# Prove that $\gcd(m+m', n+n') = 1$

I'm stuck trying to solve this problem:

"Given positive integers $$m, n, m', n'$$ such as $$m/n < m'/n'$$ and $$m'n - mn' = 1$$, we define $$a/b = (m+m')/(n+n').$$ Check that $$m/n < a/b < m'/n'$$ and prove that $$gcd(a, b) = 1."$$

The way I see it, it must be that $$a = m+m'$$ and $$b = n+n'$$. But I'm not sure how to prove the $$gcd(a,b) = 1$$ part. Any help would be appreciated.

## 2 Answers

Hint: $$\ (m+m')\,n - m\,(n'+n) = m'n-mn'= 1$$,

• Oh, now I see it. Thanks! – Da Mike Apr 27 at 13:47
• That wasn't a hint, but +1 anyway since you've helped the OP! – Toby Mak Apr 27 at 13:48

It is clear as a column operation preserving a determinant

$$\left|\begin{array}{} m'+m & m\\ n'\,+\,n & n\end{array}\right|\,=\,\left|\begin{array}{} m' & m\\ n' & n\end{array}\right|\qquad$$

Remark  This Farey mediant operation has a natural geometric interpretation as a change of basis:

$$\quad \dfrac{m}n,\,\dfrac{m'}{n'}$$ are Farey adjacent \!\iff\! \begin{align}u &= (n,\ m)\\ v &= (n',m')\end{align} are a basis of $$\,\Bbb Z^2$$ $$\!\iff\! |\det(u,v)| = 1$$

This has a pretty proof via Pick's Area theorem, e.g. see here and here. The mediant operation yields the diagonal $$\,u\!+\!v\,$$ of the fundamental parrallelogram with sides $$\,u,v\,$$ and we obtain another basis by replacing one side by the diagonal, i.e. $$\, u,v \to u\!+\!v,\,v$$, depicted below (from Wikipedia)

$$\qquad\qquad$$ 