Discrete Math Help: Is x Rational?

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Homework Help Overview

The problem involves determining whether the variable x is rational given a set of equations involving integers a, b, c and nonzero real numbers x, y, z. The equations relate these variables through specific fractional forms.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to manipulate the equations to express x in terms of other variables and questions the rationality of x based on potential irrational values of y and z.
  • Some participants suggest modifying the equations to isolate x and substituting expressions for y and z to derive a form for x solely in terms of a, b, and c.
  • Concerns are raised about the validity of algebraic manipulations, particularly regarding the cancellation of terms and the implications of solving for z instead of x.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on how to approach the problem. There is a focus on clarifying the relationships between the variables and ensuring that the manipulations adhere to algebraic rules. Multiple interpretations of the equations are being explored, and no consensus has been reached regarding the rationality of x.

Contextual Notes

Participants note the importance of correctly interpreting the equations and the implications of rational versus irrational values in the context of the problem. There is an emphasis on ensuring that the derived expressions accurately reflect the relationships defined in the original equations.

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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.
 
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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
 
Last edited:
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?
 
Last edited:
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
 
Last edited:

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