Isl 2004 a3
by Wolstenholme, Aug 7, 2014, 5:32 AM
Does there exist a function 
such that if 
and 
are distinct rational numbers satisfying 
or 
, then 
? Justify your answer.
Solution:
Every positive rational number has a unique continued fraction of the form
for some 
where 
for all 
. From now on this will be denoted as ![$ [a_0; a_1, a_2, \dots, a_m] $](//latex.artofproblemsolving.com/1/c/4/1c47e13850fd92c24cf30707900dc53e12c32418.png)
.
For every
, let 
. Then let 
and for all 
let 
. I claim that this 
satisfies the desired properties. To prove this, consider two rational numbers 
.
If 
then by definition 
as desired.
If 
and 
then by definition as desired since 
.
If and 
then to prove that it suffices to show that 
. Let ![$ -x = [a_0; a_1, a_2, \dots, a_m] $](//latex.artofproblemsolving.com/1/0/e/10e2dcb8e4ad502ecb43d7e1c2744f3995658d81.png)
. Then ![$ y = [a_0 + 1; a_1, a_2, \dots, a_m] $](//latex.artofproblemsolving.com/2/c/e/2ce601c354188bdad5e03d9833f5a0efd26eb275.png)
and so 
as desired.
If and 
we can let ![$ y = [0; 1, a_2, \dots, a_m] $](//latex.artofproblemsolving.com/e/6/b/e6b1ac208f52b35a47ba274b63b6aede8c05b3b5.png)
so that ![$ x = [0; a_2 + 1, a_3, \dots, a_m] $](//latex.artofproblemsolving.com/5/c/a/5ca981e9723d9f4c0ee0a619a2f05b17f18fee09.png)
so 
and 
which implies that as desired.
If 
, we can assume WLOG that 
. Then letting ![$ x = [0; a_1, a_2, \dots, a_m] $](//latex.artofproblemsolving.com/f/a/4/fa41f1dfed52008255c6fe47ad8f150591dc820c.png)
we have that ![$ y = [a_1; a_2, a_3, \dots, a_m] $](//latex.artofproblemsolving.com/1/6/6/166a8243036f6a8f2362c13f93282945d869aff5.png)
so 
and 
which implies that as desired.
such that if 
and 
are distinct rational numbers satisfying 
or 
, then 
? Justify your answer.Solution:
Every positive rational number has a unique continued fraction of the form
for some 
where 
for all 
. From now on this will be denoted as ![$ [a_0; a_1, a_2, \dots, a_m] $](http://latex.artofproblemsolving.com/1/c/4/1c47e13850fd92c24cf30707900dc53e12c32418.png)
.For every
, let 
. Then let 
and for all 
let 
. I claim that this 
satisfies the desired properties. To prove this, consider two rational numbers 
.
If 
then by definition 
as desired.
If 
and 
then by definition as desired since 
.
If and 
then to prove that it suffices to show that 
. Let ![$ -x = [a_0; a_1, a_2, \dots, a_m] $](http://latex.artofproblemsolving.com/1/0/e/10e2dcb8e4ad502ecb43d7e1c2744f3995658d81.png)
. Then ![$ y = [a_0 + 1; a_1, a_2, \dots, a_m] $](http://latex.artofproblemsolving.com/2/c/e/2ce601c354188bdad5e03d9833f5a0efd26eb275.png)
and so 
as desired.
If and 
we can let ![$ y = [0; 1, a_2, \dots, a_m] $](http://latex.artofproblemsolving.com/e/6/b/e6b1ac208f52b35a47ba274b63b6aede8c05b3b5.png)
so that ![$ x = [0; a_2 + 1, a_3, \dots, a_m] $](http://latex.artofproblemsolving.com/5/c/a/5ca981e9723d9f4c0ee0a619a2f05b17f18fee09.png)
so 
and 
which implies that as desired.
If 
, we can assume WLOG that 
. Then letting ![$ x = [0; a_1, a_2, \dots, a_m] $](http://latex.artofproblemsolving.com/f/a/4/fa41f1dfed52008255c6fe47ad8f150591dc820c.png)
we have that ![$ y = [a_1; a_2, a_3, \dots, a_m] $](http://latex.artofproblemsolving.com/1/6/6/166a8243036f6a8f2362c13f93282945d869aff5.png)
so 
and 
which implies that as desired.
August 2016
June 2016