Computing Ancestors of # for Stern-Brocot Tree

A well known property of the Stern-Brocot tree is that there exists a one to one correspondence between the fractions in the interval $[0,1]$ and the fractions in the interval $[1,\infty]$. But it's much simpler to initialize the tree with the interval $[0,1]$. Doing so we get the (Pascal) algorithm below for travelling two paths in the Stern-Brocot tree that lead us to the fraction $5/7$ by enclosing it between a lower bound $m1/n1$ and an upper bound $m2/n2$ with each step. The algorithm halts because it is known that any rational number is in the tree. 
program Kevin;

procedure Stern_Brocot(teller,noemer : integer);
{
  Walk through Stern-Brocot tree
  until fraction = teller/noemer
 (English: numerator/denominator)
}
var
  m1,m2,n1,n2 : integer;
begin
{ Initialize tree }
  m1 := 0; n1 := 1;
  m2 := 1; n2 := 0;
  while true do
  begin
  { if teller/noemer < (m1+m2)/(n1+n2) then }
    if (n1+n2)\*teller < (m1+m2)\*noemer then
  { Tightening }
    begin
    { Upper Bound }
      m2 := m1+m2;
      n2 := n1+n2;
    end else begin
    { Lower Bound }
      m1 := m1+m2;
      n1 := n1+n2;
    end;
    Writeln(m1,'/',n1,' < ',teller,'/',noemer,' < ',m2,'/',n2);
    if (teller/noemer = m1/n1)
    or (teller/noemer = m2/n2) then Break;
  end;
end;

begin
  Stern_Brocot(5,7);
end.
Output: 
0/1 < 5/7 < 1/1
1/2 < 5/7 < 1/1
2/3 < 5/7 < 1/1
2/3 < 5/7 < 3/4
5/7 < 5/7 < 3/4
From this output we derive the required sequence by hand (oh well, the whole thing could have been done by hand in this simple case, but it's useful to have such a computer program for e.g. approximating irrational numbers). 
1/1 < 7/5 < 1/0
1/1 < 7/5 < 2/1
1/1 < 7/5 < 3/2
4/3 < 7/5 < 3/2
4/3 < 7/5 < 7/5
Note that the closest smaller and larger ancestors are in the tree just before the algorithm halts.

EDIT. My favorite reference is this:

More impressive examples are obtained by calculating for irrational numbers, such as $1/\sqrt{2}$: 
if (n1+n2)\*(n1+n2) < (m1+m2)\*(m1+m2)\*2 then
Giving, after e.g. 45 iterations: $$ \frac{318281039}{225058681} < \sqrt{2} < \frac{768398401}{543339720} $$