Bijective Function from N to N^2: Examples and Help

  • Thread starter Thread starter Anzewill
  • Start date Start date
AI Thread Summary
A bijective function from N to N^2 can be constructed by arranging natural numbers in a grid, where the cell at position (m,n) corresponds to the pair (m,n). The zigzag pattern starts at (1,1) and moves horizontally, then diagonally, covering all pairs systematically. An alternative formula for this function is (m,n) → 2^m(2n+1)-1. This approach ensures that every natural number is uniquely paired with a coordinate in N^2. The discussion highlights both a visual method and a mathematical formula for establishing the bijection.
Anzewill
Messages
1
Reaction score
0
Anyone can give me a example of a bijective fuction from N to N^2?
 
Physics news on Phys.org
I can't give you a simple "formula" but here is how to get a bijective function:

Write the numbers 1, 2, 3, ... horizontally and to left of "1" and slightly below write 1, 2, 3, ... vertically so that the cell below "m" and to the right of "n" is the pair (m,n). Now start at (1, 1) and "zigzag" through that array. That is, go from (1,1) horizontally to (2, 1) then diagonally, down and left, to (1,2), down to (1, 3), diagonally up and right to (2,3) and (3, 1), right to (4, 1), diagonally down to (3,2), etc.
 
Or for a simple formula try (m,n)\rightarrow 2^m(2n+1)-1.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top