๐๏ธ Shepiwot ยท 19 points ยท Posted at 23:05:13 on November 7, 2013 ยท (Permalink)
So for n=2 formula "n edgy (n+2) me" seems fine, but beyond that is it still
n edgy (n+2) me
or
n edgy (n2) me
or
n edgy (2n) me
Also, is n limited to natural numbers?
[1]
Saved comment
glorioussideboob ยท 4 points ยท Posted at 01:49:12 on November 8, 2013 ยท (Permalink)
I'm afraid none of them are correct. The actual formula as any standard textbook will tell you is '(eiฯ +2)(ln(en )) edgy (ฯ+0.85840734641)((2n)-1/2 ) me' where n is limited to the set of edgy numbers: nโE
MPORCATO ยท 8 points ยท Posted at 02:41:17 on November 8, 2013 ยท (Permalink)*
But the set of edgy numbers is exactly the natural numbers N !
Proof: If the set of edgy numbers is not N then there exists natural numbers which are not edgy. Hence consider the non-empty set of non-edgy numbers N\E. Since < on N is a well-order, it is also a well-order on N\E. Thus there is a smallest element in N\E. This number is the smallest non-edgy number, which is a cool and edgy property. Therefore this number is edgy. But this contradicts the definition of this number being non-edgy, so we conclude that N\E must be empty. So every number is edgy. QED.
glorioussideboob ยท 2 points ยท Posted at 16:58:07 on November 8, 2013 ยท (Permalink)
Why N factorial and not just N?
MPORCATO ยท 2 points ยท Posted at 18:25:47 on November 8, 2013 ยท (Permalink)
Here ! is a plain old exclamation mark :p
We are doing shitty math anyway, so...
glorioussideboob ยท 2 points ยท Posted at 20:42:01 on November 8, 2013 ยท (Permalink)*
I know :p but back to the maths sure the set of natural numbers contains all the edgy numbers, but edgy numbers are still a set in their own right. There are edgy numbers which are irrational but there are also some irrational numbers which aren't edgy. The irrational numbers which aren't edgy numbers are those which can be written in the forms pik2 =n and 4/3pik3 =n where k is a natural number. These are not edgy numbers as circles and spheres have no edges.
MPORCATO ยท 3 points ยท Posted at 21:10:21 on November 8, 2013 ยท (Permalink)
That would only be true if we don't assume the axiom of choice. Given AC we may derive the well-ordering theorem and apply my original proof to show the edgy subset of any set is equal to the whole set.
glorioussideboob ยท 2 points ยท Posted at 20:09:28 on November 9, 2013 ยท (Permalink)
Shit dude, 3 edgy 5 me... I'm only just doing maths at A-level.