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p/q<1

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p/q<1 [#permalink]

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New post 05 Apr 2013, 08:15
Can I algebraically prove that if p/q<1 then q/(p^2) is not necessarily greater than 1? Picking numbers is easy, I am just wondering if I can algebraically solve it.
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Re: p/q<1 [#permalink]

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New post 05 Apr 2013, 08:47
Not very clear on the problem without specifics about whether p & q are positive integers

Because if p > 0 and q > 0 and integers then p/q < 1 => q/p > 1 and q/p2 can be >=< 1
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Kudos [?]: 10 [0], given: 43

Re: p/q<1 [#permalink]

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New post 05 Apr 2013, 11:01
nt2010 wrote:
Not very clear on the problem without specifics about whether p & q are positive integers

Because if p > 0 and q > 0 and integers then p/q < 1 => q/p > 1 and q/p2 can be >=< 1


Yes sorry, p and q are positive integers. When you say can q/p2 be greater or less than 1, is there a way we can reach that algebraically.
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GMAT 1: 530 Q43 V20
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Kudos [?]: 60 [1] , given: 65

Re: p/q<1 [#permalink]

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New post 05 Apr 2013, 12:32
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q/p > 1 so divide both side by p since P is a positive integer P > 0

q/p2 > 1/p. IFF q > p


//kudos, please if this explanation is good
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Manager
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Kudos [?]: 10 [0], given: 43

Re: p/q<1 [#permalink]

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New post 05 Apr 2013, 12:39
nt2010 wrote:
q/p > 1 so divide both side by p since P is a positive integer P > 0

q/p2 > 1/p. IFF q > p


//kudos, please if this explanation is good


One last question, why do you say q/p2 > 1/p if q>p? How did you get to q>p?
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GMAT 1: 530 Q43 V20
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Kudos [?]: 60 [0], given: 65

Re: p/q<1 [#permalink]

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New post 05 Apr 2013, 12:52
Remember we got to the equation q/p2 > 1/p from q/p > 1 and if you multiple both side by p you'll get the condition q > p.

Hope this is clear.
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Re: p/q<1   [#permalink] 05 Apr 2013, 12:52
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