Absorption Law states $A∪(A∩B)=A$ and $A∩(A∪B)=A$.
I can't seem to picture these with Venn diagrams can someone help me out ? Thanks
[Math] absorption laws don’t understand it with Venn diagrams
discrete mathematicselementary-set-theory
Related Solutions
I don't know exactly what you have written, but I would venture to say that anything you "prove" with Venn diagrams probably has an extremely direct translation into set theory, which would certainly be an acceptable form of proof.
The strongest reason to not let you just use a Venn diagram alone is that your teacher probably wants you to verbalize your explanation. This is a key part of mathematics. Drawing a picture can really help illustrate the idea involved, but it does not always explain the connection to the logic you are working within.
There is also a huge drawback to proving things by Venn diagram: your visual preconceptions may fool you into making a mistake. This cannot happen (or happens to a much smaller degree) when you work in the language of set theory.
$A\triangle B$ can be defined either as $(A\cup B)\setminus(A\cap B)$ or as $(A\setminus B)\cup(B\setminus A)$. The two definitions are entirely equivalent, but it’s the latter that explains why $A\triangle B$ is called the symmetric difference of $A$ and $B$: $A\setminus B$ is the difference in one direction, $B\setminus A$ is the difference in the other direction, and taking their union removes any directionality. From an intuitive point of view, however, you might be best off thinking of $A\triangle B$ simply as the set of things belonging to exactly one of $A$ and $B$, just as $A\cap B$ is the set of things belonging to exactly two of $A$ and $B$, and $A\cup B$ is the set of things belonging to at least one of $A$ and $B$.
Now let’s look at $A\triangle(B\cap C)$. $A$ is striped red in the figure below, and $B\cap C$ is solid blue. The points that are in exactly one of those two sets are exactly the points shaded in your picture.
And here they are again, in a modified version of my picture: the remaining blue points are the points that are in $B\cap C$ but not in $A$ (in symbols, in $(B\cap C)\setminus A$), and the remaining red-and-white striped region contains the points that are in $A$ but not in $B\cap C$ (in symbols, in $A\setminus(B\cap C)$). Between the two we have
$$A\triangle(B\cap C)=\underbrace{\Big(A\setminus(B\cap C)\Big)}_{\text{remaining striped region}}\cup\underbrace{\Big((B\cap C)\setminus A\Big)}_{\text{remaining blue region}}\;.$$
Alternatively, we started with the things that were in at least one of the sets $A$ and $B\cap C$ and removed the things that were in both to get the things in exactly one:
$$A\triangle(B\cap C)=\underbrace{\Big(A\cup(B\cap C)\Big)}_{\text{in at least one}}\setminus\underbrace{\Big(A\cap(B\cap C)\Big)}_{\text{in both}}\;.$$
The picture that you already have for $(A\triangle B)\cap(A\triangle C)$ is pretty good. The set $A\triangle B$ is shaded from upper left to lower right (bendwise, if you’re a herald), and the set $A\triangle C$ is shaded from upper right to lower left (bendwise sinister if you’re a herald). The intersection of these two sets consists of those points that are in both sets, so it comprises the regions that are shaded in both directions. In the picture below it’s the blue together with the red-and-white regions.
(The diagrams are a bit crude, but they should help a bit, at least.)
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Best Answer
I think it is best to combine a few diagrams to illustrate. Unfortunately my CAD skills are limited.
The following shows $A \cap B$ and $A$ and the last is the union, which is any point that is in either of the top two sets.
The second equality follows similarly.