عبدالله جلغوم
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هذه الرسالة من باحث متخصص في الرياضيات يعمل في جامعة اكسفورد ، تتضمن ورقتين من بحث له حول قانون الحالات الأربع لسور القرآن ..
وقد وصلني من قبل رسالة من فرنسا من باحث ( دكتوراة ] قام بترجمة الموضوع ، ورسالة اخرى من كندا ...
...............
أتمنى أن يصدر مثل هذا العمل عن مؤسسة دينية او جامعة إسلامية . وليكن ذلك بعد الاستماع إلى كل ما لدي وأن يتم تشكيل لجنة لمناقشة هذه المسألة يتم اختيار أعضائها من اهل الاختصاص في علوم الشريعة والرياضيات .
صحيح أنني صاحب الفكرة والمكتشف الأول إلا أن هناك من هو أقدر مني على تقديم الموضوع وتطويره ، بعد أن أقدمه له ..
From : abderrahim youssef <[email protected]>
Sent : Thursday, June 7, 2007 2:46 PM
To : [email protected]
Subject : RE: Koran's ranking
Have a look at this paper that i have written inspiring from some of your ideas ....in 1994, an article about the Tourat code was published in statistical science, so I am wondering why we can't do the same thing?
salamoualaykoum,
Abderrahim
Some analysis of the current Koran’s ranking
Abderrahim Oulhaj
May 29, 2007
1 Introduction
The KORAN is decomposed into N = 114 chapters called Sourats in arabic. Each Sourat contains a
number of verses called Ayats in arabic. One mysterious matter that made me thinking for a while is
that the actual ranking of the sourats in the Koran called MASHAF ranking is completely different from
the chronological one. For instance, sourat Albaqara has rank 2 (chapter number 2) in the MASHAF
ranking and has rank 87 (chapter number 87) in the chronological one. It is well known that the prophet
(sws), by a revelation from allah, has asked his compagnions to use this particular ranking of sourats in
the Koran and nobody knows why. This current ranking may then have some relevant properties making
it qualitatively and quantitatively different from all other possible rankings of the Koran. This is the
main object of this work. We aim to prove that the MASHAF ranking which is one possibility from
N! = 114! was not randomly established however something logical is hidding behind it. This work is a
mathematical reformulation and extension of some works that I have found in the web.
2 Notations
Let S be the set of all sourats in the Koran with card(S) = 114 (the total number of sourats in the
Koran) and R be the set of all ranking functions defined on S. A ranking function R in R is any bijective
function assigning ranks to the sourats of the Koran, more specifically:
R : S −! {1, 2, 3, ..., 114}
s −! R(s)
The number of ayats in sourat s 2 S is denoted by na(s). Let us now define 2 binary functions. The first
one indicates whether the number of ayats in a given sourat is even or not, namely:
I1(s) = 11
{na(s) is even}
=8<:
1 if na(s) is even
0 otherwise
(2.1)
and the second one indicates whether the rank of a given sourat, according to a ranking function R 2 R,is
even or not, namely:
I2(s,R) = 11
{R(s) is even}
=8<:
1 if R(s) is even
0 otherwise
(2.2)
Definition 1 A sourat s 2 S is said to be R-homogeneous (or homogeneous according to a ranking
function R) if the number of ayats of s is even and the rank of s (according to R) is even or if the
number of ayats of s is odd and the rank of s (according to R) is odd, i.e. if I1(s) = I2(s,R) = 1 or if
I1(s) = I2(s,R) = 0. A sourat s is not R-homogenous otherwise.
Let SR _ S be the set of all R-homogenous sourats. Then SR may be written as:
SR = {s 2 S | I1(s) + I2(s,R) 2 {0, 2}} (2.3)
1
3 Some results
Lemma 1 The following two properties hold whatever the ranking function R 2 R
(1) Xs2S
na(s) = 6236
(2) Xs2S
R(s) = 6555 8R 2 R
Property (1) in lemma 1 gives the total number of ayats in the Koran (6236 ayats) and property (2) gives
the sum of ranks of sourats in the Koran (6555). This sum is independent of the used ranking function.
The following theorem shows a very interesting and surprising property of Rm, the current ranking
function of the Koran also called the MASHAF ranking. More specifically:
Theorem 1 Let Rm be the MASHAF ranking function, SRm the set of all Rm-homogenous sourats and
ScR
m
its complementary. Then we have the following two properties:
card(SRm) = N
2
(3.4)
X {s2SRm}
Rm(s) = X {s2Sc
Rm}
na(s) (3.5)
where card(SRm) stands for the cardinal of SRm (i.e. the number of sourats in SRm) and N = 114 is
the total number of sourats in the Koran. Theorem 1 shows the following: The actual Koran’s ranking
function Rm splits the Koran into two equal parts SRm and ScR
m
. Furthermore, the sum of the ranks of
the sourats in the first part is exactely equal to the total number of ayats in the second part!
The question afterwards is whether these two properties are unique for the current ranking function
Rm. In order to answer this question we need to check these two properties for all the N! = 114! possible
functions R 2 R. This is definitely not possible. An estimation of the proportion of R 2 R satisfying
properties (3.4) and (3.5) in theorem 1 is therefore needed. A random sample of size n = 264000 was
drawn from the population R. The estimated proportion is 0.0003. This means that only 0.03% of ranking
functions satisfy the two properties in theorem 1. This result is estonishing and is only a begining to a
deeper one. Adding some other criterias will surely make Rm unique. Another option is to study the
relation between Rm and the 0.03% other functions and see whether they have something in common.
