Basic Operation On Sets: Examples and History of Sets
1.Basic Operation On Sets: Examples and History of Sets
In this anticle, “Basic Operation On Sets” we will also read history of sets from ancient to modern age.
“Before diving into these solutions,make sure you understand the foundational concept of “Properties of Sets Class 11 with Proof” and have reviewed the essential “Types of Sets Class 11th“. This will make following today’s step-by-step solutions much easier.”
2.Basic Operation On Sets Illustrations
Illustration: 1. Find the union of each of the following pairs of sets:
Illustration: 1 (i) X = \{1, 3, 5\}, Y = \{1, 2, 3\}
Solution: X \cup Y = \{1, 2, 3, 5\}
Illustration: 1 (ii) A = \{a, e, i, o, u\}, B = \{a, b, c\}
Solution:A = \{a, e, i, o, u\}
B = \{a, b, c\}
A \cup B = \{a, b, c, e, i, o, u\}
Illustration: 1 (iii)
A = \{x : x \text{ is a natural number and multiple of } 3\}
B = \{x : x \text{ is a natural number less than } 6\}
Solution: A \cup B = \{x : x = 1, 2, 4, 5 \text{ or a multiple of } 3\}
Illustration: 1 (iv)
A = \{x : x \text{ is a natural number and } 1 < x \le 6\}
B = \{x : x \text{ is a natural number and } 6 < x < 10\}
Solution:
A = \{x : x \text{ is a natural number and } 1 < x \le 6\}
B = \{x : x \text{ is a natural number and } 6 < x < 10\}
Solution: A \cup B = \{x : 1 < x < 10, x \in N\}
Illustration: 1 (v) A = \{1, 2, 3\}, B = \emptyset
Solution: A = \{1, 2, 3\}, B = \emptyset
A \cup B = \{1, 2, 3\}
Illustration: 2. Let A = \{a, b\}, B = \{a, b, c\}. Is A \subset B? What is A \cup B?
Solution: A = \{a, b\}, B = \{a, b, c\}
Yes, A \subset B and A \cup B = \{a, b, c\}
Illustration: 3. If A and B are two sets such that A \subset B, then what is A \cup B?
Solution: A \subset B means every element of A is also an element of B. Therefore A \cup B = B
Illustration: 4. If A = \{1, 2, 3, 4\}, B = \{3, 4, 5, 6\}, C = \{5, 6, 7, 8\} and D = \{7, 8, 9, 10\}; find
Illustration: 4 (i). A = \{1, 2, 3, 4\}, B = \{3, 4, 5, 6\} then A \cup B
Solution: A = \{1, 2, 3, 4\}, B = \{3, 4, 5, 6\}
A \cup B = \{1, 2, 3, 4, 5, 6\}
Illustration: 4 (ii). A = \{1, 2, 3, 4\}, C = \{5, 6, 7, 8\}
A \cup C
Solution: A = \{1, 2, 3, 4\}
C = \{5, 6, 7, 8\}
A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}
Illustration: 4 (iii) B \cup C
Solution: B = \{3, 4, 5, 6\}
C = \{5, 6, 7, 8\}
B \cup C = \{3, 4, 5, 6, 7, 8\}
Illustration: 4 (iv) B \cup D
Solution: B = \{3, 4, 5, 6\}
D = \{7, 8, 9, 10\}
B \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}
Illustration: 4 (v) A \cup B \cup C
Solution: A = \{1, 2, 3, 4\}
B = \{3, 4, 5, 6\}, C = \{5, 6, 7, 8\}
A \cup B = \{1, 2, 3, 4, 5, 6\}
A \cup B \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}
Illustration: 4 (vi) A \cup B \cup D
Solution: A = \{1, 2, 3, 4\}
B = \{3, 4, 5, 6\}, D = \{7, 8, 9, 10\}
A \cup B = \{1, 2, 3, 4, 5, 6\}
A \cup B \cup D = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}
Illustration: 4 (vii) B \cup C \cup D
Solution: D = \{7, 8, 9, 10\}, B = \{3, 4, 5, 6\}
C = \{5, 6, 7, 8\}
B \cup C = \{3, 4, 5, 6, 7, 8\}
B \cup C \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}
