🔢 Sets — Chapter 1

A set is a well-defined collection — we can definitively say whether an object belongs or not. "All natural numbers less than 100" is precise. Options A, C, D are subjective and cannot be determined definitively.

No natural number can simultaneously be less than 5 AND greater than 7. Therefore A = φ (empty set / null set). A set with no elements is called an empty, null, or void set.

We need x² < 40 where x is a positive integer. Check: 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36 (all <40), but 7²=49 > 40. So the set is {1, 2, 3, 4, 5, 6}.

Two sets A and B are equal (A = B) if and only if they have exactly the same elements. The order of listing elements does not matter; {1,2,3} = {3,1,2}. Equal cardinality alone is not sufficient.

For {a, b, c}, n = 3, so subsets = 2³ = 8. They are: φ, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}.

Every element of A (1, 2, 3) is also in B, but B has additional elements (4, 5). Therefore A ⊂ B (A is a proper subset of B). A proper subset means A ⊂ B and A ≠ B.

A ∪ B = set of all elements in A or B (common elements taken once). A ∪ B = {2, 4, 6, 8, 10, 12}. Note: 6 and 8 are common and appear only once.

A ∩ B = set of elements common to both A and B. Since B = {2,3,5,7} and all these are in A, A ∩ B = {2,3,5,7} = B. This confirms that B ⊂ A → A ∩ B = B.

A′ (complement of A) = all elements of U that are not in A. Since A has odd numbers, A′ = {2, 4, 6, 8, 10} (the even numbers in U). Note A ∪ A′ = U and A ∩ A′ = φ.

The complement of the union = intersection of complements. Named after the mathematician De Morgan.

A – B = elements in A but NOT in B. From A, remove 2, 4, 6 (which are in B). So A – B = {1, 3, 5}. Note: B – A = {8} (element in B not in A). A – B ≠ B – A.

Each element has numerator = n and denominator = n+1, with n ranging from 1 to 6. So set-builder form: {x : x = n/(n+1), where n is a natural number and 1 ≤ n ≤ 6}.

A → {1,2} (finite), B → {2} (finite), D → φ (finite). Option C is the set of all prime numbers, which is an infinite set (there are infinitely many primes — proven by Euclid).

If A ∩ B = φ (no common elements), A and B are called disjoint sets. For example, A = {2,4,6,8} and B = {1,3,5,7} are disjoint since they share no elements.

Option A = {2} (2 is the only even prime) — not empty. Option B = {0} — has one element. Option D = {–2, 2}. Option C: 2 is the only even prime, so no even prime > 2 exists → empty set φ.

By convention, the empty set φ is a subset of every set. This is because the condition "every element of φ is in A" is vacuously true (φ has no elements to violate the condition).

These are called Absorption laws. They show that a set "absorbs" operations with its own subsets or supersets.

If B ⊂ A, every element of B is already in A. So combining them gives nothing new: A ∪ B = A. Similarly, if B ⊂ A, then A ∩ B = B.

1 = 1², 4 = 2², 9 = 3², 16 = 4², 25 = 5²… These are perfect squares. In set-builder form: {x : x = n², where n ∈ N} or equivalently {x : x is the square of a natural number}.

Venn diagrams are named after the English logician John Venn (1834–1883). They use rectangles (for universal set) and circles (for subsets) to represent set relationships pictorially.

Verify: A′ = {1,4,5,6}, B′ = {1,2,6}, A′ ∩ B′ = {1,6} ✓ (De Morgan's Law)

MATHEMATICS has letters: M, A, T, H, E, M, A, T, I, C, S. Removing repetitions: {M, A, T, H, E, I, C, S} = 8 distinct elements. In sets, repeated elements are counted only once.

The three sets A – B, A ∩ B, and B – A are mutually disjoint — the intersection of any two of them is the null set φ. Together they form the complete "partition" of A ∪ B.

Solution set in roster form = {1, –2}

This is the set of perfect squares of natural numbers. Since there are infinitely many natural numbers, A is an infinite set.

Both sides equal {1,3,4,5,6,7,8,9}, verifying De Morgan's Second Law.

Two-digit numbers (10–99) where digit sum = 8: 17 (1+7), 26 (2+6), 35 (3+5), 44 (4+4), 53 (5+3), 62 (6+2), 71 (7+1), 80 (8+0).

For a set with 3 elements, number of subsets = 2³ = 8.

The set itself and the empty set are also counted as subsets. Total = 8 subsets.

Length of interval (–5, 7] = 7 – (–5) = 12.

Compare elements (order doesn't matter in sets):

A ≠ C (A has 2, C has 14). F and H cannot be compared without knowing 'a'. So equal pairs: B = D and E = G.

Part 1 (A ∪ B = B): Since A ⊂ B, every element of A is in B. So A ∪ B contains only elements of B → A ∪ B = B.

Part 2 (A ∩ B = A): A ∩ B = elements common to both. Since every element of A is in B (A ⊂ B), all elements of A are common → A ∩ B = A ✓

LHS = RHS = {5, 7, 9} ✓ De Morgan's First Law is verified.