2
وقد وصلني من قبل رسالة من فرنسا من باحث ( دكتوراة ] قام بترجمة الموضوع ، ورسالة اخرى من كندا ...
...............
أتمنى أن يصدر مثل هذا العمل عن مؤسسة دينية او جامعة إسلامية . وليكن ذلك بعد الاستماع إلى كل ما لدي وأن يتم تشكيل لجنة لمناقشة هذه المسألة يتم اختيار أعضائها من اهل الاختصاص في علوم الشريعة والرياضيات .
صحيح أنني صاحب الفكرة والمكتشف الأول إلا أن هناك من هو أقدر مني على تقديم الموضوع وتطويره ، بعد أن أقدمه له ..
From : abderrahim youssef <[email protected]>
Sent : Thursday, June 7, 2007 2:46 PM
To : [email protected]
Subject : RE: Koran's ranking
Have a look at this paper that i have written inspiring from some of your ideas ....in 1994, an article about the Tourat code was published in statistical science, so I am wondering why we can't do the same thing?
salamoualaykoum,
Abderrahim
Some analysis of the current Koran’s ranking
Abderrahim Oulhaj
May 29, 2007
1 Introduction
The KORAN is decomposed into N = 114 chapters called Sourats in arabic. Each Sourat contains a
number of verses called Ayats in arabic. One mysterious matter that made me thinking for a while is
that the actual ranking of the sourats in the Koran called MASHAF ranking is completely different from
the chronological one. For instance, sourat Albaqara has rank 2 (chapter number 2) in the MASHAF
ranking and has rank 87 (chapter number 87) in the chronological one. It is well known that the prophet
(sws), by a revelation from allah, has asked his compagnions to use this particular ranking of sourats in
the Koran and nobody knows why. This current ranking may then have some relevant properties making
it qualitatively and quantitatively different from all other possible rankings of the Koran. This is the
main object of this work. We aim to prove that the MASHAF ranking which is one possibility from
N! = 114! was not randomly established however something logical is hidding behind it. This work is a
mathematical reformulation and extension of some works that I have found in the web.
2 Notations
Let S be the set of all sourats in the Koran with card(S) = 114 (the total number of sourats in the
Koran) and R be the set of all ranking functions defined on S. A ranking function R in R is any bijective
function assigning ranks to the sourats of the Koran, more specifically:
R : S −! {1, 2, 3, ..., 114}
s −! R(s)
The number of ayats in sourat s 2 S is denoted by na(s). Let us now define 2 binary functions. The first
one indicates whether the number of ayats in a given sourat is even or not, namely:
I1(s) = 11
{na(s) is even}
=8<:
1 if na(s) is even
0 otherwise
(2.1)
and the second one indicates whether the rank of a given sourat, according to a ranking function R 2 R,is
even or not, namely:
I2(s,R) = 11
{R(s) is even}
=8<:
1 if R(s) is even
0 otherwise
(2.2)
Definition 1 A sourat s 2 S is said to be R-homogeneous (or homogeneous according to a ranking
function R) if the number of ayats of s is even and the rank of s (according to R) is even or if the
number of ayats of s is odd and the rank of s (according to R) is odd, i.e. if I1(s) = I2(s,R) = 1 or if
I1(s) = I2(s,R) = 0. A sourat s is not R-homogenous otherwise.
Let SR _ S be the set of all R-homogenous sourats. Then SR may be written as:
SR = {s 2 S | I1(s) + I2(s,R) 2 {0, 2}} (2.3)
1
3 Some results
Lemma 1 The following two properties hold whatever the ranking function R 2 R
(1) Xs2S
na(s) = 6236
(2) Xs2S
R(s) = 6555 8R 2 R
Property (1) in lemma 1 gives the total number of ayats in the Koran (6236 ayats) and property (2) gives
the sum of ranks of sourats in the Koran (6555). This sum is independent of the used ranking function.
The following theorem shows a very interesting and surprising property of Rm, the current ranking
function of the Koran also called the MASHAF ranking. More specifically:
Theorem 1 Let Rm be the MASHAF ranking function, SRm the set of all Rm-homogenous sourats and
ScR
m
its complementary. Then we have the following two properties:
card(SRm) = N
2
(3.4)
X {s2SRm}
Rm(s) = X {s2Sc
Rm}
na(s) (3.5)
where card(SRm) stands for the cardinal of SRm (i.e. the number of sourats in SRm) and N = 114 is
the total number of sourats in the Koran. Theorem 1 shows the following: The actual Koran’s ranking
function Rm splits the Koran into two equal parts SRm and ScR
m
. Furthermore, the sum of the ranks of
the sourats in the first part is exactely equal to the total number of ayats in the second part!
The question afterwards is whether these two properties are unique for the current ranking function
Rm. In order to answer this question we need to check these two properties for all the N! = 114! possible
functions R 2 R. This is definitely not possible. An estimation of the proportion of R 2 R satisfying
properties (3.4) and (3.5) in theorem 1 is therefore needed. A random sample of size n = 264000 was
drawn from the population R. The estimated proportion is 0.0003. This means that only 0.03% of ranking
functions satisfy the two properties in theorem 1. This result is estonishing and is only a begining to a
deeper one. Adding some other criterias will surely make Rm unique. Another option is to study the
relation between Rm and the 0.03% other functions and see whether they have something in common.
2