Even more infinity

My plan right now is to first show that the rational numbers are just as big as the natural numbers. I am going to do this a little more formally so that I can practice using LaTeX (a math coding language).

Last time I defined the natural numbers \mathbb{N} to be \{1,2,3,...\},
but I think I made a mistake as I should have also included 0. So now let \mathbb{N} = \{0,1,2,3,...\}. To define the rational numbers in a more precise way than “all numbers that can be written as a fraction” I just have to let the rational numbers be \mathbb{Q} = \{\frac {a} {b} | a,b \in \mathbb{N} with b \ne 0 \}.
In plain English this is saying, let \mathbb{Q} be the set of numbers “a divided b”, where “a” and “b” are natural numbers, but “b” is not allowed to be 0. So for example \frac {1} {2} is in the set when a = 1 and b = 2

Now, how do we show that this set is actually the same size as \mathbb{N}? Remember that to do this we have to find a way to match up everything from \mathbb{N} with \mathbb{Q} in a pairwise fashion. So only one thing from one set gets matched with one thing from the other set. To do this I need to somehow list ALL of the rational numbers. For now it is simpler to worry just about the positive ones. I do it in the following way:

First we can write out all possible rationals in a table. Notice that every rational number will be contained in this table if you go out far enough. For the general case, if you want to find the number \frac {a}{b} you just need to find where the a’th row intersects the b’th column.

\frac {1}{1}

\frac {1}{2}

\frac {1}{3}

\frac {1}{4}

\frac {2}{1}

\frac {2}{2}

\frac {2}{3}

\frac {2}{4}

\frac {3}{1}

\frac {3}{2}

\frac {3}{3}

\frac {3}{4}

\frac {4}{1}

\frac {4}{2}

\frac {4}{3}

\frac {4}{4}






Next, I need to match this up with my regular natural numbers. This is now very simple.
Using that \mathbb{N} = \{0,1,2,3,...\}, we can match each thing in the table with each natural number using the method below that I’ve done with color. For notation, I’ve used double arrows pointing from the natural number to the entry, meaning that I am pairing up that number with that entry.

We match the first entry, which is the top left, with 0. From there we move to the right and match 1 with that entry. Then we go down, but since \frac {2}{2} = 1 we skip this number (although skipping isn’t really necessary). Then we go back to the left and match that entry with 2, then down again and so on in a snaking fashion so that we do not miss any entries, while skipping numbers that are repeated. Using colors, just follow the pattern of Pink, Red, Blue, Green, Cyan, Orange, Grey, Pink.

0 \gg \frac {1}{1}

1 \gg \frac {1}{2}

6 \gg\frac {1}{3}

7 \gg\frac {1}{4}

2 \gg \frac {2}{1}

Skip \frac {2}{2}

5 \gg \frac {2}{3}

Skip \frac {2}{4}

3 \gg \frac {3}{1}

4 \gg \frac {3}{2}

Skip \frac {3}{3}

8 \gg \frac {3}{4}

10 \gg \frac {4}{1}

Skip \frac {4}{2}

9 \gg \frac {4}{3}

Skip \frac {4}{4}






This shows that the size of the rational numbers is equivalent to the size of the natural numbers. When I started today, I also wanted to prove that the stuff I talked about yesterday having to do with uncountable infinity, but since I decided to code for the first time this took way longer than I expected. So that’s all for today.

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