May 29th, 2018, 04:48 AM  #1 
Senior Member Joined: Nov 2011 Posts: 197 Thanks: 2  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 04:52 AM. 
May 29th, 2018, 07:20 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,183 Thanks: 870 
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, 08:29 AM  #3 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 196 Thanks: 59 Math Focus: Algebraic Number Theory, Arithmetic 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, 12:54 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,919 Thanks: 533 
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 12:57 PM. 

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