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Conjecture Time!

  1. Jun 30, 2013 #1
    I don't know if this is the proper thing to call it, but I haven't used any mathematical terminology in a while so I think I will try :P

    The number of imperfect roots between any two consecutive perfect roots will always be twice the preceding root number.

    for example there is 2 imperfect roots between ## \sqrt1 and \sqrt4## and 6 in between ##\sqrt9 and \sqrt16## This pattern seems to work I have tried this up for large numbers and it seems to hold, it seems like it should very easy to prove and I am working on this proof as we speak, but I think I could use a little help...
     
  2. jcsd
  3. Jun 30, 2013 #2

    mfb

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    Hint: if n^2 is a perfect square, the next perfect square is (n+1)^2. How many integers (edit: better) are between those two numbers?
     
    Last edited: Jun 30, 2013
  4. Jun 30, 2013 #3
    I don't understand the question mfb? there are countably infinite number, correct?

    My question is strictly for natural number D:
     
  5. Jun 30, 2013 #4

    mfb

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    I mean integers only, of course.
     
  6. Jun 30, 2013 #5
    Well, then it would always be a odd number. unless you count the number you start on then it should be even every time?
     
  7. Jun 30, 2013 #6

    mfb

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    How is 2 (as an example, the number of integers between 1 and 4) odd?
    You can expand (n+1)^2 and get the number of integers between n^2 and (n+1)^2...
     
  8. Jun 30, 2013 #7
    Sorry, I was counting like this 1...2*...3*...4*, I counted each element, like the distance >.< LOL! But you are right that is true, so it works for both squares and square roots, oh that is interesting, I should have checked that case... but now the real question is "Can I make a generalization for this for the n-th root of any "consecutive" numbers like##\sqrt[n]m## and ##\sqrt[n]k## are there n times the previous number where m and k are integers?" But I think you answered my question.
     
  9. Jun 30, 2013 #8
    Let ##n## be a positive integer. You are looking to see if ##2n+1 \stackrel{?}{=} (n+1)^2-n^2##.

    $$2n+1 \stackrel{?}{=} (n+1)^2-n^2 \\ 2n+1 \stackrel{?}{=} ((n+1)-n)((n+1)+n) \\ 2n+1 = 2n+1$$

    I don't understand your confusion.
     
    Last edited: Jun 30, 2013
  10. Jun 30, 2013 #9

    mfb

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    Not n times the previous number, but you can get similar (just more complicated) formulas with the same approach as for the square root (n=2).
     
  11. Jun 30, 2013 #10
    So, they become more complicated than just n times the previous number? I mean it works for consecutive square roots of perfect squares, and it would be a little more complicated for consecutive cubes of perfect cubes and all the way forward, This is awesome!
     
  12. Jul 1, 2013 #11

    mfb

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    Just calculate some examples, and you'll see it.
     
  13. Jul 1, 2013 #12
    I did, and thanks, so I guess it busted that hypothesis lol
     
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