More Islamic math

Al Karaji (?- 1019) wrote "The Marvelous" and studied algebra of exponents. So
1:x= x:x^2 =x^3: x^2.....

He also gave a limited algorithm for square roots of a polynomials. Al Karaji also proved
1^3 +2^3 +....+10^3 = (1+...+10)^2 (which you may have seen in some "intro to proof" class) by inverse induction.

His proof as follows: Using the diagram above
Area(red square) =  (1 + 2 +  · · · + 9)²
Area (yellow rectangle) =  10(1 + 2 +  · · · + 9)
Area(blue square) = 10² 

Then,
Area(2 yellow rectangle + blue square) = 2* 10(1 + 2 +  · · · + 9) + 10²
                                                                   = 2*10(9*10/2)  + 10² 
                                                                  = 10 * 10²
                                                                  = 10³      (1)

Area (of Entire gnomon) = (1 + 2 +  · · · + 9)² +2*10(1 + 2 +  · · · + 9) +10²

 Plugging (1) into the above equation yields
Area (of Entire gnomon) = (1 + 2 +  · · · + 9)² + 10³
                                           = (1 + 2 +  · · · + 9+ 10)²


By repeating a similar argument for the rest of the number yields Al karaji's results
 1^3 +2^3 +....+10^3 = (1+...+10)^2

QED

Al Samaw al (1112-1174) wrote "Shining Book of Calculation" and performed long division.

Unanswered Question

1. Was there any practical purpose for Al karaji discovery of  1^3 +2^3 +....+10^3 = (1+...+10)^2 ?

Evaluation:
Interesting: 8
Quality: 7
Complexity: 7

Mathematics of Islam

A library in Baghdad was established by the caliph Harun al Rashid (786-809) and books that were collected where translated in Arabic. Rashid successor , his son, Al Mamun (813-833) established a research institute call Bayt al Hikmat or House of wisdom; the institute lased 200 years. Despite being a center of learning, the House of wisdom was restricted by belief. Scholars were invited to the institute to translate work and conduct research. They translated many work by the end of the 9th century such as Euclid, Archimedes and other Greek Mathematicians. At the institute they also learn Babylonian mathematics.

Al Khwarizmi(~825) worte "The condensed book on calculations of al-jabra and al-muqabola.
Some now and then:
"al- jabra"====> "algebra
"Sunya"====>sifr===>Zephirum====> zero
sifr====> cifra===>?(guess)

Al Khwarizmi solved 6 types of equatioins

1. square equal to roots (such as ax^2 = bx)
2. square equal to numbers
3. Roots equal to number
4. squares and roots equal to numbers (ax^2 + bx = c)
5. squares and numbers equal to roots
6. roots and numbers equal to squares

This showed a move from concrete math to abstract math


Unanswered Question:

1. How heavily was the House of wisdom restricted by religion? Did it interfere with any mathematical development?

Evaluation:
Interesting: 7
Quality: 8
Complexity:  5

Chinese remainder theorem

One of the most famous technique that comes from china is know as the Chinese remainder theorem. The theorem is used to solve problems dealing with system of linear congruence and its earliest uses is dated back in the book Mathematical Classic Master sun.

Example:
"We have things of which we do not know the number: if we count them by threes, the remainder is 2; if we count them by fives, the remainder is 3; if we count them by sevens , the remainder is 2. How many things are there?"

This is translated into
N= 3x+2
N=5y+3
N=7z+2

or for those with a background in number theory
N== 2(mod 3)
N== 3(mod 5)
N== 2 (mod 7)

where == stands for congruence. For simpler purpose, we will use the earlier notation. So,
1) 3*7=21
     21 /5 give you a remainder 1
    so 3*21= 63

2) 3*5= 15
    15/ 7 gives you a remainder 1
  so 2* 15= 30

3) 5*7 = 35
35*2= 70   we multiply again because 35/ 3 gives you a remainder 2. We want a remainder of 1
70/3 gives you a remainder 1
70*2 = 140


Then adding 63 + 30 + 140 = 233
so, 233-210 = 23                             210= 2(3*5*7)
thus N= 23

When plunging in our answer we see that it satisfies the system of congruences.
Did you notice any patterns during the steps? 
In number theory terms:

• We multiplied two mods together and then divided by the third and we did that for each of the mod
• When we divide the product by the third mod and got a remainder of 1 we multiplied the product by the remainder of that mod

Unanswered Question:

1. Was the Chinese remainder theorem very applicable? I find it hard to find a situation where the objects are unable to be counted

Evaluation:
Interesting: 7
Complexity: 6
Quality: 8