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Square root of a square plus 1

  1. Dec 31, 2008 #1
    1. The problem statement, all variables and given/known data
    prove that if n is a natural number greater than 0, then sqrt(n^2 + 1) is not a natural number.


    2. Relevant equations



    3. The attempt at a solution
    I can't tell if I am right, which probably means that I am not.

    Assume sqrt(n^2+1) is a natural number. Then there is a natural number j such that j = sqrt(n^2+1). Thus
    j^2 = n^2 + 1. Then, I am stuck.
    It seems too easy to prove, is there some little trick I am missing.
     
  2. jcsd
  3. Dec 31, 2008 #2

    D H

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    What is (j+1)2?
     
  4. Dec 31, 2008 #3

    HallsofIvy

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    If [itex]\sqrt{n^2+1}= m[/itex] then [itex]n^2+ 1= m^2[/itex]. Now, what can you say about [itex]n^2- m^2[/itex]?
     
  5. Dec 31, 2008 #4
    [itex] n^2 - m^2 = -1 [/itex] Which gives me a contradiction because this can only happen when n=0 and m =1, and the problem assumes n is strictly > 0.

    Did I get it?

    If so, then thank you!
     
  6. Dec 31, 2008 #5
    Well, yeah, i think that should be right.

    However, i would be inclined to proceede this way:

    [tex]m^2-n^2=1=>(m-n)(m+n)=1[/tex] Then since both m, n are supposed to be integers(natural nrs) then the only possible way for this to be true is if:

    [tex]m-n=m+n=1[/tex]

    or

    [tex]m-n=m+n=-1[/tex]

    Because, we know that the only numbers that have inverses in Z are 1 and -1.

    From the firs eq we get:m=1, n=0, which is a contradiction since the hypothesis says, n>0

    From the second: m=-1 and n=0.

    Which is again not possible.
     
  7. Dec 31, 2008 #6

    jgens

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    Possibly another way of going about it:

    We begin with some number n∈N, we may then claim n^2∈N. We now want to prove that √(n^2+1) ∉N. We do this by defining the next natural number m by m=(n+1); hence, m^2=(n+1)^2= n^2+2n+1. Therefore n^2<n^2+1<m^2 ∀n>0, proving that √(n^2+1) ∉N. Q.E.D.
     
    Last edited: Dec 31, 2008
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