I am sure most of u have come across "dudes" who try too hard to be wat they are not and vice versa...
But strangely these "dudes" have sum properties that are so similar that one can put them into a set which is a subset of "fake" individuals. Mathematically
P(dudes)
now dudes group/set follows a few axioms
a. Dudes are Dudes( which is self evident)
b. Dudes + Dude = Dude
c. Dude+ intelligent = Dude+ Frustrated Intelligent(this is a transformation matrix)
Dude Dude [ intelligent intelligent] = [ DI DI]
Dude Dude X
Here the dude is the transformation matrix and not the intelligent, In lay man terms dudes transform the intelligent guys and never vice versa.
d. Dudes always spew gibberish.
Ie. [dude]X ( any signal)= Total White noise
This is sumthing like the infinite series property.
Infinity X any finite Qnty = infinity
Hence Dude is Human equivalent of infinity
e. Dudes don’t always understand each other
Ie. Dude1 --à Dude2 doesn’t necessarily mean
Dude 2--à Dude 1
Which basically means that Dudes are not commutative
f. Dudes deny their culture even while staying inside it
g. Dudes do not have any weight hence the total mean of Dudes is zero
ie. Sum (Dudes) = 0
Using these basic Axioms we can build a new branch of mathematics and use it to study “dudes” . Any scientific papers on the subject are welcome on this blog
2 comments:
following the same spirit of dude matics let us now have a brief session on "dude calculus"
d(dude)/dt >= 0
(once a dude, remains a dude or becomes a bigger dude)
d(k*dude)/dt >= k*d(dude)/dt
('k' dudes hanging out together would result in greater dudeness in time then when the dudes are saperated)
d2(dude)/dt2 >= 0
the rate at which the dude becomes a bigger dude is also positive.
d(dude)/d(money) >= 0
it cost's money for the dude to become bigger dude. which they become anyways.
d(dude)/d(girl) = girls^(her intelligence)
The more intelligent the girl is the more dude the dude becomes... (which in turn costs him money)
dude has no additive inverse!
dude + (-dude) = 2*dude
Therefore,
d(dude1 + or - dude2)/d(same girl) = 2*girls^(her intelligence)
(an intelligent girl takes advantage of both the dudes, irrespective of whether the dudes are friends or enemies)
integral(dude).dt = area covered by the dude roaming behind girls
integral(integral(dude)).dt.d(girl) = area covered by the dude roaming around the same girl
dude + bike = bigger dude
therefore
integral(d(dude+bike)/d(girls)).d(girl) = girl + shopping + multiplex + restaurants
a dude with bike roaming around girls, ends up as a paying driver
dude + mobile = bigger dude
basically
dude + anything = bigger dude
actually
dude +/- anything = bigger dude
wait that proves my original axiom that
d(dude)/dt >= 0
- thanks and regards,
"intelligent dude"
Pradeep ..ur work is original and intelligent . However we have to sit down and solve it on pen and paper. I have a few axioms on graph theory of dudes. will require your comments. Infact we can make a branching tree of dudes and non dudes and model a markov chain on that basis. (the memoryless property of Dudes will be an asset in this analysis)
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