# Given two ratios $\frac{p_i}{q_i}$, what is $\frac{p_1+p_2}{q_1+q_2}$ in their terms

I am ashamed to say that I cannot figure this one out: I am given two ratios $\dfrac{p_i}{q_i}$ where $i=1$, $2$. (We just know the ratios and not the numbers $p_i, q_i$. What I mean by this is simply that in the expression $\psi$ I cannot separate the $p_i$ and $q_i$)

I want an elementary expression $\psi:\mathbb R \to \mathbb R$ such that

$$\dfrac{p_1+p_2}{q_1+q_2}=\psi \left(\dfrac{p_1}{q_1},\dfrac{p_1}{q_2}\right)$$

### NOTE

• $p_i, q_i \in \mathbb Z$
• $q_2 \nmid q_1$, since otherwise we cannot find such a function.
• $p_2 \nmid p_1$ since this would make things trivial.
• Let us assume $(p_i,q_i)=1$.
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Do you really mean to assume $q_1=1$? – Chris Eagle Mar 7 '13 at 12:29
Are you saying $q_1 = 1?$ Also have these been reduced to their lowest terms? – muzzlator Mar 7 '13 at 12:29
@Martin, I don't think clarification is needed, I think you've simply demonstrated that what OP wants is impossible. Never mind, OP has edited to rule out your example. – Gerry Myerson Mar 7 '13 at 12:34
BTW this is called mediant. – Martin Sleziak Mar 7 '13 at 12:39
If you do know the ratios, don't you know the numbers? – hjpotter92 Mar 7 '13 at 12:42

As Martin Sleziak said in his comment, the function you defined is called the mediant.

As others have suggested, there is no reason to expect there to be a simpler formula than the definition you give. There is however a natural and obvious way to represent $(p_1 + p_2)/(q_1 +q _2)$ as a weighted average of $p_1/q_1$ and $p_2/q_2$, where the weights depend on $q_1,q_2$:

$$\frac{p_1+p_2}{q_1+q_2} = \frac{q_1}{q_1+q_2}\frac{p_1}{q_1} + \frac{q_2}{q_1+q_2}\frac{p_2}{q_2}.$$

This shows, for instance, that $(p_1+p_2)/(q_1+q_2)$ always lies between $p_1/q_1$ and $p_2/q_2$.

In fact a simple argument shows that the function $\psi:\mathbb{Q}^2\to\mathbb{Q}$ so defined is discontinuous (with respect to the usual order topology) everywhere except the diagonal. Indeed, consider replacing $p_2/q_2$ with a nearby rational but with much larger denominator. The value of $(p_1+p_2)/(q_1+q_2)$ is then changed to pretty much $p_2/q_2$. This puts a lower bound on the simplicity of a formula for $\psi$.

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It's unlcear what your question is referencing, @Puzzled. $\mathbb Q$ has the usual order topology, and $\mathbb Q^2$ has the product topology. The point is, most "nice" formulas on $\mathbb R^2$ are continuous at most points under these topologies, but this function is at minimum discontinuous in a dense subset of $\mathbb R^2$. So that rules out "nice functions" on $\mathbb R^2$. – Thomas Andrews Mar 7 '13 at 13:14
@ThomasAndrews: Thanks for clearing that up. – Puzzled Mar 7 '13 at 13:17