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Venn Diagrams- Easiest way to understand sets

Mathematics is the most beautiful creation of the human spirit-Stefan Banach 

Venn diagrams the relations between sets can be illustrated by certain diagrams. Venn Diagrams are very important in all aspects. It provides an understanding of the concept visually.  Under this article understand the concept of union, intersection, the difference of two sets and symmetry of two sets, and compliment of a set. Also, we discuss the algebraic laws on sets and solve a few examples to understand all these concepts. So, are you ready to understand the concept of the Venn diagram?

The relationships between sets can be represented by means of diagrams which are known as Venn diagrams. These diagrams consist of rectangles and closed curves usually circles. The universal set is represented usually by a rectangle and its subsets by circles. 

Operations on Sets
We will now define certain operations on sets and examine their properties. Henceforth, we will refer all our sets as subsets of some universal set. 

1. Union of Sets
Let `A` and `B` be any two sets. The union of `A` and `B` is the set which consists of all the elements of `A` and all the elements of `B`, the common elements being taken only once. The symbol `\cup` is used to denote the union. 
Symbolically, we write `A \cup B` and usually read as `A` union `B`


Example: Let `A={1, 2, 4, 6, 7}` and `B={4, 7, 8}`. Find `A \cup B`. 
Solution: Given `A={1, 2, 4, 6, 7}` and `B={4, 7, 8}`. We have to find `A \cup B`. Thus `A \cup B = {1, 2, 4, 6, 7, 8}`. 
We have to write all the elements in set `A` and `B` but the common elements have been taken only once while writing `A \cup B. `
Thus, we can define the union of two sets as follows:
 The union of two sets `A` and `B` is the set `C` which consists of all those elements which are either in `A` or in `B` (including those which are in both). In symbols, we write `A \cup B = { x : x \in A ; or ; x \in B}`. 

2. Intersection of Sets 
The intersection of two sets `A` and `B` is the set of all those elements which belong to both `A` and `B`. Symbolically, we write `A \cap B = {x : x \in A ; and; x \in B}`. 


Example: Let `A={1, 2, 4, 6, 7}` and `B={4, 7, 8}`. Find `A \cap B`. 
Solution: Given `A={1, 2, 4, 6, 7}` and `B={4, 7, 8}` . We have to find `A \cap B`. Thus `A \cap B = {4, 7}`. We have to write all the common elements in set `A` and `B`. 

If `A` and `B` are two sets such that `A \cap B = \phi`, then `A` and `B` are called disjoint sets. For example, let `A = { 2, 4, 6, 8 }` and `B = { 1, 3, 5, 7 }`. Then `A` and `B` are disjoint sets, because there are no elements which are common to `A` and `B`. The disjoint sets can be represented by means of Venn diagram. 

3. Difference of Sets
The difference of the sets `A` and `B` in this order is the set of elements which belong to `A` but not to `B`. Symbolically, we write `A - B` and read as `"A` minus `B`”. Using the set-builder notation, we can rewrite the definition of difference as `A - B = { x : x \in A ; and; x \not \in B}`. The difference of two sets `A` and `B` can be represented by Venn diagram. 



Example: Let `A = { 1, 2, 3, 4, 5, 6}`, `B = { 2, 4, 6, 8 }`. Find `A - B` and `B - A`. 
Solution: We have, `A - B = { 1, 3, 5 }`, since the elements `1, 3, 5` belongs to `A` but not to `B` and `B - A = { 8 }`, since the element `8` belongs to `B` and not to `A`. We note that `A - B \not = B - A`. Example: Let `P = { a, e, i, o, u }` and `Q = { a, i, k, u}`. Find `P - Q` and `Q - P`. 
Solution: We have, `P- Q = { e, o }`, since the elements `e, o` belong to `P` but not to `Q` and `Q - P = { k }`, since the element `k` belongs to `Q` but not to `P`. We note that `P- Q \not = Q - P`. 

4. Compliment of Sets 
Let `A` be a subset of universal set `U`. Then the complement of `A` is the set of all elements of `U` which are not the elements of `A`. We denote compliment of `A` by `A^c` or `A'`. 
We can write `A'=\{x: x \in U \; \mbox{and}\; x \not \in A\}`


Example: Let `U= {1, 2, 3, \cdots, 10}` and `A={2, 4, 6, 8, 10}. Find `A'`. 
Solution: We note that `1, 3, 5, 7, 9` are the only elements of `U` which do not belong to `A`. Hence `A'={1, 3, 5, 7, 9}`. 

