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Mathematics with Minimum Raw Material, Part 2

Quark Boy December 21, 1998

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listing 16-32   1 2

#17 Posted by Goga on January 2, 1999 10:30:53 am
Shafqat:

My memory is that the general solution of quardratic equations was devised by Omar Khayyam by geometric means. Khwarizimi is known for introducing the numerals and most importantly the number ``zero`` which according to some has hindu origins and also the methods for division and multiplication. The term ``algorithm``, which is step by step application of operations to obtain a solution, was introduced by Khwarizimi.

In any case, he is deserves to be called a great mathematician.



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#18 Posted by Fozia Qazi on January 3, 1999 3:21:29 pm
Re: Wasiq(16)

Very nice proof! Now I must find the first one.

Yes, I did check the pi algorithm. For me though, the digit hunt itself is not very interesting ( I lost interest in it after two decimals places :) ). What I find fascinating about pi is its unexpected occurrence in myriad mathematical situations.

Fozia

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#19 Posted by shafqat on January 4, 1999 3:59:03 pm
RE: Goga.

I finally found my source for the statement about Al-Khwarizmi and the solution to quadratic equations. It is in a book by William Dunham entitled ``Journey Through Genius,`` (John Wiley & Sons, 1990), according to which Al-Khwarizimi is indeed credited with being the first one to propose the general solution to quadratics (minus b plus minus root b squared minus 4ac over 2a).

As to your other point, the term `algorithm` was not coined by Al-Khwarizmi. It is simply the English corruption of his name, and was employed by later mathematicians.

Interestingly, the term `algebra` is also indirectly attributable to Al-Khwarizimi. It is the English corruption (through Latin) of part of the name of his major treatise, Hisab al-jabr.

Saad

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#20 Posted by Fozia Qazi on January 5, 1999 10:36:05 pm
Re: Wasiq and the dissection of a square

Not sure if you`ll read this since this article is off the front page now, but I saw a dissection of the square today in the hallway leading to my office! Talk about coincidences :) It was posted on the wall by a colleague of mine who is away at present. He has a habit of posting such problems on the board for our amusement and must have done that before the semester break. I didn’t get a chance to look at it until today. Anyway, it is a clever dissection with a total of 24 squares. To get you started call the square abcd
and divide each side into 175 units.
Along side ab are squares of areas 64, 33, 35 and 43 ( in that order)
Along side bc ….43, 51 and 81
Along cd 81, 39 and 55
Along da 55, 56 and 64

The gap in the middle is filled by 15 other squares having areas 1, 2, 3, 4, 5, 8, 9, 14, 16, 18, 20, 29, 30, 31and 38

I don’t have the source, but I can check with my colleague once he comes back.

Fozia


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#21 Posted by Fozia Qazi on January 6, 1999 9:35:06 pm
In my last reply I should have said ‘sides’ instead of ‘areas’ . Oops!

Fozia


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#22 Posted by Goga on January 10, 1999 12:55:57 am
Some thing elegant that I found on the cover of a book on number theory is the so called golden rectangle. The ``note on the cover`` explains that if the largest square from a golden recangle is cut away the remaining rectangle is again a golden rectangle (a sort of a fractal). A golden rectangle is the one with length to width ratio of (1+SQRT(5))/2. It also says that the ancient Egyptian might have used this ratio to construct the pyramids. This ratio recures quite often in the number theory for example this is the ratio of n+1th and nth Fibonacci number as n approaches infinite.

shafqat:

Omar Khayyam is credied with geometrically finding solutions to cubic equations not quardratic as i said previously. I stand corrected.



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#23 Posted by Inquirer on August 13, 2003 11:13:17 am
Is Quark Boy same as Wasiq?

Anyway here is a much clearer proof of the Pythagorean theorem.
Let there be any right angled triangle with base b, perpendicular p and hypotenuse h and an acute angle theta. Drop a perpendicular from vertex to hypotenuse dividing the triangle into two parts.

Area of the triangle equals the sum of areas of two right angle parts. Hence,

0.5 h sin theta x h cos theta = 0.5 b sin theta x b cos theta + 0.5 p sin theta x p cos theta

Thus, hsquare = bsquare + psquare

QED

PS. It is interesting to note that this proof of the theorem was developed by Einstein at the age of eleven!!!!!
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listing 16-32   1 2

Interact Index

    #23 Inquirer
    #22 Goga
    #21 Fozia Qazi
    #20 Fozia Qazi
    #19 shafqat
    #18 Fozia Qazi
    #17 Goga
    #16 wasiq
    #15 Fozia Qazi
    #14 Fozia Qazi
    #13 Goga
    #12 wasiq
    #11 Goga
    #10 shafqat
    #9 wasiq
    #8 Goga
    #7 Goga
    #6 Goga
    #5 Fozia Qazi
    #4 khan
    #3 khan
    #2 wasiq
    #1 Fozia Qazi

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