Discrete Math Help: Is x Rational?

[SpatialVacancy]

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.

 

[MathStudent]

hint:

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
Regards,
-MS

 

[SpatialVacancy]

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 canceled 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?

 

[MathStudent]

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?
edit:
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

-MS