]
    For everything positive.

    The consequence of the Well-Ordering principle is that any two positive integers a,b has the greatest common divisor d.*)

    In that case we can write,
    \frac{a'd}{b'd}=\frac{a'}{b'}
    Where, \gcd(a',b')=1.

    *)Let S=\{ax+by>0|x,y \in \mathbb{Z} \}.
    Then by Zorn's lemma this set has a minimial element (this is greatest common divisor).
    Thus, any two positive integers have a greatest common divisor.
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  4. #4
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    I think that the well ordering principle referred here is the axiom “Any non-empty set of positive integers has a first term”. Then consider \left\{ {k \in Z^ +  :kr \in Z} \right\}. That set is not empty and has a first term N.
    Now \left( {\exists M \in Z} \right)\left[ {r = \frac{M}{N}} \right], so show that M & N cannot have a common divisor other than 1.
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  5. #5
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    Quote Originally Posted by Plato View Post
    “Any non-empty set of positive integers has a first term”.
    I know. I do not like to take as an axiom, it makes me want to vomit. Using AC is the most natual approach and hence this is a theorem.
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  6. #6
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    It may well make you want to vomit. But I will bet 2 to 1 that that is the approach required by this instructor/text. It is the natural way to do it in a basic real analysis course.
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