Illustration 5. Find the intersection each pair of sets of question 1 above
Solution: (i) X = \{1, 3, 5\}, Y = \{1, 2, 3\}
X \cap Y = \{1, 3\}
(ii) A = \{a, e, i, o, u\}, B = \{a, b, c\}
A \cap B = \{a\}
(iii) A = \{x : x \text{ is a natural number and multiple of } 3\}
B = \{x : x \text{ is a natural number less than } 6\}
A \cap B = \{3\}
(iv) A = \{x : x \text{ is a natural number and } 1 < x \le 6\}
B = \{x : x \text{ is a natural number and } 6 < x < 10\}
Solution: A \cap B = \emptyset
(v) A = \{1, 2, 3\}, B = \emptyset
A \cap B = \emptyset
Illustration: 6. If A = \{3, 5, 7, 9, 11\}, B = \{7, 9, 11, 13\}, C = \{11, 13, 15\} and D = \{15, 17\}, find
Illustration: 6 (i) A \cap B
Solution: A = \{3, 5, 7, 9, 11\}, B = \{7, 9, 11, 13\}
A \cap B = \{7, 9, 11\}
Illustration: 6 (ii) B \cap C
Solution: B = \{7, 9, 11, 13\}, C = \{11, 13, 15\}
B \cap C = \{11, 13\}
Illustration: 6 (iii) A \cap C \cap D
Solution: A = \{3, 5, 7, 9, 11\}, C = \{11, 13, 15\}, D = \{15, 17\}
A \cap C = \{11\}
A \cap C \cap D = \emptyset
Illustration: 6 (iv). A \cap C
Solution: A = \{3, 5, 7, 9, 11\}, C = \{11, 13, 15\}
A \cap C = \{11\}
Illustration: 6 (v) B \cap D
Solution: B = \{7, 9, 11, 13\}, D = \{15, 17\}
B \cap D = \emptyset
Illustration: 6 (vi). A \cap (B \cup C)
Solution: A = \{3, 5, 7, 9, 11\}, B = \{7, 9, 11, 13\}, C = \{11, 13, 15\}
B \cup C = \{7, 9, 11, 13, 15\}
A \cap (B \cup C) = \{7, 9, 11\}
Illustration: 6 (vii) A \cap D
Solution: A = \{3, 5, 7, 9, 11\}, D = \{15, 17\}
A \cap D = \emptyset
Illustration: 6 (viii) A \cap (B \cup D)
Solution: A = \{3, 5, 7, 9, 11\}, B = \{7, 9, 11, 13\}, D = \{15, 17\}
B \cup D = \{7, 9, 11, 13, 15, 17\}
A \cap (B \cup D) = \{7, 9, 11\}
Illustration: 6 (ix) (A \cap B) \cap (B \cup C)
Solution: A = \{3, 5, 7, 9, 11\}, B = \{7, 9, 11, 13\}, C = \{11, 13, 15\}
A \cap B = \{7, 9, 11\}, \quad B \cup C = \{7, 9, 11, 13, 15\}
(A \cap B) \cap (B \cup C) = \{7, 9, 11\}
Illustration: 6 (x). (A \cup D) \cap (B \cup C)
Solution: A = \{3, 5, 7, 9, 11\}, B = \{7, 9, 11, 13\}, C = \{11, 13, 15\}, D = \{15, 17\}
A \cup D = \{3, 5, 7, 9, 11, 15, 17\}
B \cup C = \{7, 9, 11, 13, 15\}
\star (A \cup D) \cap (B \cup C) = \{7, 9, 11, 15\}
Illustration: 7. If A = \{x : x \text{ is a natural number}\}, B = \{x : x \text{ is an even natural number}\}, C = \{x : x \text{ is an odd natural number}\} and D = \{x : x \text{ is a prime number}\}; find
Illustration: 7 (i). A \cap B
Solution: A \cap B = \{x : x \text{ is an even natural number}\}
\Rightarrow A \cap B = B
Illustration: 7 (ii). A \cap C
Solution: A \cap C = \{x : x \text{ is an odd natural number}\}
\Rightarrow A \cap C = C
Illustration: 7 (iii). A \cap D
Solution: A \cap D = \{x : x \text{ is a prime number}\}
\Rightarrow A \cap D = D
Illustration: 7 (iv). B \cap C
Solution: B \cap C = \emptyset
[B \text{ and } C \text{ have no common element}]
Illustration: 7 (v). B \cap D
Solution: B \cap D = \{2\}
Illustration: 7 (vi). C \cap D
Solution: C \cap D = \{x : x \text{ is an odd prime number}\}
\Rightarrow C \cap D = \{3, 5, 7, 11, \dots\}
Illustration: 8. Which of the following pairs of sets are disjoint.
