Limited time utterance

[Kathegetes]

Originally Posted by Uncwilly

I would like to inject something of a cure here. The BOTtom line maybe the cause of this ill. We may find something TURnING out to be AIling.

Alan Touring.
Poor man.
I feel sorry for him.

[Kathegetes]

Originally Posted by R.D. Silverman

The level of technical discourse in this forum is already too low.

Was it really necessary to add to the pollution by posting the spam/garbage
that you posted above??

Tech? Somebody solve my "u" mass and see me move my top, bottom, strange beauty and charm up and down.
Raqs Sharki

[jasong]

Originally Posted by Kathegetes

Tech? Somebody solve my "u" mass and see me move my top, bottom, strange beauty and charm up and down.
Raqs Sharki

Kathegetes, who created you? Are there more of you on the Internet, perhaps on other forums under other names?

[Kathegetes]

Originally Posted by Kathegetes

Tired of not being noticed?
Try Kathegetes Dynami Sophrosyne's Prime Shampoo. (for 3 and older).

n = 1, 2, 3 ...
6n - 1 = Silky
2 + Silky = Shiny

If Silky or Shiny greater than 5 or 7 is divisible by 5 or 7, it has split ends and should by yanked out to composite wigs for bald lost lambs of logic.

Lather, rinse, repeat

K.D.S.

Skip zero primes. Find all the twins. If composite is falsely classed, buy Kathegetes Dynami Sophrosyne's conditioner!

Who's looking pretty?

Combine three infinity symbols into a flower placing I in the top petal, and lower case s,t,u,d,y in remainder clockwise. First composite 4 goes to set of all s. All n follows in set rotation. All members of s,u,y are even; divisible by oddest half of Siamese twins. All members of I are divisible by the other. All primitives greater than Siamese are members of set t, or d. If what is true for n is clockwise true for n+1, the proof by induction is complete. Combine the members of t and d to form the set of all h. The composite members of h have a lowest divisor in h itself. All t*t,t*d, and d*d as with the squares thereof form the members of set "c" the composites not I,s,u,y. Removal of c from h leaves set j, the primitive members of n...not Siamese. In the set of all p are members of K. When 2^K-1 is primitive, it is a member of uppercase D as some d. All D triangulated is a member of uppercase S, some s; being the sum of its immoral detractors. Pretty as K.D.S

[f1pokerspeed]

Originally Posted by Kathegetes

Combine the members of t and d to form the set of all h. The

...sound of silence, as we reflect on the utter babble transferred from your brain to the muscles in your hands. Oh, nevermind - I doubt anyone can make head or tailbone of this. If you can, a cookie for you.

How's about demonstrating your point (if such a thing exists) with some real mathematical statements, instead of forcing us to draw pictures? The words should speak for themselves... after all, mathematical papers are supposed to be written with the utmost clarity! (it is a forum, but who cares)

[kladner]

Originally Posted by f1pokerspeed

...sound of silence, as we reflect on the utter babble transferred from your brain to the muscles in your hands. Oh, nevermind - I doubt anyone can make head or tailbone of this. If you can, a cookie for you.

How's about demonstrating your point (if such a thing exists) with some real mathematical statements, instead of forcing us to draw pictures? The words should speak for themselves... after all, mathematical papers are supposed to be written with the utmost clarity! (it is a forum, but who cares)

Reading these posts is an exercise similar to reading James Joyce. (No offense is intended to Joyce fans.)