1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Natural number
  1. Nature of this number (Replies: 13)

Loading...