Sol: (i) \{1, 2, 3, 4\} and \{x : x \text{ is a natural number}\} \text{ natural number and } 4 \le x \le 6
(ii) \{a, e, i, o, u\} and \{c, d, e, f\}
(iii) \{x : x \text{ is an even integer}\} and \{x : x \text{ is an odd integer}\}
Solution: Who have no common elements \rightarrow disjoint sets.
Hence (iii) is disjoint sets.
Illustration: 9. If A = \{3, 6, 9, 12, 15, 18, 21\}, B = \{4, 8, 12, 16, 20\}, C = \{2, 4, 6, 8, 10, 12, 14, 16\}, D = \{5, 10, 15, 20\}; find
Illustration: 9 (i) A - B
Solution: A = \{3, 6, 9, 12, 15, 18, 21\}
B = \{4, 8, 12, 16, 20\}
A - B = \{3, 6, 9, 15, 18, 21\}
Illustration: 9 (ii). A - C
Solution: A = \{3, 6, 9, 12, 15, 18, 21\}
C = \{2, 4, 6, 8, 10, 12, 14, 16\}
A - C = \{3, 9, 15, 18, 21\}
Illustration: 9 (iii) A - D
Solution: A = \{3, 6, 9, 12, 15, 18, 21\}
D = \{5, 10, 15, 20\}
A - D = \{3, 6, 9, 12, 18, 21\}
Illustration: 9 (iv) B - A
Solution: B = \{4, 8, 12, 16, 20\}
A = \{3, 6, 9, 12, 15, 18, 21\}
B - A = \{4, 8, 16, 20\}
Illustration: 9 (v). C - A
Solution: C = \{2, 4, 6, 8, 10, 12, 14, 16\}
A = \{3, 6, 9, 12, 15, 18, 21\}
C - A = \{2, 4, 8, 10, 14, 16\}
Illustration: 9 (vi). D - A
Solution: D = \{5, 10, 15, 20\}
A = \{3, 6, 9, 12, 15, 18, 21\}
D - A = \{5, 10, 20\}
Illustration: 9 (vii). B - C
Solution: B = \{4, 8, 12, 16, 20\}
C = \{2, 4, 6, 8, 10, 12, 14, 16\}
B - C = \{20\}
Illustration: 9 (viii) B - D
Solution: B = \{4, 8, 12, 16, 20\}
D = \{5, 10, 15, 20\}
B - D = \{4, 8, 12, 16\}
Illustration: 9 (ix). C - B
Solution: C = \{2, 4, 6, 8, 10, 12, 14, 16\}
B = \{4, 8, 12, 16, 20\}
C - B = \{2, 6, 10, 14\}
Illustration: 9 (x). D - B
Solution: D = \{5, 10, 15, 20\}, B = \{4, 8, 12, 16, 20\}
D - B = \{5, 10, 15\}
Illustration: 9 (xi). C - D
Solution: C = \{2, 4, 6, 8, 10, 12, 14, 16\}
D = \{5, 10, 15, 20\}
C - D = \{2, 4, 6, 8, 10, 12, 14, 16\}
Illustration: 9 (xii). D - C
Solution: D = \{5, 10, 15, 20\}
C = \{2, 4, 6, 8, 10, 12, 14, 16\}
D - C = \{5, 15, 20\}
Illustration: 10. If X = \{a, b, c, d\} and Y = \{f, b, d, g\}, find
Illustration: 10 (i). X - Y
Solution: X = \{a, b, c, d\}, Y = \{f, b, d, g\}
X - Y = \{a, c\}
Illustration: 10 (ii). Y - X
Solution: X = \{a, b, c, d\}, Y = \{f, b, d, g\}
Y - X = \{f, g\}
Illustration: 10 (iii). X \cap Y
Solution: X = \{a, b, c, d\}, Y = \{f, b, d, g\}
X \cap Y = \{b, d\}
Illustration: 11. If R is the set of real numbers and Q is the set of rational numbers, then what is R - Q?
Solution: R - Q = \{\text{set of irrational numbers}\}
Illustration: 12. State whether each of the following statement is true or false. Justify your answer.
Illustration: 12 (i) \{2, 3, 4, 5\} and \{3, 6\} are disjoint sets.
Solution: \{2, 3, 4, 5\} \cap \{3, 6\} = \{3\}
These two sets have common element \{3\}. Hence this is not disjoint sets. Therefore this statement is false.
Illustration: 12 (ii). \{a, e, i, o, u\} and \{a, b, c, d\} are disjoint sets.
