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Homework Help: If a and b divide n prove or disprove that a.b divides n.

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data
    If a and b divide n prove or disprove that a.b divides n.[a,b,n are positive intergers]

    2. Relevant equations

    3. The attempt at a solution
    Suppose a.b does not divide n.
    Then [itex]\frac{n}{a.b}[/itex]=I [I must not be a postitive interger]
    [itex]\frac{n}{a.b}[/itex]=[itex]\frac{1}{a}[/itex].[itex]\frac{n}{b}[/itex] since a divides n it follows that n=a.x where x is a positive interger.

    This implies that the quotient of two positive intergers cannot be an interger and this is a contradiction.

    I have a feeling this proof is not sufficient since it is a only a contradiction in certain cases where b divides x. If anybody could tell me if this proof is sufficient and/or give me a better one that would be appreciated.
  2. jcsd
  3. Aug 25, 2012 #2


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    Think of a number with a lot of divisors, say 12 :

    Divisors are : 2,3,4,6, 12...

    Can you spot an issue with the products?
  4. Aug 25, 2012 #3
    Is it that the products can be more than the original number 12 and therefore clearly don't divide 12.
  5. Aug 25, 2012 #4


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    Right. 4|12, and 6|12 , but (4)(6)=24. Your statement, (I think) , would be true,if

    you restricted yourself to divisors of n smaller than Sqr(n) , the square root of n.
  6. Aug 25, 2012 #5
    That does makes sense that would work and I also think it would work if the a and b were relative primes such that (a,b)=1.
  7. Aug 25, 2012 #6


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    You're right, 6|300 and 15|300, but 90 does not. This is also a counter to the

    original of a different type--since 90<300 ,so that it could in theory divide it.

    Maybe this counterargument would also work:

    If a|n and b|n implied ab|n without qualification,then we could extend this to :

    a|n and ab|n would imply a(ab)|n . Then a(ab)|n and b|n would imply a(ab)b|n ,....
  8. Aug 25, 2012 #7
    however, if the G.C.D of a and b is 1, then ab does divide n.
    In general, the L.C.M of a and b does divide n by definition.
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