Prove that $x$ is rational where $x$ is mixed with other variables

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:

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

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:

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

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

• 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? – coffeemath Aug 14 '16 at 15:41
• Thanks for spotting the mistake. I have corrected my post. – 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.

• I don't understand why solution has to be rational since coefficients are rational – LanceHAOH Aug 14 '16 at 16:48
• 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. – 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

• How does being able to invert indicate that the expressions are rational? – LanceHAOH Aug 14 '16 at 17:51
• @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. – Bill Dubuque Aug 14 '16 at 17:54