What Is A Intersection B Union C?
A intersection B union C is the set of elements that belong to set A and also belong to at least one of set B or set C, written $A \cap (B \cup C)$. The union $B \cup C$ collects everything found in B or C or both. The intersection with A then keeps only the members of that union that are also in A.
Read the symbols as words. $\cup$ means "or", so $B \cup C$ holds an element that sits in either set. $\cap$ means "and", so $A \cap (\ldots)$ keeps an element only when it is also in A.
Table: What the two symbols ask for.
Symbol | Reads as | Keeps an element when |
|---|---|---|
$\cup$ (union) | "or" | it is in either set, or in both |
$\cap$ (intersection) | "and" | it is in both sets at once |
The brackets matter as much as the symbols. They tell you which operation runs first, exactly the way brackets work in ordinary arithmetic.
How Do You Solve A Intersection B Union C?
To solve $A \cap (B \cup C)$, work the bracket first, then intersect. The brackets set the order of operations: union inside, intersection outside.
Take three sets:
$$A = {1, 2, 3, 4, 5}, \quad B = {4, 5, 6, 7}, \quad C = {5, 6, 8}$$
Example 1: Find $A \cap (B \cup C)$
Step 1, union inside the brackets:
$$B \cup C = {4, 5, 6, 7} \cup {5, 6, 8} = {4, 5, 6, 7, 8}$$
Step 2, intersect that result with A:
$$A \cap (B \cup C) = {1, 2, 3, 4, 5} \cap {4, 5, 6, 7, 8} = {4, 5}$$
Final answer: $A \cap (B \cup C) = {4, 5}$
Only 4 and 5 survive: each one sits inside A, and each one also appears in B or C. The number 6 is in the union but not in A, so it drops out.
What Is The Distributive Law For A Intersection B Union C?
The distributive law says intersection distributes over union: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. You can intersect A with each set first, then take the union, and the answer does not change.
This mirrors how multiplication distributes over addition in algebra, where $a \times (b + c) = (a \times b) + (a \times c)$. Swap $\times$ for $\cap$ and $+$ for $\cup$, and the shape is identical.
Example 2: Verify the law for $A = {2, 4, 6, 8, 10}$, $B = {1, 2, 3, 4, 5}$, $C = {4, 5, 6, 7}$
Left side, brackets first:
$$B \cup C = {1, 2, 3, 4, 5, 6, 7}$$
$$A \cap (B \cup C) = {2, 4, 6, 8, 10} \cap {1, 2, 3, 4, 5, 6, 7} = {2, 4, 6}$$
Right side, intersect each part, then unite:
$$A \cap B = {2, 4}, \qquad A \cap C = {4, 6}$$
$$(A \cap B) \cup (A \cap C) = {2, 4} \cup {4, 6} = {2, 4, 6}$$
Both sides give ${2, 4, 6}$, so the law holds. For each building-block operation on its own, see intersection of sets and union of sets.
What Does A Intersection B Union C Look Like On A Venn Diagram?
On a three-circle Venn diagram, $A \cap (B \cup C)$ is the part of circle A that overlaps circle B or circle C. Everything inside A that touches either of the other two circles is shaded; the slice of A that touches neither stays blank.
A quick way to read the picture: shade $B \cup C$ lightly first, then darken only the portion that also falls inside A. The dark region is the answer. A fuller tour of these diagrams lives at Venn diagram.
Is A Intersection B Union C The Same As (A Intersection B) Union C?
No. Moving the brackets changes the answer, because it changes which operation runs first. $A \cap (B \cup C)$ keeps everything inside A; $(A \cap B) \cup C$ can reach outside A, since the whole of C is added at the end.
Using $A = {1, 2, 3, 4, 5}$, $B = {4, 5, 6, 7}$, $C = {5, 6, 8}$ from earlier:
Table: The same three sets, two different bracket placements.
Expression | What runs first | Result |
|---|---|---|
$A \cap (B \cup C)$ | union $B \cup C = {4,5,6,7,8}$, then $\cap, A$ | ${4, 5}$ |
$(A \cap B) \cup C$ | intersection $A \cap B = {4,5}$, then $\cup, C$ | ${4, 5, 6, 8}$ |
The two results are different sets. The 6 and 8 appear in the second expression only, because adding all of C at the end pulls in elements that A never contained. Brackets are not decoration; they decide the answer.
Why Does The Distributive Law Work?
The distributive law is not a coincidence of these examples. It follows from what "and" and "or" mean when you check a single element.
Follow one element at a time. Pick any element $x$. Saying $x \in A \cap (B \cup C)$ means "$x$ is in A, and $x$ is in B or C."
Split the "or". That statement is true exactly when "$x$ is in A and in B" is true, or "$x$ is in A and in C" is true. Which is the meaning of $(A \cap B) \cup (A \cap C)$.
Same members, same set. Every element that qualifies on one side qualifies on the other, so the two sets hold precisely the same members.
This is the same logic that runs Boolean algebra, the algebra of true-and-false that sits under every search box and database query. "A and (B or C)" and "(A and B) or (A and C)" pick out the same records, which is why the rule reaches far beyond the classroom.
Who Discovered The Algebra Of Sets?
Long before sets had a formal notation, people reasoned about groups that overlap. The rules for combining them became an algebra in the 1800s, led by the self-taught logician George Boole.
Two more names shaped how these operations are written and pictured:
John Venn (1834–1923, England) gave us the overlapping-circle diagram in 1880, the picture that makes $A \cap (B \cup C)$ easy to see at a glance.
Giuseppe Peano (1858–1932, Italy) introduced the $\cap$ and $\cup$ symbols in 1888, the compact notation used in every set-theory course since.
Where Is A Intersection B Union C Used In The Real World?
The pattern "A, and either B or C" turns up any time a filter narrows one group by a choice between others.
Databases and search: a query for customers in region A who bought product B or product C is exactly $A \cap (B \cup C)$, and the brackets around the "or" are what keep the result correct.
Probability: the chance of event A happening together with B or C depends on the region $A \cap (B \cup C)$ inside the sample space.
Streaming and shopping filters: choosing titles tagged A that are also tagged B or C returns this same set.
Access control: a person in department A who also belongs to team B or team C matches the rule, deciding who can open a shared file.
One small expression describes a database filter, a probability, and a permissions check. The same set logic quietly runs systems that look nothing alike.
What Are The Most Common A Intersection B Union C Mistakes?
Three errors cause most of the lost marks on this topic, and each one is easy to avoid once you name it. They match the misconceptions flagged in order-of-operations guides and set-theory error notes.
Ignoring the brackets and working left to right.
Where it slips in:
A student reads $A \cap (B \cup C)$ from the left and computes $A \cap B$ first, then unions C, landing on $(A \cap B) \cup C$.
Don't do this:
Do not treat the expression as a left-to-right chain. The brackets, not the reading direction, decide the order.
The correct way:
Do the bracket first every time. Build $B \cup C$, then intersect the result with A.
Swapping the meaning of $\cap$ and $\cup$.
Where it slips in:
Under time pressure, a student treats the intersection like a union and collects every element of A, B, and C instead of the overlap.
Don't do this:
Do not read $\cap$ as "or". Intersection is the stricter operation; it keeps only what is shared.
The correct way:
Say the words while you work: $\cup$ is "or" (in either set), $\cap$ is "and" (in both). An element survives $A \cap (B \cup C)$ only when it is in A.
Shading the wrong region on the Venn diagram.
Where it slips in:
A student shades all of B and C, or shades the part of A that misses both circles, instead of the overlap of A with the union.
Don't do this:
Do not shade anything outside circle A. Nothing beyond A can belong to $A \cap (B \cup C)$.
The correct way:
Shade $B \cup C$ first, then keep only the part that also lies inside A. The final shaded piece sits entirely within circle A.
A Real-World Version Of The Bracket Mistake
The bracket slip is a well-known source of bugs in database queries. In SQL, "and" binds tighter than "or", so a filter written as A and B or C is read by the computer as (A and B) or C, not the intended A and (B or C). A support team searching for "priority-A tickets that are open or escalated" can pull in every escalated ticket from every priority when they forget the parentheses. The fix is the same one students learn on paper: put the brackets around the "or" so the union runs first.
Practice Problems On A Intersection B Union C
Use $A = {1, 2, 3, 4, 5, 6}$, $B = {2, 4, 6, 8}$, and $C = {3, 6, 9}$ unless a problem says otherwise. Answers follow each line.
Find $B \cup C$.
(Answer: ${2, 3, 4, 6, 8, 9}$.)Find $A \cap (B \cup C)$.
(Answer: ${2, 3, 4, 6}$.)Verify the distributive law by finding $(A \cap B) \cup (A \cap C)$.
(Answer: $A \cap B = {2, 4, 6}$ and $A \cap C = {3, 6}$, so the union is ${2, 3, 4, 6}$, matching Problem 2.)Find $(A \cap B) \cup C$ and say whether it equals $A \cap (B \cup C)$.
(Answer: ${2, 3, 4, 6, 9}$; not equal, because the brackets change the result.)Let $A = {M, A, T, H}$, $B = {A, E, I}$, and $C = {H, I, T}$. Find $A \cap (B \cup C)$.
(Answer: $B \cup C = {A, E, I, H, T}$, so $A \cap (B \cup C) = {A, H, T}$.)If $A = {1, 2}$, $B = {3, 4}$, and $C = {5, 6}$, find $A \cap (B \cup C)$.
(Answer: the empty set $\varnothing$, since A shares nothing with B or C.)
Where Should You Go Next After A Intersection B Union C?
Set operations open into some of the most useful ideas in algebra and probability, and a few natural doors lead onward.
Operations on sets. See how intersection and union sit alongside difference and complement in one place.
Venn diagram. Practice reading and shading two- and three-set pictures until the regions become automatic.
Sets. Go back to the foundations: what a set is, how membership works, and the language the rest of this topic is built on.
If your child is building these foundations, a live Bhanzu trainer teaches set operations starting from the questions they answer, not just the symbols, in the Bhanzu algebra program.
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