Why Neither Tea Nor Coffee Means the Same as Not Tea and Not Coffee

Once you have sets, you can combine them: join them, find what they share, remove one from another, or take everything outside one. These operations follow laws much like the rules of ordinary algebra, and Venn diagrams let you see them at a glance.

Throughout this part:

$$U=\{1,2,\dots ,10\}A=\{2,4,6,8,10\}B=\{3,6,9\}$$

This part covers universal sets and Venn diagrams, union, intersection and difference, complements and De Morgan's laws, and the laws of set algebra.

**The universal set $U$ contains every element under discussion and is drawn as a rectangle; each set is a circle inside it, overlapping where the sets share elements.

Choosing $U$.** For even numbers and multiples of $3$ up to $10$, $U=\{1,2,\dots ,10\}$ is suitable. For sets of triangles, $U$ could be all triangles in a plane.

**Regions for $A$ and $B$:

- In $A$ only**: $\{2,4,8,10\}$
- In both: $\{6\}$
- **In $B$ only**: $\{3,9\}$
- In neither: $\{1,5,7\}$

Worked check. $4+1+2+3=10$ elements, which is $n(U)$.

Other shapes of diagram:

- **$A\subset B$** — circle $A$ drawn inside circle $B$
- Disjoint sets — two circles that do not touch

An everyday example. A school where some students study Hindi, some Sanskrit and some both can show all four groups in one Venn diagram.

The substance. **$U$ must contain every set in the problem**; changing $U$ changes complements, though not unions or intersections.

**The union $A\cup B$ contains elements in either set, the intersection $A\cap B$ contains elements in both, the difference $A-B$ contains elements of $A$ not in $B$, and disjoint sets have an empty intersection.

With $A$ and $B$ above:**

$$A\cup B=\{2,3,4,6,8,9,10\}A\cap B=\{6\}$$
$$A-B=\{2,4,8,10\}B-A=\{3,9\}$$

Counting formula:

$$n(A\cup B)=n(A)+n(B)-n(A\cap B)=5+3-1=7$$

Three sets. With $C=\{1,2,3\}$: $A\cup B\cup C=\{1,2,3,4,6,8,9,10\}$ and $A\cap B\cap C=\varnothing$.

Disjoint sets. The odd numbers and the even numbers in $U$ have no element in common.

An everyday example. **In a class of $40$, $25$ students like cricket, $18$ like football and $8$ like both.**

$$n(C\cup F)=25+18-8=35\text{liking neither}=40-35=5$$

The substance. **$A-B$ and $B-A$ are usually different**, and they are always disjoint from each other.

**The complement $A^{\prime }$ contains every element of $U$ that is not in $A$, and De Morgan's laws state that $(A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$ and $(A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }$.

Complements:**

$$A^{\prime }=\{1,3,5,7,9\}{B}^{\prime }=\{1,2,4,5,7,8,10\}$$

First law.

$$(A\cup B{)}^{\prime }=U-\{2,3,4,6,8,9,10\}=\{1,5,7\}$$
$$A^{\prime }\cap B^{\prime }=\{1,5,7\}$$

Second law.

$$(A\cap B{)}^{\prime }=U-\{6\}=\{1,2,3,4,5,7,8,9,10\}$$
$$A^{\prime }\cup B^{\prime }=\{1,2,3,4,5,7,8,9,10\}$$

Both laws hold.

Useful results: $A\cup A^{\prime }=U$, $A\cap A^{\prime }=\varnothing$, $U^{\prime }=\varnothing$ and ${\varnothing }^{\prime }=U$.

An everyday example. A student who drinks neither tea nor coffee is outside the union of tea drinkers and coffee drinkers — the same as being a non-tea drinker and a non-coffee drinker.

The substance. Taking a complement swaps union and intersection, which is the whole content of De Morgan's laws.

Rewrite expressions step by step using the commutative, associative, distributive, identity and complement laws, justifying each step by name, until both sides match.

The laws:

- Commutative: $A\cup B=B\cup A$, $A\cap B=B\cap A$
- Associative: $(A\cup B)\cup C=A\cup (B\cup C)$, and the same for $\cap$
- Distributive: $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$ and $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$
- Identity: $A\cup \varnothing =A$, $A\cap U=A$
- Double complement: $(A^{\prime }{)}^{\prime }=A$

Worked example 1 — verify distributive law with $C=\{1,2,3\}$:

$$A\cap (B\cup C)=\{2,4,6,8,10\}\cap \{1,2,3,6,9\}=\{2,6\}$$
$$(A\cap B)\cup (A\cap C)=\{6\}\cup \{2\}=\{2,6\}$$

**Worked example 2 — prove $A-B=A\cap B^{\prime }$.** Both sets contain exactly the elements in $A$ and not in $B$; with our sets, $A\cap B^{\prime }=\{2,4,8,10\}=A-B$.

Worked example 3 — simplify.

$$(A\cup B)\cap (A\cup B^{\prime })=A\cup (B\cap B^{\prime })=A\cup \varnothing =A$$

An everyday example. Filters in an online shopping page — size M and colour blue or black — follow the distributive law exactly.

The substance. Unlike numbers, union also distributes over intersection, so both distributive laws hold for sets.