Earlier, we made mention of a father who gave his three children, ‘A’, ‘B’ and ‘C’ £120 to share in proportions of 1/3, 1/6 and ½ respectively. Of course the LCM of 3, 6 and 2 is 6 which is also the base number; so £120 is divided into 6 portions.
‘A’: 6 × 1/3 = 2 portions
‘B’: 6 × 1/6 = 1 portion
‘C’: 6 × ½ = 3 portions
Sum of the portions = 2 + 1 + 3 = 6 portions; equivalent to the base number. This is an example of a perfect distribution of an estate. However, supposing the share of ‘C’ was 2/3, what happens? Their shares will then be 1/3, 1/6 and 2/3. The base number is still 6. But ‘C’ shall be entitled to 4 portions! How?
‘C’: 6 × 2/3 = 4 portions
New sum of portions = 2 + 1 + 4 = 7 portions; which is greater than the base number. What is the implication of this? Let’s examine it critically.
1 portion of £120 = £120 ÷ 6 = £20
‘A’ has 2 portions = £20 × 2 = £40
‘B’ has 1 portion = £20 × 1 = £20
‘C’ has 4 portions = £20 × 4 = £80
Summation = £40 + £20 + £80 = £140
Something must be wrong somewhere. The possibilities are:
a) The share of at least one of the children (i.e. 1/3, 1/6 or 2/3) is wrong.
b) The father erroneously gave them £120 instead of £140.
But in inheritance, none of these assumptions hold water. That is to say, the shares of the children which denote the shares of heirs are correct. Also, the amount the father gave; a figurative expression meaning the total asset of a deceased is equally correct. Actually, this is a practical example whereby a deceased leaves behind 2 uterine brothers, mother and 2 full sisters with a total estate worth £120. To solve this problem, the base number is increased from 6 to 7. This procedure is called Increment of Base Number. Therefore,
1 portion of £120 = £120 ÷ 7 = £17.14
‘A’ (2 Uterine brothers): £17.14 × 2 portions = £34.28
‘B’ (Mother): £17.14 × 1 portion = £17.14
‘C’ (2 Full sisters): £17.14 × 4 portions = £68.56
Summation = £34.28 + £17.14 + £68.56 = £119.98; approximately £120
Note that the amount each category of heir finally gets reduces in proportion to its share. For instance, full sisters with the largest share have the highest reduction. Thus,
|Heirs||Original value of portion||New value of portion||Reduction|
Rule H: If the result of summation of portions is greater than the base number, such result becomes the base number. Yet, each heir retains his/her number of portions originally allotted to him/her. Though, their shares reduce proportionately.
|Increased base number||8|
Mother: 6 × 1/3 = 2 portions
Husband: 6 × ½ = 3 portions
Full sister: 6 × ½ = 3 portions
Summation = 2 + 3 + 3 = 8 portions; which is greater than the base number. Hence, the base number is increased to 8.
|Heirs||Wife||2 Consanguine sisters||2 Uterine sisters|
|Portions||3||4 apiece||2 apiece|
|Increased base number||15|
Wife: 12 × ¼ = 3 portions
2 consanguine sisters: 12 × 2/3 = 8 portions; each one gets 4 portions
2 uterine sisters: 12 × 1/3 = 4 portions; each one gets 2 portions
Total = 3 + 8 + 4 = 15 portions; which is greater than the base number. Accordingly, the base number is increased to 15.
|Increased base number||27|
Observe that the father’s share suppose to be “1/6 + residue,” but the base number, 24, is not even enough to share among the heirs, so there will be no question of any residue.
This is a celebrated case of inheritance called MIMBARIYYA for the reason that Caliph ‘Ali solved it while delivering a sermon on the mimbar (i.e. pulpit) in a Mosque at Kufa, in present day Iraq. He was asked what a wife’s share will be if the surviving heirs of a deceased are wife, both parents and 2 daughters. There and then, he answered, “The wife’s 1/8 becomes 1/9.” Let’s examine this.
Using the original base number, wife has 24 × 1/8 = 3 portions
With increment of base number and considering ‘Ali’s response, wife gets 27 × 1/9 = 3 portions
This further buttresses the point that whenever the base number is increased, an heir’s share reduces (in this case from 1/8 to 1/9) but his/her number of portions remains intact.
Increment of base number only applies when ALL categories of heirs have fixed shares. If residuaries are present, increment will not be necessary because they are given whatever remains. Residuaries cannot force those with fixed heirs to reduce their shares to enable them have something. However, in exceptional cases whereby increment has to be done and a residuary is among the heirs, he/she most likely receives nothing. For example,
|Heirs||Husband||Mother||2 Daughters||Full uncle|
|New base number||13|
Notice that even though full uncle is an heir, despite the increment, he still gets nothing.
- Male heirs
- Female heirs
- Non heirs
- Impediments to inheritance
- Exclusion – Part 2
- Exclusion – Part 3
- Note on difference of opinion
- Inheritance of children
- Inheritance of spouses
- Inheritance of parents
- Inheritance of grandparents
- Inheritance of siblings
- Residuaries (‘Asabah)
- Partial exclusion
- Inheritance arithmetic (“inherithmetic”)
- Procedure of solving inheritance problems
- Levels of inheritance problems (Level one)
- Lowest Common Multiple (LCM)
- Level one – continued
- Highest Common Factor (HCF)
- Prime numbers
- YOU ARE HERE: Increment of base number (‘Awl)
- Level two – Part 1
- Level two – Part 2
- Level two – Part 3
- Level two – Part 4
- Level three
- Inheritance of grandfather along with siblings
- Inheritance of grandfather along with siblings in the presence of other heirs
- Special cases
- Summary of rules
- Further reading
- Solutions to exercises
Your Questions, Our Answers
We have received a number of emails from those who visited this website or downloaded and read INHERITANCE IN ISLAM. Almost all of them were questions on either aspects of inheritance not covered in the book or clarifications needed regarding specific cases. Hence, we thought it wise to reproduce the emails so that others may benefit as well. As always, we welcome suggestions, criticisms and of course, more questions!