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!

K|n!+k from k=2,3,4 .n

  1. Jul 8, 2014 #1
    1. The problem statement, all variables and given/known data

    prove by induction that K|n!+k for all integers [tex]n \ge 2 [/tex] and k = 2,3,4,......,n



    this one is different than other proofs I have had because Ive only dealt with n changing but now n and k change.

    k goes up to n, does this mean k and n have the same values?

    for instance,

    P(2)

    [tex]2|2! + 2 \Rightarrow 2|4[/tex]

    so p(k)

    is [tex]k|n!+k; n=k[/tex]

    [tex]k|k!+k[/tex]

    ??
     
    Last edited: Jul 8, 2014
  2. jcsd
  3. Jul 8, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You ask: "does this mean k and n have the same values?". No, that is not what it says and not what it means. Look at some small-n examples. If n = 3 it says that 2|3!+2 and 3|3!+3. So, k is not 3; it is any number in the set {2,3}. If n = 4 it says that all three of the statements 2|4!+2, 3|4!+3 and 4|4!+4 are true.
     
  4. Jul 8, 2014 #3
    ah okay

    whatever n is, then k takes on each value in the set {2,...,n}
     
  5. Jul 8, 2014 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The fact that if k< n the n k divides n! makes this pretty close to trivial.
     
  6. Jul 8, 2014 #5
    [tex]2|n!+2[/tex]
    [tex]3|n!+3[/tex]
    [tex]4|n!+4[/tex]
    [tex]\downarrow[/tex]
    [tex]k|n!+k[/tex]

    where k takes on the values of 2,3,4...n and n is an integer greater than or equal to 2

    and because for all k<n, k|n and when n=k, k|n

    so my p(k) is

    [tex]k|n!+k[/tex]

    and

    p(k+1)

    [tex](k+1)|n!+(k+1)[/tex]
     
    Last edited: Jul 9, 2014
  7. Jul 9, 2014 #6
    I have to make a correction. its not asking for induction. all the other problems on this problem set were induction but I confirmed with my prof this is not an induction proof question
     
  8. Jul 9, 2014 #7

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Can you see a common factor in the numerator?
     
  9. Jul 9, 2014 #8
    k is a factor of k, clearly and k is a also a factor of n! by the given information


    [tex]2 \le k \le n[/tex] and by definition of a factorial k had to be used somewhere along the product

    [tex] \prod_{k=2}^{n} k[/tex]

    case 1

    k = 2, n > or equal 2

    [tex] \frac{2*.....*n + 2}{2}[/tex]

    case 2

    2 < k <n

    [tex]\frac{2*...*k*....*n + k}{k}[/tex]

    case 3

    n=k

    [tex]\frac{2*......*k + k}{k}[/tex]
     
    Last edited: Jul 9, 2014
  10. Jul 9, 2014 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Those case are not necessary. WWDG's point is that if [itex]k\le n[/itex] then k is a factor of n!.
     
  11. Jul 9, 2014 #10
    I understand that but I am expected to show a proof. Seems a tad bit too easy to say if

    [tex] k \le n[/tex] then k is a factor of n!
     
  12. Jul 9, 2014 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    But that is absolutely correct, so IS a proof! Sometimes proofs are easy---after you "see" them.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted