Results 1 to 6 of 6

Math Help - Real Numbers

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    145

    Real Numbers

    Hi,

    Someone give me a hand in checking these please.

    a.) Given that x > y , and k < 0 for the real numbers x, y and k, show that kx < ky.

    kx>y...1

    x>y x - y is +ve...2

    k<0...3

    If kx>ky then kx - ky is +ve

    Putting values using lines 2 and 3

    x=6, y=4, k= -2

    kx>ky

    -2(6) > -2(4) OR kx-ky=+ve

    ...................OR -12 + 8 \not = +ve
    -12 > 8
    therefore kx < ky must be true

    b.) Show that if x, y ∈ R, and x<y, then for any real number k<0, kx>ky.

    kx<ky...1

    x<y, x-y is -ve

    k<0...3

    If  kx<ky then kx-ky is -ve

    Putting in values using lines 2 and 3

    x=2, y=3, k=-4

    kx<ky .......................OR kx -ky=-ve

    -4(2)<-4(3)................OR -8-(-12) \not = -ve

    -8<-12

    therefore  kx>ky must be true
    Last edited by Hellbent; November 26th 2009 at 02:15 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by Hellbent View Post
    a.) Given that x > y , and k < 0 for the real numbers x, y and k, show that kx < ky.

    kx>y...1
    If x = 3, y = 2, and k = -1, then kx = -3, which is less than y. So your last line above cannot be assumed always to be true.

    Try starting like this:

    Let x > y, and let k < 0. Assume that kx > ky. Then kx - ky > 0. Also, kx - ky = k(x - y). Since x > y, then x - y > 0.

    See where this leads....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    145
    Am I right this time?

    a.) Since x > y, so x - y is positive and k is negative.
    Product of a negative and positive number is negative, kx - ky
    Hence it follows that kx < ky.

    b.) Since x < y, so x - y is negative and k is negative.
    Product of two negative numbers is equal to a positive number.
    Hence it follows that kx > ky.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2009
    Posts
    145
    Right...wrong?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by Hellbent View Post
    Right...wrong?
    it seems right for me.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    x> y\Rightarrow x-y is strictly positive.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Set of 13 real numbers
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: July 5th 2011, 09:37 AM
  2. Replies: 1
    Last Post: September 27th 2010, 04:14 PM
  3. Real Numbers
    Posted in the Algebra Forum
    Replies: 6
    Last Post: February 21st 2010, 03:31 AM
  4. Real Numbers - Real Anaylsis
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 3rd 2008, 11:54 AM
  5. The Sum of all real numbers
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 22nd 2005, 02:17 PM

Search Tags


/mathhelpforum @mathhelpforum