Tagged: algebra

Algebra and Successional Legitimes

It’s almost funny how taboo math is in law studies. As much I hate to admit it, I do miss it. Although, what I remember is probably a nostalgic, sanitized memory of doing calculations and solving practical problems (in reality it probably was a lot of raging and frustration and cursing). The thing about those problems is that it had a bottom line – you were either right or wrong, that theorem was proved or you were just screwed. All this ruminating sprung from our past lessons on successional legitimes. Prof Balane asked us to watch out in case legitimate children (LCs) concur with illegitimate children (ICs) since there might be a case where the legitimes of the LCs will be impaired (i.e. they will take less than what they’re entitled to). The question is: how do you know they’re impaired? He said we have to be careful applying the 2:1 proportion in relation to legitimate children and illegitimate children right away unless you have an instinct for finding if the legitimes of legitimate children will be impaired. As can be seen later, instinct is not required.

First of all there are three cases that involve LCs and ICs who survive the decedent.

Case 1: There is an LC and ICs and a surviving spouse (SS), i.e., LC = 1
Case 2: There are LCs and ICs and a SS, where there are more than 1 LCs (LC > 1)
Case 3: There are LCs and ICs and no SS

According the Civil Code the legitimes are calculated as follows:

Case 1: LC = 1/2 of the estate, IC = 1/2 of that of a LC (i.e. 1/4 of the estate divided equally among them), and SS = 1/4
Case 2: LCs = 1/2 divided equally among them, ICs = 1/2 of that of a LC, and SS = same as that of a LC
Case 3: LCs = 1/2 divided equally among them, ICs = 1/2 of that of a LC

Next we note that the only legitimes that can be reduced are those of the ICs since those of the LCs and SS are preferred. Also we have to note that Case 1 does not provide an issue since there is only one variable (the number of ICs), LC and SS are constant at 1. That being said, Case 2 and 3 have two variables (the number of legitimate and illegitimate children).

So how can you tell if the legitimes of the LCs are impaired?

First let’s define the variables:

Let X = the number of legitimate children, and Y = the number of illegitimate children
Let LC = the share of a legitimate child, SS = share of surviving spouse, IC = share of an illegitimate child

Since Case 3 is the simpler one, that will be first.

Solution 1: First, you can calculate how much LC is. Since it is 1/2 divided equally between the legitimate children, LC =1/(2X). Next for IC, since the share of an illegitimate child is 1/2 of a legitimate child, IC = 1/2 of LC, or IC = 1/(4X). Then you multiply IC with Y (Y/4X)to get the total share of the ICs in the pie (the estate). Adding 1/2 (legitime of legitimate children) to that you get the total dispositions. If that exceeds 1 (i.e. exceeds the total estate), IC must be reduced.

This is a bit cumbersome since there are so many steps, each depending on the previous one to be correct. A mistake in the shares of legitimate children will screw up the whole thing. Here’s another solution.

Solution 2: Start with the scenario that a reduction will happen. We can state it in this way: if the total fractional dispositions exceed 1, then there is not enough free portion to cover all the heirs. It can be stated like this:

If (Legitime of legitimate children) + [Y(legitime of an illegitimate child)] > 1, then there is not enough free portion and consequently, legitimes of the illegitimate children have to be reduced. Using the notation above, the inequality can be expressed in this way: IF 1/2 + Y/(4X) > 1, THEN NOT ENOUGH FREE PORTION. Therefore, we merely have to simplify the inequality to determine what combination of X and Y will result to a case where the free portion is not enough.

(1) 1/2 + Y/4X > 1 (X,Y > 0)

(2) Y/4X > 1 – 1/2

(3) Y/4X > 1/2

(4) (4X)(Y/4X) > (4X)(1/2)

(5) Y > 2X

Thus, if Y > 2X then there is not enough free portion and the legitimes of the illegitimate children must be reduced. Otherwise state, if the number of illegitimate children are greater than twice the number of legitimate children, then the 1:2 share will not apply. Reductions must be made.

As regards Case 2, the method is the same except except that since SS = LC, line (1) must be modified in this manner:

1/2 + [1/2X] + Y/4X > 1

simplifying, we arrive at

Y > 2(X – 1)

Therefore, in Case 2 if the number of illegitimate children are greater than twice the number of legitimate children minus two, reductions must be made.

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I’ve only worked with fractions and I don’t really understand why we have to go beyond fractional amounts. Each estate is different anyway. I’ve tried the test many times and all yielded successes. It saves some time calculating legitimes and it’s an easy inequality to memorize along with providing a memory aid to legitimes.