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If n is a positive integer n then sqrt(4n-2) is irrational.

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data

    if n is a positive integer than √(4n-2) is irrational.




    2. Relevant equations



    3. The attempt at a solution

    √(4n-2) Assume is rational

    then by definition of rationality

    √(4n-2)=p/q for some integers p,q where q≠0

    so √(2(2n-1))=p/q by factoring out the 2 as common

    √2 *√(2n-1) =p/q

    but 2n-1 is always odd

    so √(2n-1) is always odd

    now let u=√(2n-1)

    but √2*u cannot be factored since √2 is irrational and u is odd.

    so √(4n-2)≠p/q

    Therefore our assumption must have been wrong therefore

    √(4n-2) must be irrational

    Is this proof ok??
     
  2. jcsd
  3. Apr 30, 2012 #2

    Curious3141

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    Homework Helper

    Not OK.

    Problem here. How do you know 2n-1 is a perfect square? Only integers can be meaningfully described as even or odd, so you're making an unfounded assumption that 2n-1 is a square.

    There's a simple and instructive way to solve this by considering the cases where 2n-1 is a perfect square, and when it isn't, and, in the second case, the prime factorisation of 2n-1. However, you're probably more comfortable solving this in the typical mould you're used to.

    So begin by stating your assumption that [itex]\sqrt{4n-2} = \frac{p}{q}[/itex], where p and q are coprime (no factors in common, because the fraction is reduced).

    Now, [itex]\frac{p^2}{q^2} = 4n - 2 \Rightarrow p^2 = 2(2n-1)q^2[/itex].

    and draw the obvious conclusion about p's evenness.

    Now rearrange to: [itex]p^2 + 2q^2 = 4nq^2[/itex].

    Clearly, the RHS is divisible by 4, which means the LHS also has to be divisible by 4. Since p is even, [itex]4 | p^2[/itex]. Can you now derive a contradiction with regard to q?
     
  4. Apr 30, 2012 #3
    Hi.

    You should look at even or odd character of p and q.
    p/q is already reduced so not both p and q are even.
    But you will see in another way that both p and q are even. Contradiction.

    Regards.
     
    Last edited: Apr 30, 2012
  5. Apr 30, 2012 #4
    Well as shown p2 and q2 both have a common factor of 2 but p2/q2 was already in it's most reduced form.

    Contradiction, therefore √(4n-2) must be irrational.

    ??
     
  6. Apr 30, 2012 #5

    Curious3141

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    Yes, but how did you deduce that q is even?
     
  7. Apr 30, 2012 #6
    sorry, from the line

    p2=2(2nq2-q2)

    Since LHS is even therefore p is even.

    ie p2/2 is also even

    so p2/2=2nq2-q2

    so p2/2=q(2n-1)

    Since LHS is even then the RHS must be to.

    clearly 2n-1 is not even

    so obviously q has to be even if q(2n-1) is even.
     
  8. Apr 30, 2012 #7

    Curious3141

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    That should be q2(2n-1), but yes, that reasoning is sound.
     
  9. Apr 30, 2012 #8
    Ok thanks a bunch Curious!
     
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