Results 1 to 3 of 3

Math Help - Range of x,y for which x^2 > y^2 holds true

  1. #1
    Newbie
    Joined
    Feb 2011
    From
    India
    Posts
    2

    Range of x,y for which x^2 > y^2 holds true

    The following is the problem on 63rd page of Zakon's "Basic Mathematics"

    (iii) If x > y ≥ 0, then
    x2 > y2 and x3 > y3 ≥ 0 (where x3 = x2x);
    Similarly,
    x4 > y4 ≥ 0 (where x4 = x3x).
    Which (if any) of these propositions remain valid also if x or y is negative? Give proof.

    I am able to solve the first part of the problem. Got stuck in the second. If somebody can give me a hint of how to determine whether the proposition is valid if x or y is negative.

    Regards...............D
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by dawoodvora View Post
    The following is the problem on 63rd page of Zakon's "Basic Mathematics"

    (iii) If x > y ≥ 0, then
    x2 > y2 and x3 > y3 ≥ 0 (where x3 = x2x);
    Similarly,
    x4 > y4 ≥ 0 (where x4 = x3x).
    Which (if any) of these propositions remain valid also if x or y is negative? Give proof.

    I am able to solve the first part of the problem. Got stuck in the second. If somebody can give me a hint of how to determine whether the proposition is valid if x or y is negative.

    Regards...............D


    For example,

    x^2>y^2\Longrightarrow (x-y)(x+y)>0\Longrightarrow (1)\,x<y\,\,and\,\, x>-y\,\,or\,\, (2)\,x>y\,\,and\,\,x<-y

    So if x<0 say, then (1) above is true for any y s.t. |x|<|y| ,and

    (2) is true for any y<x.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    From
    India
    Posts
    2

    My understanding of Tonio's solution

    Hello everybody,
    Please you all have a look at it and let me know if I understood what Tonio tried to explain me. Please let me know if my understanding of Tonio's solution is Okay or Not.


    x2 > y2 => (x-y)(x+y) > 0

    => (x-y) > 0 & (x+y) > 0 OR (x-y) < 0 & (x+y) < 0
    (Both the terms +ve) OR (Both the terms –ve)
    => x > y & x > -y OR => x < y & x < -y
    => x > |y| OR => x < |y|
    => x > 0 OR => x < 0

    Thus x2 > y2 holds true only when:
    1. x > |y| if x > 0
    2. x < |y| if x < 0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. True Radius - range of values?
    Posted in the Geometry Forum
    Replies: 2
    Last Post: September 17th 2010, 12:12 AM
  2. Inequality holds?
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 7th 2010, 08:02 AM
  3. Inequality holds
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 7th 2010, 01:58 AM
  4. something new?holds?why?is it possible to prove?
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: April 16th 2009, 01:51 AM
  5. Showing A Identity Holds
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 22nd 2008, 09:53 AM

Search Tags


/mathhelpforum @mathhelpforum