Results 1 to 6 of 6

Math Help - Using a direct method of proof.

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    9

    Using a direct method of proof.

    Hi guys,

    Was hoping someone could help with this question - I'd appreciate it.

    The question;

    "Use a direct method of proof to show that if x and y are odd integers, then xy is also odd.

    The given solution;

    First, notice that if x is odd then x = 2m + 1 where m is an integer. Similarly,
    y = 2n + 1 for some integer n. <- I understand up to this point.

    Then get lost around here.
    Then, xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2 (2mn + m + n) + 1 which is an odd integer."

    Could someone explain what exactly is happening in that section?
    Thankyou,
    - Jamie.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by Skeith View Post
    Hi guys,

    Was hoping someone could help with this question - I'd appreciate it.

    The question;

    "Use a direct method of proof to show that if x and y are odd integers, then xy is also odd.

    The given solution;

    First, notice that if x is odd then x = 2m + 1 where m is an integer. Similarly,
    y = 2n + 1 for some integer n. <- I understand up to this point.

    Then get lost around here.
    Then, xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2 (2mn + m + n) + 1 which is an odd integer."

    Could someone explain what exactly is happening in that section?
    Thankyou,
    - Jamie.
    Let u=2mn+m+n, if xy=2u+1, and since u\in\mathbb{Z}, it follows that xy is odd.

    In other words, anything of the form 2u+1 is odd if u is an integer.
    Last edited by VonNemo19; December 6th 2009 at 10:38 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    Quote Originally Posted by Skeith View Post
    "Use a direct method of proof to show that if x and y are odd integers, then xy is also odd.

    The given solution;

    First, notice that if x is odd then x = 2m + 1 where m is an integer. Similarly,
    y = 2n + 1 for some integer n. <- I understand up to this point.

    Then get lost around here.
    Then, xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2 (2mn + m + n) + 1 which is an odd integer.
    Two times any integer is an even integer.
    So {\color{blue}2}(2mn+m+n) is an even integer.
    Add one to that to get an odd integer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2009
    Posts
    9
    Thankyou both Von and Plato - I feel I understand now. However could you check my working just to be sure?

    Lets say i assign m and n values of 4 and 5 respectively. this means x = 9 and
    y = 11 since it is a matter of x = 2m + 1 and 2n + 1 for y?

    Then moving to the section at which I got lost,

    xy = (2m + 1)(2n + 1) <- I understand now...

    xy = (2m + 1)(2n + 1)
    -or-
    2 (2mn + m + n) + 1 <- Am I right in thinking either one or the other can be used?

    Example,

    4mn (m x n x 4 = 80) + 2m (8) + 2n (10) + 1 = 99
    - or -
    2(2mn (m x n x 2 = 40) + m + n = 49) then multiplying by 2 = 98 + 1 = 99?

    Am I on the right lines or have I got lost?

    Thankyou again.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,561
    Thanks
    785
    Yes, you are on the right track.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2009
    Posts
    9
    Great - thanks emakarov.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. by direct method
    Posted in the Number Theory Forum
    Replies: 13
    Last Post: August 29th 2011, 11:45 PM
  2. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 11:07 PM
  3. Lyapunov's direct method
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: October 25th 2009, 10:16 PM
  4. Direct Proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 28th 2009, 06:25 PM
  5. need help with Direct proof
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 1st 2009, 02:53 PM

Search Tags


/mathhelpforum @mathhelpforum