Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - if p|a^2, then p|a. True

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    if p|a^2, then p|a. True

    if p|a^2, then p|a.

    This is true but how to prove it?

    p|a^2\rightarrow p|a*a now what?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by dwsmith View Post
    if p|a^2, then p|a.

    This is true but how to prove it?

    p|a^2\rightarrow p|a*a now what?
    For a prime  p, \;\; p\mid ab\implies p\mid a or  p\mid b .

    What can you conclude from this?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by chiph588@ View Post
    For a prime  p, \quad p\mid ab\implies p\mid a or  p\mid b .

    What can you conclude from this?
    Then p|a \ \mbox{or} \ p|a; therefore, p|a
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    I'm not sure I buy the theorem, unless p is a prime. Is that a convention in number theory? If so, I would probably write out the prime factorization of a and go from there.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    If I wanted to generalize it, I could then say:

    if p|a^n, then p|a.
    p|a^n\rightarrow p|a_1*a_2\dots *a_n\rightarrow p|a_1 \ \mbox{or} \ \dots \ p|a_n where a=a_1=a_2=\dots=a_n; hence, p|a
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by dwsmith View Post
    If I wanted to generalize it, I could then say:

    if p|a^n, then p|a.
    p|a^n\rightarrow p|a_1*a_2\dots a_n\rightarrow p|a_1 \ \mbox{or} \ \dots \ p|a_n where a=a_1=a_2=\dots=a_n; hence, p|a
    I think induction would provide a more elegant proof.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by chiph588@ View Post
    I think induction would provide a more elegant proof.
    P(n): \ p|a^n, \ n\geq 2
    P(2): \ p|a^2\rightarrow p|a*a\rightarrow p|a \ \mbox{or} \ p|a
    P(k): \ p|a^k
    P(k+1): \ p|a^{k+1}
    Assume P(k) is true.
    p|a^{k} multiple by a a*p|a*a^k \rightarrow a*p|a^{k+1}\rightarrow p*(a*m)=a^{k+1}\rightarrow p|a^{k+1}\rightarrow p|a^k \ \mbox{or} \ p|a
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dwsmith View Post
    if p|a^2, then p|a.

    This is true but how to prove it?

    p|a^2\rightarrow p|a*a now what?
    Consider the prime decompositions of \displaystyle a and a^2

    CB
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by dwsmith View Post
    P(n): \ p|a^n, \ n\geq 2
    P(2): \ p|a^2\rightarrow p|a*a\rightarrow p|a \ \mbox{or} \ p|a
    P(k): \ p|a^k
    P(k+1): \ p|a^{k+1}
    Assume P(k) is true.
    [tex]p|a^k[\math] multiple by a a*p|a*a^k \rightarrow a*p|a^{k+1}
    Having a problem since I now have a*p
    We have our base case of  k=2 down.

    Now assume  p\mid a^k\implies p\mid a for any  a

    So now  p\mid a^{k+1}\implies p\mid a^k or  p\mid a . Apply the inductive hypothesis are we're done.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by chiph588@ View Post
    We have our base case of  k=2 down.

    Now assume  p\mid a^k\implies p\mid a for any  a

    So now  p\mid a^{k+1}\implies p\mid a^k or  p\mid a . Apply the inductive hypothesis are we're done.
    I edited my original post.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by dwsmith View Post
    I edited my original post.
    Your inductive hypothesis is still incomplete.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by chiph588@ View Post
    Your inductive hypothesis is still incomplete.
    How I was able to show p|a^{k+1}\mbox{?}
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by dwsmith View Post
    How I was able to show p|a^{k+1}\mbox{?}
    You showed that, but that's not what you want to prove. You assume  p\mid a^{k+1} and want to conclude  p\mid a .
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Suppose  p|a is false , then one must be coprime to the other , so there exists integers  x,y such that  ax + py = 1

     a^n x  + p (a^{n-1}y) = a^{n-1}

    Let  a^n = pm  because we have  p|a^n

     pmx + pa^{n-1}y = a^{n-1}

     p (mx + a^{n-1}y ) = a^{n-1} , we find that  a^{n-1} is the multiple of  p say  a^{n-1} = pm' , followed by  a^{n-1}x + p a^{n-2} y = a^{n-2} ,   p(m'x + a^{n-2}y) = a^{n-2} so we have  a^{n-2} = pm'' ~ .... it eventually goes to  a = p m^{(n-1)} , a contradiction .
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Nothing against simplependulum's proof, but if we write a^{k+1}=a\cdot a^k then we can complete the induction step as was done in the original problem (first few posts of the thread).

    Edit: Oh I guess the OP already did that. Well here's how I would write it then.

    Induction step: Assume (p | a^k \Rightarrow p | a).

    We want to show that (p | a^{k+1} \Rightarrow p | a).

    p | a^{k+1} \Leftrightarrow p | a\cdot a^k \Rightarrow (p|a \lor p|a^k) \Rightarrow p|a. QED.

    Edit 2: Sorry sorry, I've been illiterate. chiph588@ already posted basically the same thing except for the very last step. But maybe my typesetting will be easier for the OP to follow..

    Edit 3 (!): By the way, it's not necessary to use base case n=2. We can use n=1 since p|a^1 \Rightarrow p|a.
    Last edited by undefined; June 21st 2010 at 09:11 AM.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: October 5th 2011, 01:45 PM
  2. Replies: 1
    Last Post: October 4th 2011, 03:19 AM
  3. help which one is not true
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 11th 2010, 11:53 PM
  4. True or False. Prove if true show counter example if false.
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 2nd 2010, 11:54 AM
  5. is that true?
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 2nd 2009, 08:00 PM

/mathhelpforum @mathhelpforum