Results 1 to 3 of 3

Math Help - Prove that adding one to itself is always different from 0

  1. #1
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1

    Prove that adding one to itself is always different from 0

    Hi,

    problem:

    \mathbb{Z}_p=\{0,1,\cdots,p-1\}.
    If \alpha and \beta are in \mathbb{Z}_p, let \alpha+\beta be the least positive remainder obtained by dividing the (ordinary) sum of \alpha and \beta by p,
    and similarly, let \alpha\beta be the least positive remainder obtained by dividing the (ordinary) product of \alpha and \beta by m

    Prove that if \mathbb{F} is a field, then either the result of repeatedly adding 1 to itself is always different from 0,
    or else the first time that it is equal to 0 occurs when the number of summands is a prime.

    attempt:
    I look at n\cdot 1 where n is the number of times 1 is added to itself.

    n\cdot 1 (mod\;p), p is prime since \mathbb{Z}_p is a field.

    n\cdot 1 (mod\;p)=0 when n is a multiple of p, n=\alpha p,\;n\in\mathbb{N}.

    For n=1, 1 (mod\;p)\neq 0.
    For n=\alpha p+1, \alpha p(mod\;p)+1(mod\;p)\neq 0

    I am really bad at this so any suggestion is very welcome!
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    Quote Originally Posted by Mollier View Post
    Hi,

    problem:

    \mathbb{Z}_p=\{0,1,\cdots,p-1\}.
    If \alpha and \beta are in \mathbb{Z}_p, let \alpha+\beta be the least positive remainder obtained by dividing the (ordinary) sum of \alpha and \beta by p,
    and similarly, let \alpha\beta be the least positive remainder obtained by dividing the (ordinary) product of \alpha and \beta by m

    Prove that if \mathbb{F} is a field, then either the result of repeatedly adding 1 to itself is always different from 0,
    or else the first time that it is equal to 0 occurs when the number of summands is a prime.

    attempt:
    I look at n\cdot 1 where n is the number of times 1 is added to itself.

    n\cdot 1 (mod\;p), p is prime since \mathbb{Z}_p is a field.

    n\cdot 1 (mod\;p)=0 when n is a multiple of p, n=\alpha p,\;n\in\mathbb{N}.

    For n=1, 1 (mod\;p)\neq 0.
    For n=\alpha p+1, \alpha p(mod\;p)+1(mod\;p)\neq 0

    I am really bad at this so any suggestion is very welcome!
    Thanks
    search "Character of field".
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Thanks for the heads up. I did a bit of reading and came up with the following:

    char(\mathbb{F}) is the smallest positive number n such that 1+\cdots+1(n\;summands)=0.

    If char(\mathbb{F})=n=ab\;(composite),
    then 1+\cdots+1(ab\;times)=0,
    which means that ab=0.
    Then either a=0 or b=0, so char(\mathbb{F}) is smaller than ab Contradicting the fact that n is the smallest number..(minimality)

    Is this any good?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Adding exponents/thorough way.
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: November 8th 2010, 06:38 PM
  2. Adding
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 20th 2009, 02:29 AM
  3. mgf adding two binomials
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: February 16th 2009, 04:55 AM
  4. Adding Pi.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 14th 2008, 12:48 PM
  5. adding numbers !!!!
    Posted in the Math Topics Forum
    Replies: 7
    Last Post: January 14th 2007, 12:25 AM

Search Tags


/mathhelpforum @mathhelpforum