May 29th, 2018, 05:48 AM  #1 
Senior Member Joined: Nov 2011 Posts: 248 Thanks: 3  bijection function
How I prove this statement: bijection f : f(rational) > f(positive rational)? Or how bijection function is a function of rational numbers to positive rational numbers? to prove by: Arithmetic of powers[of the functions] and not by: Schröder–Bernstein theorem Last edited by shaharhada; May 29th, 2018 at 05:52 AM. 
May 29th, 2018, 08:20 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
To prove what statement? "bijection f : f(rational) > f(positive rational)?" simply introduces a function, f. What do you want to prove about such a function? Do you want to prove such function exists? Do you just want to find an example? The rational numbers are countable. That is, we can order the set of all rational numbers. say $\displaystyle a_1, a_2, a_3\cdot\cdot\cdot$, and order the set of all positive rational numbers, sat $\displaystyle b_1, b_2, b_3, \cdot\cdot\cdot$. Map $\displaystyle a_i$ to $\displaystyle b_i$. 
May 29th, 2018, 09:29 AM  #3 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Not quite. For example, the set of real numbers is naturally ordered but is not countable. Countable actually means "can be put in bijection with the naturals", which is much, much stronger than "can be ordered". In fact, "can be ordered" says nothing at all  any set can be ordered or even wellordered. (This is at least true in ZFC. You certainly need choice to be able to wellorder all sets, but you don't quite need it just to order them. It's sufficient for this, anyway.)

May 29th, 2018, 01:54 PM  #4 
Senior Member Joined: Aug 2012 Posts: 2,156 Thanks: 630 
I think I see an explicit bijection between the rationals and the positive rationals. You just take the standard snaking diagonals argument but extend the rows and columns infinitely in both directions. So you have basically the integer lattice in the plane. Now you start at the origin and you run the standard snake in the positive direction, and also a mirror snake in the negative direction. Then the algorithm is to go back and forth alternating between the positive and the negative snakes. In this way we can enumerate all the rationals, not just the positive ones. Then you can just pair up the nth term of the double snake with the nth term of the standard (positive) snake and you have an explicit bijection. I'm sure an enterprising individual could work out the explicit formula for the nth term, by extending the idea of the Cantor pairing function. And "Today I Learned" that it's a theorem that this is the only quadratic pairing function; and that it's an open question as to whether this is the only polynomial pairing functions. Now that's interesting! Last edited by Maschke; May 29th, 2018 at 01:57 PM. 

Tags 
bijection, function 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Is there a bijection from R^n to R?  Magnitude  Topology  1  September 14th, 2017 02:09 AM 
pq bijection  Elwardi  Calculus  1  July 8th, 2016 05:39 AM 
show that the function f is bijection  zodiacbrave  Applied Math  1  February 23rd, 2012 09:02 PM 
Bijection  rebecca  Applied Math  2  January 21st, 2010 08:23 AM 
bijection  fakie6623  Abstract Algebra  3  October 17th, 2007 10:51 PM 