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Math language class

 For fun.
  (+6, -1)
(+6, -1) 		  [vote for,
against]

Just like you can take an english class to learn how to write and speak better, schools should offer a math language class where you learn the terms and symbols that go along with the math class you are taking. It would be supplemental of course. Students would not have to solve any equations but would be graded on symbol placement and quality of problem layout.
—Antegrity, Oct 10 2005

Not your ordinary k http://www.kelloggs.../specialk/index.asp
Bolztmann? Who he?
[coprocephalous, Oct 11 2005]

[link]






       I'm skeptical of the "for fun" part, but it maybe wouldn't be too hard so... (+)
—PollyNo9, Oct 10 2005
   

       It would be like a technical writing class.
—Antegrity, Oct 10 2005
   

       Would they teach you to spell "maths" correctly?   

       Flippancy aside, this is actually something that should be incorporated into all maths courses but isn't. I'm fairly sure that if I had been taught a bit more about problem layout in my maths A-level then I wouldn't have got a D.   

       Delta(score)=Pastry.
—wagster, Oct 10 2005
   

       How can you score zero in an English school system exam? Shirley you must've spelled you own name correctly? (From the person who managed to get only an 'O' level pass at 'A' level, despite already having had an A grade 'O' level for the past two years - go figure)
—coprocephalous, Oct 10 2005
   

       What sort of maths class are you taking where you *don't* learn the terms and symbols? They strike me as being somewhat fundamental to the subject.
—angel, Oct 10 2005
   

       I had a great teacher for higher maths. Some of my class might disagree.
—Jinbish, Oct 10 2005
   

       My grade 3 in 'O'-level maths is almost entirely due to Harry Whelding. Great teacher.
—angel, Oct 10 2005
   

       Note on spelling: maths : uk = math : us.
(Bonus question: What's the value of "k"?)
—jutta, Oct 10 2005
   

       maths = harder to pronounths. math = easier
—phundug, Oct 10 2005
   

       uk = m; (micro(kilo)=milli);   

       us = microsecond;   

       csea
—csea, Oct 10 2005
   

       Not only can 1k be either 10^3 or 2^8, 1M can be 1k (either version) multiplied by 1k (not necessarily the same version). Which is why my new 80G hard drive is not as big as I thought it would be. But really, who's counting?
Bonus answer: -273.15 (not expecting any marks for that, just being smart-assed).
—angel, Oct 10 2005
   

       Correct bonus answer: k = 1.38066*10^-23 J*K^-1 :-)   

       All smartass remarks aside, I think it would be difficult to find enough material to fill even one semester, unless, perhaps, you added in Mathematical Markup Language.   

       I agree with angel, as anytime a new concept is introduced, any concept-specific symbols are explained. Additionally, I think you would have to understand the concept first before learning how to write it down. I know what the integral symbol looks like but I won't know its function until the start of the next semester.
—Cuit_au_Four, Oct 11 2005
   

       The math language class would explain wtf all you are talking about means.
—Antegrity, Oct 11 2005
   

       //Not only can 1k be either 10^3 or 2^8// sp. "2^10"

//-273.15 (not expecting any marks for that, just being smart-assed).// Ah, no, that would be "0 K", not "k" - no marks for being a half-assed smart-ass.

According to my 20 year old Casio fx-570, the value of k is 1.380662x10^-23 JK^-1 - I claim my bonus pastry, for getting one extra place of decimals over [cuit-au-four].
—coprocephalous, Oct 11 2005
   

       surely k=sqrt(k^2)
—reap, Oct 11 2005
   

       Well, in [jutta]'s question, I'm vaguely confident that the value of k is "kingdom".
—hidden truths, Oct 11 2005
   

       "Nautical Almanac Office used two separate values for k in their predictions. The larger value (k = 0.2724880), representing a mean over topographic features, was used for all penumbral (exterior) contacts and for annular eclipses. A smaller value (k = 0.272281), representing a mean minimum radius, was reserved exclusively for umbral (interior) contact calculations of total eclipses [Explanatory Supplement, 1974]. Unfortunately, the use of two different values of k for umbral eclipses introduces a discontinuity in the case of hybrid or annular-total eclipses"   

       don't know much about math or maths, but I can use google [jutta] Do I get a prize?
—dentworth, Oct 11 2005
   

       K = "contract", does k = little less than a contract?
—reensure, Oct 11 2005
   
      
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