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.


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

  • $\begingroup$ Oh, now I see it. Thanks! $\endgroup$ – Da Mike Apr 27 at 13:47
  • $\begingroup$ That wasn't a hint, but +1 anyway since you've helped the OP! $\endgroup$ – 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 $ enter image description here


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