1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Discrete Math Help

  1. Jan 27, 2005 #1
    Discrete Math Help!!!

    Here is the problem:

    Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the following equations:

    [tex]\dfrac{xy}{x+y}=a [/tex] and [tex]\dfrac{xz}{x+z}=b [/tex] and [tex]\dfrac{yz}{y+z}=c [/tex].

    Is x rational? If so, express it as a ratio of two integers.

    I have calculated that [tex]x=\dfrac{-(bz-xz)}{b}[/tex]. I am inclined to answer no, since x, y, or z could be irrational.

    Any help would be appreciated.
  2. jcsd
  3. Jan 27, 2005 #2

    first modify your equation to solve for x explicitly in terms of b and z ( x on one side , b and z terms on the other ).
    then use the other two equations to come up with an expression for z and substitute that into your original equation.
    Your goal is to try to solve for x in terms of a,b,c only... eliminating any z or y

    Post your steps if you run into more trouble
    Last edited: Jan 27, 2005
  4. Jan 27, 2005 #3
    Ok, I solved and got this:

    [tex]x=\dfrac{ay}{y-a}[/tex] and [tex]x=\dfrac{bz}{z-b}[/tex] and [tex]y=\dfrac{cz}{z-c}[/tex].

    I then solved for z and got [tex]z=\dfrac{aby}{(a-b)y+ab}[/tex]. I just cancelled the y's and was left with [tex]z=\dfrac{ab}{(a-b)+ab}[/tex].

    I plugged this in for z in the second equation, and got [tex]z=\dfrac{b \dfrac{ab}{(a-b)+ab}}{\dfrac{ab}{(a-b)+ab}-b}[/tex]. This simplifies to [tex]\dfrac{ab}{b-ab}[/tex].

    If this answer is correct, I would say that yes, x is indeed rational. Am I correct?
    Last edited: Jan 27, 2005
  5. Jan 27, 2005 #4
    Where did the x come from in the third equation (top right) , and where did the y go? That completely changes the problem so the rest is all wrong.

    Also, for
    [tex]z=\dfrac{aby}{(a-b)y+ab}[/tex] you can't cancel the y's, because y is not a factor of the denominator.
    If the denominator was instead [itex](a-b)y + aby[/itex] then you could factor out the y which could become [itex]y((a-b) + ab)[/itex] and the y's could cancel
    (do you see the difference? )

    Another thing: in your result,,, you solved for z (although incorrectly). The equation asks if X is rational... so how does knowing whether or not Z is rational help you?
    based on your equation relating x to z you could use the fact that z is rational to prove that x is rational providing you knew that theorem, but this seems like more trouble than its worth why not just solve for x. /edit

    Your going to have to solve for x in terms of y or z, and then solve for either y or z using the other equations to come up with an equation in terms of x, a, b, and c only

    Last edited: Jan 27, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Discrete Math Help
  1. Discrete Math Help (Replies: 1)