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Hi I am trying to complete the following question in my practice:

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

\begin{equation} \cfrac{xy}{x+y} = a \quad and \quad \cfrac{xz}{x+z} = b \quad and \quad \cfrac{yz}{y+z} = c \end{equation}

Prove that $ x $ is rational

Source: Discrete Mathematics with Applications 4th edition

I tried combining the equations and logically conclude that $ x $ is rational but to no avail as I do not know how to separate $ x $ from the rest of the variables. The only starting point that I have is the theorem for rational numbers:

\begin{equation} if \; x \in \mathbb{Q}, \; x = \cfrac{a}{b} \; where \; a,b \in \mathbb{Z} \; and \; b \neq 0 \end{equation}

I have no other idea how to carry on from here. Could someone please advise me?

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  • $\begingroup$ The question you quote doesn't go on to say to prove $x,y,z$ are rational, but I assume that's what you meant. Did you just forget to include that question at the end? $\endgroup$ – coffeemath Aug 14 '16 at 15:41
  • $\begingroup$ Thanks for spotting the mistake. I have corrected my post. $\endgroup$ – LanceHAOH Aug 14 '16 at 16:28
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By rewriting we get

$$\frac{xy}{x+y} = \frac{1}{\frac{1}{x}+\frac{1}{y}}$$

so we can rewrite the equations as $$\frac{1}{x}+\frac{1}{y} = \frac{1}{a}, ~~\frac{1}{x}+\frac{1}{z} = \frac{1}{b}, ~~\frac{1}{y}+\frac{1}{z} = \frac{1}{c}.$$

This is a regular system of linear equations in $\frac{1}{x}, \frac{1}{y}$ and $\frac{1}{z}$, and since all the coefficients are in $\mathbb{Q}$, the solutions must be in $\mathbb{Q}$ as well. Feel free to solve that last system if you like.

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  • $\begingroup$ I don't understand why solution has to be rational since coefficients are rational $\endgroup$ – LanceHAOH Aug 14 '16 at 16:48
  • $\begingroup$ If we interpret the LES as a linear equation system over the field of rational numbers, we have nonzero determinant and get a unique rational solution. If we interpret the LES as a linear equation system over the field of real numbers, we still have nonzero determinant and get a unique real solution. Since the rational solution is also a real solution, both solutions must be identical. $\endgroup$ – Anon Aug 14 '16 at 16:59
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Since $x,y,z$ are nonzero one can define $u=1/x,v=1/y,w=1/z,$ and it also follows from the given equations that none of $a,b,c$ are zero. So we can also put $d=1/a,e=1/b,f=1/c.$ Now the equations, after taking reciprocals, are $u+v=d,u+w=e,v+w=f$ which can be solved rationally for $u,v,w.$

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Suppose $x$ is irrational. Then

$$ \frac{\frac{xy}{x+y}}{\frac{xz}{x+z}} = \frac ab $$, since $b \ne 0$. Now

$$ \frac{xy(x+z)}{xz(x+y)} = \frac ab \implies \frac{x+z}{x+y} = \frac{az}{by} = q \in \Bbb Q$$, since $y \ne 0$

So $x+z = qx+qy \implies (1-q)x = qy - z \in \Bbb Q$, but then $1-q \notin \Bbb Q$ ! Thus, leading to a contradiction.

Note that this same argument could be applied to $y, z$ as well, so it shows at once that all of them have to be rational.

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Inverting the equations shows $\,x^{-1}\!+y^{-1},\,\ z^{-1}\!+x^{-1},\,\ \color{#c00}{y^{-1}\!+z^{-1}}\in \Bbb Q,\ $ i.e. are all rational.

Adding them $\ 2(x^{-1}\!+y^{-1}\!+z^{-1}) =:q\in \Bbb Q\ $ so $\ x^{-1} = q/2-(\color{#c00}{y^{-1}\!+z^{-1}}) \in\Bbb Q.\ \ $ QED

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  • $\begingroup$ How does being able to invert indicate that the expressions are rational? $\endgroup$ – LanceHAOH Aug 14 '16 at 17:51
  • $\begingroup$ @Lance Inverting equation $\,1\,$ yields $\,\dfrac{x+y}{xy} = \dfrac{1}a,\ $ i.e. $\,y^{-1}\!+x^{-1} = a^{-1}\in \Bbb Q,\,$ etc for others. $\endgroup$ – Bill Dubuque Aug 14 '16 at 17:54

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