Difference between revisions of "2013 IMO Problems/Problem 5"
Illogical 21 (talk | contribs) (added problem) |
|||
| (2 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
Let <math>\mathbb Q_{>0}</math> be the set of all positive rational numbers. Let <math>f:\mathbb Q_{>0}\to\mathbb R</math> be a function satisfying the following three conditions: | Let <math>\mathbb Q_{>0}</math> be the set of all positive rational numbers. Let <math>f:\mathbb Q_{>0}\to\mathbb R</math> be a function satisfying the following three conditions: | ||
| Line 7: | Line 8: | ||
Prove that <math>f(x)=x</math> for all <math>x\in\mathbb Q_{>0}</math>. | Prove that <math>f(x)=x</math> for all <math>x\in\mathbb Q_{>0}</math>. | ||
| − | + | ==Solution== | |
| + | {{solution}} | ||
| + | |||
| + | ==See Also== | ||
| + | *[[2013 IMO]] | ||
| + | {{IMO box|year=2013|num-b=4|num-a=6}} | ||
Latest revision as of 00:32, 19 November 2023
Problem
Let 
be the set of all positive rational numbers. Let 
be a function satisfying the following three conditions:
(i) for all 
, we have 
;
(ii) for all , we have 
;
(iii) there exists a rational number 
such that 
.
Prove that 
for all 
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See Also
| 2013 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
