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Homework Help: Natural number

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Hello! I am new user on this forum.

    I have one problem with Algebra.

    Here it is:

    For which [tex]n \in \mathbb{N}[/tex], this fraction is also natural number?

    [tex]\frac{n^2-n+2}{n+1}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I don't have any idea, where to start from. Thanks for the help.
     
  2. jcsd
  3. Mar 7, 2008 #2

    tiny-tim

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    Hi Theofilius! Welcome to PF! :smile:

    Whenever you have a nasty fraction …
    :smile: turn it into a nice one! :smile:

    So … do the long division … try expressing [tex]\frac{n^2-n+2}{n+1}[/tex] as a sum of an ordinary polynomial (that isn't a fraction), and a number over (n + 1).

    And then … :smile:
     
  4. Mar 8, 2008 #3
    Like this?

    [tex](n^2-n+2)(n+1)^-^1[/tex]

    And then what?
     
  5. Mar 8, 2008 #4

    tiny-tim

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    No.

    Like this: (n+7)/(n - 3) = 1 + 10/(n - 3).

    Try again! :smile:
     
  6. Mar 8, 2008 #5
    How did you find this? And what did we prove with this equation?
     
  7. Mar 8, 2008 #6

    tiny-tim

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    Ah!

    Obviously, 1 = (n - 3)/(n - 3).

    So 1 + 10/(n - 3)
    = (n - 3)/(n - 3) + 10/(n - 3)
    = (n - 3 + 10)/(n - 3)
    = (n - 7)/(n - 3).

    Similarly, you can use n = n(n - 3)/(n - 3), n^2 = n^2(n - 3)/(n - 3), and so on …

    Have a go! :smile:
     
  8. Mar 8, 2008 #7
    I think you have some error. We can't write (n+7)/(n - 3) = 1 + 10/(n - 3), because if I write n=1,
    (1+7)/(1-3)=1+10/(1-3)
    8/-2=1+10/-2
    -4=-4

    And if we get n=1 in the fraction, we will have [tex]\frac{1^2-1+2}{1+1}=1[/tex]
     
  9. Mar 8, 2008 #8

    tiny-tim

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    But -4=-4 is correct!

    Theofilius, you are somehow going to have to convince yourself that [tex](n+7)/(n - 3) = 1 + 10/(n - 3)[/tex] , or you'll never be able to deal with fractions of polynomials.

    First, try some numbers other than 1.

    Then go though my proof again:
    and tell me which bit of the proof you don't follow. :smile:
     
  10. Mar 8, 2008 #9
    Look, we need to find out for which number n which belongs to the natural numbers, we get also natural number for the fraction. Not an equation -4=-4 or whatever.
     
  11. Mar 8, 2008 #10

    tiny-tim

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    Theofilius, I repeat:
    (a) That's very true!

    (b) Take my word for it - once you've rewritten [tex]\frac{n^2-n+2}{n+1}[/tex] as a sum of an ordinary polynomial (that isn't a fraction), and a number over (n + 1), the answer will be obvious. :smile:
     
  12. Mar 8, 2008 #11
    And what have I done with it? With every n, number I get natural number?
     
  13. Mar 8, 2008 #12

    tiny-tim

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    No, you'll find that with most n, you don't. But with some you do …

    Try, and you'll see …
     
  14. Mar 8, 2008 #13
    And my task says, for which n we get natural number for the fraction?
     
  15. Mar 8, 2008 #14

    HallsofIvy

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    And, as you have already been told, your fraction is the same as 1+ 10/(n-3).

    1 is, of course, a natural number. For what values of n is 10/(n-3) a natural number?

    If you can't see it immediately try n= 1, 2, 3, 4, 5, etc.
     
  16. Mar 8, 2008 #15
    [tex]\frac{n^2-n+2}{n+1} \neq 1+ \frac{10}{n-3}[/tex]

    let's get n=2

    [tex]\frac{2^2-2+2}{2+1} \neq 1+ [/tex] [tex]\frac{10}{2-3}[/tex]

    [tex]\frac{4}{3} \neq 1+ [/tex] [tex]\frac{10}{-1}[/tex]

    [tex]\frac{4}{3} \neq -9[/tex]
     
    Last edited: Mar 8, 2008
  17. Mar 8, 2008 #16

    Shooting Star

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    Why? Have you tried to simplify the RHS and see whether you get back (n+7)/(n - 3)?

    What tiny-tim is trying to do is teach you algebraic division again. :smile:

    When you divide (n^2-n+2) by (n+1), what is the quotient and the remainder? In the example he has given, when you divide n+7 by n-3, you get a quotient of 1 and a remainder of 10. So, (n+7)/(n - 3) = 1 + 10/(n - 3). He wants you to express your original fraction in this way.
     
  18. Mar 8, 2008 #17
    Oh my god! It was its example? I thought it was connected with my example. Lol!
    When I'll divide n^2-n+2:(n+1)=n-2
    So I will get natural number for every n-2 number, right?
     
  19. Mar 8, 2008 #18

    Shooting Star

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    No. Write it as quotient and remainder form.
     
  20. Mar 8, 2008 #19
    So, I've done it.

    [tex](n+1)(n-2)+0=n^2-n+2[/tex]

    [tex](n+1)(n-2)+0=n^2-n+2[/tex] /: [tex]\frac{1}{n+1}[/tex]

    [tex]n-2=\frac{n^2-n+2}{n+1}[/tex]
     
    Last edited: Mar 8, 2008
  21. Mar 8, 2008 #20

    Shooting Star

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    Not correct...

    What you have to do is write (n^2-n+2)/(n+1) as Q + R/n, where Q would be a polynomial in n whose degree here is 1, and R is a polynomial whose degree is less than that of n-1.

    Go back to the example again.
     
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