4 ms·
You cannot bend the rules in this way and still equate it to the question at hand - you're introducing characters outside of base2 in your example. If you stuck
by aesthetics1 10y ago
You cannot bend the rules in this way and still equate it to the question at hand - you're introducing characters outside of base2 in your example. If you stuck within the realm of base2, '2' in its binary form certainly appears in the pattern.
Not only that, but Pi is normal (all digits distributed evenly), and your sequence clearly is not.
I understand what you are saying - indeed there are infinite finite sequences - but it just does not apply here.
- JamilD 10y agoYou can trivially modify their example to be 1.02003000400005... cycling through the digits and the argument still holds. As another commenter pointed out, pi is not proven to be normal, but is suspected to be.
- SomeStupidPoint 10y agoThey already showed 21 (in binary, 10101) doesn't appear anywhere, even as a binary pattern.
- evanb 10y agoThe example isn't base two. It's base 10. It just happens to be a number that only has 0s and 1s. Like eleven. Or 1000.
- TOMDM 10y agoTo phrase it another way, there are infinitely many sizes of Infinity. Just because you have one infinite sequence, does not mean it is the same size as another Infinity and as such may include or exclude elements of the other Infinity.
- monochromatic 10y agoPi is not known to be normal.