5. Symmetric Difference of Two Sets
If `A` and `B` are any two sets, then the set `(A-B)\cup (B-A)` is called symmetric difference of `A` and `B` and is denoted by `A\Delta B`. We can write `A\Delta B= {x: (x \in A; and; x\not \in B); or; (x\in B ; and; x \not \in A)}` 



Example: Let `A={1, 2, 4}` and `B={1, 2, 3, 5, 6}`. Find `A \Delta B`. 
Solution: Here `A={1, 2, 4}` and `B={1, 2, 3, 5, 6}`. Thus `A-B= {4}` and `B-A={3, 5, 6}`. Thus, `A \Delta B= {3, 4, 5, 6}` 

Some Fundamental Laws of Algebra of Sets
Idempotent Law If `A` is any set, then 
1. `A \cup A = A` 
2. `A\cap A=A` 
Solution: 
1. L.H.S. 
`A \cup A ={x: x \in A\cup A} = {x: x \in A ; or; x \in A}`
                                               ` = {x: x \in A} = A =` R.H.S. 
2. LHS 
`A \cap A ={x: x \in A\cap A} = {x: x \in A; and; x \in A}' 
                                              ` = {x: x \in A} = A =` R.H.S. 
Identity Laws 
If `A` is any sub-set of universal set `U`, then 
1. `A \cup \phi = A`
2. `A\cap U=A` 
Solution: L.H.S.: `A \cup \phi ={x: x \in A\cup \phi } = {x: x \in A \; or ; x \in \phi }` 
                                                                                ` = {x: x \in A} = A = ` R.H.S.
L.H.S.: `A \cap U ={x: x \in A\cap U} = {x: x \in A ; and ; x \in U}`
                                                        ` = {x: x \in A} = A =` R.H.S. 
Commutative Laws
If `A` and `B` are any two sets, then 
1. `A \cup B =B \cup A` 
2. `A\cap B= B \cap A` 
Solution: LHS ` A \cup B ={x: x \in A\cup B } = {x: x \in A ; or; x \in B }`
                                        ` = {x: x \in B ; or; x \in A} = {x: x \in B\cup A}= B \cup A = ` RHS. 
LHS `A \cap B ={x: x \in A\cap B} = {x: x \in A ; and ; x \in B} ` 
                        `= {x: x \in B ; and; x \in A} ={x: x \in A \cap B} = B \cap A = `RHS 
Associative Laws
If `A`, `B` and `C` are any three sets, then 
1. `A \cup (B \cup C) = (A \cup B) \cup C` 
2. `A \cap (B \cap C) = (A \cap B) \cap C` 
Solution: LHS `A \cup (B \cup C) =\{x: x \in A\cup (B \cup C) \} = \{x: x \in A \; or \; x \in (B \cup C) \}`
 ` = \{x: x \in A \; or \; (x \in B \; or \; x \in C)\} ` 
` = \{x: (x \in A \; or \; x \in B) \; or \; x \in C\} `
 `= \{x: x \in (A \cup B)\; or \; x \in C \}= \{x: x \in (A \cup B)\cup C\}= (A \cup B) \cup C = `RHS 
LHS `A \cap (B \cap C) =\{x: x \in A\cap (B \cap C) \} = \{x: x \in A \; and \; x \in (B \cap C) \}`
 ` = \{x: x \in A \; and \; (x \in B  and x \in C)\}` 
` = \{x: (x \in A and x \in B) and x \in C\} ` 
`= \{x: x \in (A \cap B) and x \in C \}= \{x: x \in (A \cap B)\cap C\}= (A \cap B) \cap C =` RHS Distributive Laws
If `A`, `B` and `C` are any three sets, then
1.  `A \cup (B \cap C) = (A \cup B) \cap (A \cup C)` 
2. `A \cap (B \cup C) = (A \cap B) \cup (A \cap C)` 
