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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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 $u,v,w.$


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.


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.