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Cardinal number of a set is the range of unique elements or members of a finite set. Essentially, cardinality is used to define the size of a set. n(A), where A is any set and n(A) is the number of individuals in set A, is the cardinal number of that set.
Consider the prime numbers less than ten in set A.
Set A = {2, 3, 5, 7}.
The cardinal number of set A is expressed as n(A) = 4 because it contains four items.
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Key Takeaways: Intersection, Sets, Cardinal Numbers, Disjoint Sets, Venn Diagram, Cardinal Successor, Ordinal Successor
Union of Disjoint Sets
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If A and B are two finite sets, and A ∩ B = ∅, then
n(A ∪ B) = n(A) + n(B)
Simply put, if A and B are finite sets that are disjoint, the cardinal number in union of sets A and B equals the sum of the cardinal numbers of sets A and B.
Disjoint Set
A ∪ B denotes a union of disjoint sets, A and B depicted by the Venn diagram, and it can be observed that;
A ∩ B = ∅, because no element is shared by both sets.
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Union of Two Sets
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If there are two finite sets A and B, then
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Simply said, the sum of the cardinal numbers of the sets A and B, minus the intersection, equals the number of elements in the union of the two sets.
Union of Two Sets
The different coloured portions in the diagram above indicate separate disjoint sets, i.e. A–B, B–A, and A ∩ B, and the sum of them represents A ∪ B. Hence,
n (A ∪ B) = n (A – B) + n(B – A) + n(A ∩ B)
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Union of Three Sets
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If A, B, and C are three finite sets, the follows is truly the case:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
The union of the three sets will be the total of the cardinal numbers of sets A, B, and C, as well as the common elements of the three sets, excluding the common elements of sets taken in pairs of two, as shown in the Venn diagram.
Union of Three Sets
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Successor Cardinal
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Successor operation on cardinal numbers can be defined in the same way as a successor operation on ordinal numbers in set theory. For finite cardinals, the cardinal successor and the ordinal successor are the same, but for infinite cardinals, they deviate because every infinite ordinal and its successor have the same cardinality (a bijection between the two can be set up by simply sending the last element of the successor to 0, 0 to 1, etc., and fixing and all the elements above; in the style of Hilbert's Hotel Infinity).
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Things to Remember
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- If A and B are two finite sets, and A ∩ B = ∅, n(A ∪ B) = n(A) + n(B)
- If there are two finite sets A and B, then, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
- If A, B, and C are three finite sets, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
- The union of the three sets will be the total of the cardinal numbers of sets A, B, and C, as well as the common elements of the three sets, excluding the common elements of sets taken in pairs of two.
- A successor operation on cardinal numbers can be defined in the same way as a successor operation on ordinal numbers in set theory.
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Sample Questions
Ques. If A = { a,b,c,d,e,f,g,h,i} and B = {a,e,i,o,u}, then Find the value n(A ∩ B). (2 Marks)
Ans. Set A = {a,b,c,d,e,f,g,h,i} and Set B = {a,e,i,o,u} are given.
As a result, n(A ∩ B) = a, e, i. (common elements of the sets A and B). As a result, n(A ∩ B) = 3.
Ques. Let P = 1, 2, 3, 5, 7, 11 be the first five even natural numbers, and Q = the first five odd natural numbers. P ∩ Q and n(P ∩ Q) are to be found. (2 Marks)
Ans. Given P = {1, 2, 3, 5, 7, 11} and Q = {2, 4, 6, 8, 10} (first five even natural integers).
As a result, P ∩ Q = {2} (common elements of sets P and Q).
As a result, n(P ∩ Q)= 1 and P ∩ Q = {2}
Ques. If A = {1,3,5,7,9} B = {0,5,10,15}, and U = {0,1,3,5,7,9,10,11,15,20}, then Examine A ∩ B and A ∩ B'. (3 Marks)
Ans. We know that A = {1,3,5,7,9}, B = {0,5,10,15} and U= {0,1,3,5,7,9,10,11,15,20} . Then A ∩ B = {5}
⇒ A ∩ B’ = {0,1,3,7,9,10,11,15,20}
As a result, A ∩ B = {5} and (A ∩ B)’ = {0,1,3,7,9,10,11,15,20}
Ques. What is the cardinal number of the composite integers within 10 and 20 in set A? (2 Marks)
Ans. A = {12,14,15,16,18}; B = {12,14,15,16,18}; C = {12,14,15,16,18};
Set A contains five elements.
n(A) equals 5.
Ques. If C = x | x is not a prime or composite number. Locate n (C). (1 Mark)
Ans. x is neither a prime nor even a composite number, hence C = { x |
C = {1}
∴ n(C) = 1
Ques. If B = {x|, where x is a letter from the word PENINSULA}. Find n (B). (2 Marks)
Ans. In the word PENINSULA, B = {x| x is a letter}.
B = { P, E, N, I, S, U, L, A}
n(B) = 8
Ques. Define the difference between cardinal and ordinal numbers? (3 Marks)
Ans. The numbers we employ to count and describe quantities are known as cardinal numbers. Ordinal numbers, on the other hand, are used to rank or position an object on a list. Ordinal numbers provide solutions to questions such as 'where?' and 'how many?'. Cardinal numbers provide answers to questions such as 'where?' and 'how many?'. The numbers 1 to infinity are examples of cardinal numbers. 1st, 3rd, 7th, 9th, and so on are the ordinal numbers.
Ques. What is a Venn diagram and how do you interpret one? (3 Marks)
Ans. The Venn diagram is a logical depiction of all potential connections between a finite number of distinct sets. The elements are shown as points on a plane, and sets are represented as regions enclosed within circles in Venn diagrams. The dots inside the circle represent elements from that set, whereas the dots outside the circle represent elements from other sets. They're used to appropriately arrange elements and display the components that are exclusive to one set and the elements that are shared by two or more sets.
Ques. When determining the cardinal number of the union of two crossing sets, why do we subtract the shared elements? (3 Marks)
Ans. The common elements are added twice when the separate cardinal numbers of 2 intersecting sets are put together. They come up again and again. As a result, we remove the common elements to ensure that no items are repeated. If this is too complicated, you can calculate the cardinal number of the union of two intersecting sets by adding up the number of unique and unusual items, as well as the number of common elements of the two sets, in each part of the Venn diagram separately. This eliminates the need for common elements to be repeated.
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