3 ms·
You have an infinite number of both since the counting never ends. In your example, if you've counted n positive integers, you've counted 2n even integers assoc
by pavas 12y ago
You have an infinite number of both since the counting never ends. In your example, if you've counted n positive integers, you've counted 2n even integers associated with those (by going both ways). But if you continue counting, eventually you will count 2n positive integers (at which point you'll have counted 4n even integers). For any number of even integers that you count, you can just keep counting the positive integers and you'll eventually reach that number.
Here's another way to think about it: are there more positive integers than there are positive integers? (Any "correct" approach should say no, or at least not yes.)
Using your approach, however, here is a proof that there are more positive integers than there are positive integers.
Lets rearrange the positive integers in this manner:
X--->
1, 2, 3, 4, ...
..., 7, 5, 3, 1, 2, 4, 6, 8, ...
<--X-->
- malandrew 12y agoSo an odd counting vector and and even counting vector, both in the positive direction would be equivalent to the approach I suggested. That makes sense. Thanks for the clarification.