Solution: \{a, e, i, o, u\} \cap \{a, b, c, d\} = \{a\}
These sets have common element. Hence this statement is false.
Illustration: 12 (iii) \{2, 6, 10, 14\} and \{3, 7, 11, 15\} are disjoint sets.
Solution: \{2, 6, 10, 14\} \cap \{3, 7, 11, 15\} = \emptyset
These sets have no common element. Hence the statement is true.
Illustration: 12 (iv). \{2, 6, 10\} and \{3, 7, 11\} are disjoint sets.
Solution: \{2, 6, 10\} \cap \{3, 7, 11\} = \emptyset
These sets have no common elements. Hence the statement is true.
3.Practice Problems of Basic Operation On Sets for Students
(1.) If A = {2, 4, 6, 8, 10} and B = {6, 7, 8, 9, 10, 11} then find A \cup B and A \cap B.
(2.) If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 4, 6} then find A'.
By solving the above questions,you can understand the Basic Operation On Sets: Examples and History of Sets well because the concept is well understood when you solve it practically.
Also Read This Article:- Set (mathematics)
4.A Historical Look at the Evolution of Sets Origin & Early Work
(1.)Creator: Modern set theory was largely originated by German mathematician Georg Cantor (1845–1918) through papers published between 1874 and 1897.
(2.)Initial Motivation: Cantor developed set theory while studying trigonometric series.
(3.)Key Finding (1874): He proved that real numbers cannot be put into a one-to-one correspondence with integers.
(4.)Reception & Early Development Support: Famous mathematician Richard Dedekind welcomed Cantor’s work. Criticism: Kronecker criticized Cantor for treating infinite sets the same as finite sets.
(5.)Logic Integration: Gottlob Frege presented set theory as logical principles based on the assumption of a “set of all sets.”
(6.)Russell’s Paradox & Need for Axioms The Contradiction (1902): Bertrand Russell showed that assuming the existence of a “set of all sets” leads to a logical contradiction, creating Russell’s Paradox.
(7.)Core Insight: As Paul R. Halmos noted in Naïve Set Theory, “nothing contains everything.”
(8.)Axiomatization Timeline 1908: Ernst Zermelo published the first axiomatization of set theory to resolve paradoxes. 1922: Abraham Fraenkel proposed further refinements. 1925: John von Neumann explicitly introduced the axiom of regularity. 1937 & 1940: Paul Bernays improved the system, which Kurt Gödel modified in 1940, leading to the Von Neumann–Bernays (VNB) or Gödel–Bernays (GB) set theory.
(9.)Modern Significance Despite early logical paradoxes and restructuring, Cantor’s set theory remains fundamental to modern mathematics, providing the language used to express most mathematical concepts today. On the Basis of above key points,you can know about Basic Operation On Sets: Examples and History of Sets.
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5.BasicFrequently Asked Questions Related to Operation On Sets
Q:1.Describe special properties of Sets
Ans:The sets A-B, and B-A are mutually disjoint sets,i.e.,the intersection of any of these two sets is the null set.
Q:2.What is meant by disjoint sets?
Ans:If A and B are two sets such that A \cap b=\phi ,then A and B are called disjoint sets.
Q:3.How to represent difference set by set builder form?
Ans:If A - B \neq B - A then we can rewrite the definition of difference as A - B = {x : x \in A \text{ and } x \notin B}
By answering the above questions,you can know about the primary terms of Basic Operation On Sets: Examples and History of Sets.
6.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students)
*”निम्नलिखित संख्या श्रृंखला में प्रश्नचिन्ह (?) के स्थान पर क्या आएगा?
70,68,74,68,78,?,68
(What will come in place of the question mark (?) in the following number series?
70,68,74,68,78,?,68)”*
🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*पिछली प्रश्नोत्तरी का हल*
वर्ग की भुजा=वृत्त का व्यास
28 = 2r \Rightarrow r = \frac{28}{2} = 14
वृत्त की त्रिज्या = 14 cm
अतः वृत्त का क्षेत्रफल = \pi r^2 = \frac{22}{7} \times 14 \times 14
= 616\text{ cm}^2
*Previous Quiz Solution*
Side of square = Diameter of the circle
\Rightarrow 28 = 2r \Rightarrow r = \frac{28}{2} = 14 \text{ cm}
Radius of circle latex = 14 cm
Hence the area of the circle = \pi r^2
= \frac{22}{7} \times 14 \times 14 = 616 \text{ cm}^2
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*
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