Solution: LHS ` A \cup (B \cap C) =\{x: x \in A\cup (B \cap C) \} = \{x: x \in A or x \in (B \cap C) \}` `= \{x: x \in A or (x \in B and x \in C)\} ` ` = \{x: (x \in A or x \in B)and (x \in A or x \in C)\} ` 
`= \{x: x \in (A \cup B) and x \in (A \cup C) \}= \{x: x \in (A \cup B)\cap (A \cup C)\}` 
` = (A \cup B) \cap (A \cup C) = `RHS
LHS ` A \cap (B \cup C) =\{x: x \in A\cap (B \cup C) \} = \{x: x \in A \; \mbox{and} \; x \in (B \cup C) \}` `= \{x: x \in A \; \mbox{and} \; (x \in B \; \mbox{or} \; x \in C)\} ` ` = \{x: (x \in A \; \mbox{and} \; x \in B) \; \mbox{or} \; (x \in A \; \mbox{and} \; x \in C)\} ` `= \{x: x \in (A \cap B)\; \mbox{or} \; x \in (A \cap C) \}= \{x: x \in (A \cap B)\cup (A \cap C)\}` ` = (A \cap B) \cup (A \cap C) = ` RHS De-Morgan's Laws If `A` and `B` are any two sub-sets of universal set `U`, then `(A \cup B)' = A' \cap B'` `(A \cap B)'= A' \cup B'` Solution: LHS `(A \cup B)' =\{x: x \in (A\cup B)' } = \{x: x \not \in (A\cup B)\}` `=\{x: x \not \in A \; \mbox{and} \; x \not \in B \}` ` = \{x: x \in A' \; \mbox{and} \; x \in B'\} = \{x: x \in A'\cap B'\}= A' \cap B' = `RHS LHS `(A \cap B)' =\{x: x \in (A\cap B)' \} = \{x: x \not \in (A\cap B)\}` `=\{x: x \not \in A \; \mbox{or} \; x \not \in B \}` ` = \{x: x \in A' \; \mbox{or} \; x \in B'\} = \{x: x \in A'\cup B'\}= A' \cup B' = `RHS Let us discuss a few examples depending on these laws to understand these concepts. Example: Verify the following identities: 1 `A \cup (B \cap C) =(A \cup B) \cap (A \cup C)` 2 `A \cap (B \cup C) =(A \cap B) \cup (A \cap C)` where `A, B, C` are three sets defined by `A={1, 2, 4, 5}; B={2, 3, 5, 6, 7}; C={4, 5, 6, 7, 8}`. Solution: 1. L.H.S. `B \cap C= \{5, 6, 7\}`, now `A \cup (B \cap C)= \{1, 2, 4, 5, 6, 7\}` R.H.S. first find `A \cup B =\{1, 2, 3, 4, 5, 6, 7\}` and `A \cup C=\{1, 2, 4, 5, 6, 7, 8\}`, now, `(A \cup B) \cap (A \cup C)= \{1, 2, 4, 5, 6, 7\}` Therefore L.H.S=R.H.S 2. L.H.S. `B \cup C= \{2, 3, 4, 5, 6, 7, 8\}`, now `A \cap (B \cap C)= \{2, 4, 5\}` R.H.S. first find `A \cap B =\{2, 5\}` and `A \cap C=\{ 4, 5\}`, now, `(A \cap B) \cup (A \cap C)= \{2, 4, 5\}` Therefore L.H.S=R.H.S Exercise: 1. Verify the De-Morgan's Law if $A=\{1, 2, 4, 5, 8\},\; B=\{1, 3, 5, 7\}$ 2. Verify Associative Law if $A=\{1, 2, 4, 5, 8\},\; B=\{1, 3, 5, 7\}, \; C=\{2, 4, 6, 8\}$ 3. If $A=\{1, 2, 4, 5, 8\},\; B=\{1, 3, 5, 7\}, \; C=\{2, 4, 6, 8\}$ and $U=\{1, 2, 3, 4, \cdots, 10\}$, then find $A\cup B$, \; $A-B$, \; $(A-B) \Delta (B-A)$, \; $(A' \cap B')$,\; $(A \cap B)'$ [1]B. S. Grewal, \emph{Higher Engineering Mathematics}, Khanna Publishers, $40^{th}$ Edition. [2] H. Rosen Kenneth, \emph{Discrete Mathematics and its Applications}, McGraw-Hill. [3] J.P. Tremblay, R. Manohar, \emph{Discrete Mathematics Structure with Applications}, McGraw-Hill.

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