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stupid idiot

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How much time do mathematicians spend on memorizing axioms, definitions, theorems, lemmas ect... ?

Do they spend more time on learning the precise definition of a theory than doing computations and learning algorithms?

I'm curious to know...
 

Affinity

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definitins come naturally.. coz when you use those definitions, things generally turn out nicely
 

stupid idiot

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I'm not sure what you mean by definitions come naturally. Surely you cannot not know any of the definitions without learning it first. If definitions did come naturally then math as a "subject" would not exist.
 

flyin'

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Without definitions there is no Mathematics. Even the addition operator is a definition.. Like what is 1 + 1, if you don't define what + is. :p
 

stupid idiot

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A Field F is any set of objects with two operations (+) and (x) (usually called "addition" and "multiplication") defined in it, provided that these objects and operations satisfy the six Axioms of addition and multiplication.

2 is defined as 1+1. :p
 

Affinity

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learning terminology is for social sciences..

anyway.. when I say definitions come naturally.. it means.. things are defined in a way that makes things simpler and not complicated, if you find definitions hard to remember that's because it's not fully appreciated.

some mathematicians spend most of their time devising new theory and some use exisiting theories to find results..
both are important. And definitions and theoretical stuff is important for both camps..

oh by the way, do this:
integrate {-1 to 1} 1/x^2 dx
 

stupid idiot

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Originally posted by Affinity
anyway.. when I say definitions come naturally.. it means.. things are defined in a way that makes things simpler and not complicated, if you find definitions hard to remember that's because it's not fully appreciated.
True that. Definitions are there to make math work and not hard. Appreciating definitions is another matter, but i get what you mean.

the answer to the question:

integrate {-1 to 1} 1/x^2 dx = 2 x infinity = infinity
 

stupid idiot

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Hey Affinity, i need your help with one more thing before i leave this place. I joined this forum for one main reason really, to find out what is going on here after browsing through a 3unit thread by someone asking to do some probability questions. This was one of the questions and the solutions given by this guy:

Originally posted by nike33
7. In how many ways can 3-letter combinations be made from the word AUSTRALIA? (Answer: 60480)

again excuse my poor wording/expalations..
there are 9P3 ways of getting the 3 values..

and then theres 6!/3! (accounting for the triple a) ways of arranging these.. .:. 9P3 x 6!/3!...(sum1 check this..no calculator :)

edit: another way..consider the full word AUSTRALIA...there are 9!/3! ways of arranging this .:. if each time this happens you split them in groups of 3..it would be the same...so 9!/3! is the answer
You seem very knowledgable so i hope you can clarify this for me. Am i extremely stupid or is high school math this tricky? How did the guy reason like that and obtained the apparently "correct solution" of 60480 according to the original question poser?

I got 42 "combinations" as my answer.
 
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Constip8edSkunk

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I {-1 -> 1} 1/x^2 dx is is an improper integral of the 2nd kind
=lim(r->0-)I{-1 -> r} 1/x^2 dx + lim(r->0+)I{r -> 1} 1/x^2 dx
=lim(r->0-)[-1/r - 1] + lim(r->0+)[-1 + 1/r]
both limits tend to infinity (and negative infinity),
therefor the integral is divergent and do not exist
 
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Affinity

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hmm.. should have thought of less obvious one :p
 

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