If \[ \left(\frac{x}{3}+1,\,y-\frac{2}{3}\right) = \left(\frac{5}{3},\frac{1}{3}\right) \] find the values of \(x\) and \(y\).
Concept Theory
In coordinate geometry, an ordered pair is written as \[ (x,y) \] where • the first element represents the x-coordinate • the second element represents the y-coordinate. Two ordered pairs are equal **only when their corresponding coordinates are equal**. \[ (a,b)=(c,d) \] This implies \[ a=c \quad \text{and} \quad b=d \] This principle allows us to form equations and solve for unknown variables.
Solution Roadmap
- Use the equality rule of ordered pairs.
- Equate first coordinates to find \(x\).
- Equate second coordinates to find \(y\).
- Solve the resulting linear equations.
Solution
Given \[ \left(\frac{x}{3}+1,\,y-\frac{2}{3}\right) = \left(\frac{5}{3},\frac{1}{3}\right) \] Since the ordered pairs are equal, their corresponding coordinates must be equal.
Equating the first coordinates, \[ \frac{x}{3}+1=\frac{5}{3} \] \[ \frac{x}{3}=\frac{5}{3}-1 \] \[ \frac{x}{3}=\frac{2}{3} \] \[ x=2 \]
Equating the second coordinates, \[ y-\frac{2}{3}=\frac{1}{3} \] \[ y=\frac{1}{3}+\frac{2}{3} \] \[ y=1 \]
Therefore, \[ (x,y)=(2,1) \]
Why This Question is Important
Understanding equality of ordered pairs is essential because it forms the foundation for several advanced topics such as:
- Coordinate Geometry
- Relations and Functions
- Vector coordinates
- Parametric equations
This concept frequently appears in: • CBSE Board Exams • JEE Main • CUET • Olympiad / Scholarship Exams Mastering this idea helps students quickly solve coordinate equality problems and algebraic coordinate equations.
If the set \(A\) has 3 elements and the set \(B=\{3,4,5\}\), find the number of elements in \(A \times B\).
Concept Theory: Cartesian Product
The Cartesian product of two sets \(A\) and \(B\) is the set of all possible ordered pairs where the first element belongs to \(A\) and the second element belongs to \(B\). \[ A \times B = \{(a,b) \mid a \in A, b \in B\} \] If the sets are finite, the number of ordered pairs in the Cartesian product is given by \[ n(A \times B) = n(A)\times n(B) \] This means every element of set \(A\) pairs with every element of set \(B\).
Solution Roadmap
- Determine the number of elements in each set.
- Use the Cartesian product formula.
- Multiply the sizes of the sets.
Solution
We are given that the set \(A\) has \(3\) elements and \[ B=\{3,4,5\} \] Hence, \[ n(A)=3 \] \[ n(B)=3 \]
The number of elements in the Cartesian product of two finite sets is \[ n(A\times B)=n(A)\times n(B) \] Substituting the values, \[ n(A\times B)=3\times3 \] \[ =9 \]
Therefore, the Cartesian product \(A\times B\) contains \[ 9 \text{ ordered pairs} \]
Why This Question is Important
Cartesian products are fundamental in understanding:
- Relations between sets
- Functions and mappings
- Coordinate geometry points
- Discrete mathematics and computer science
Questions involving the size of Cartesian products frequently appear in • CBSE Board Exams • JEE Main • CUET • NDA and Olympiad Exams A clear understanding of this concept helps students quickly compute the number of possible ordered pairs between sets.
If \(G = \{7,8\}\) and \(H = \{5,4,2\}\), find \(G \times H\) and \(H \times G\).
Concept Theory: Cartesian Product of Sets
The Cartesian product of two sets \(A\) and \(B\) is the set of all ordered pairs \((a,b)\) such that \(a\in A\) and \(b\in B\). \[ A\times B=\{(a,b)\mid a\in A,\, b\in B\} \] Important property: \[ A\times B \ne B\times A \] because the order of elements in an ordered pair matters. Thus, \[ (a,b)\ne (b,a) \] unless \(a=b\).
Solution Roadmap
- Write the elements of each set.
- Pair every element of the first set with every element of the second set.
- Construct ordered pairs to form the Cartesian product.
- Repeat the process for the reverse order.
Solution
We are given the sets \[ G=\{7,8\} \] \[ H=\{5,4,2\} \]
The Cartesian product \(G\times H\) consists of all ordered pairs \((g,h)\) such that \(g\in G\) and \(h\in H\). \[ G\times H \] \[ =\{7,8\}\times\{5,4,2\} \] \[ =\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\} \]
Similarly, the Cartesian product \(H\times G\) consists of ordered pairs \((h,g)\) such that \(h\in H\) and \(g\in G\). \[ H\times G \] \[ =\{5,4,2\}\times\{7,8\} \] \[ =\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\} \]
Thus, the two Cartesian products contain the same number of ordered pairs but the order of elements is different.
Why This Question is Important
This problem highlights an important property of Cartesian products: \[ A\times B \ne B\times A \] Understanding this concept is essential for topics such as:
- Relations and Functions
- Coordinate Geometry
- Discrete Mathematics
- Computer Science set operations
Questions involving Cartesian products frequently appear in • CBSE Board Exams • JEE Main • CUET • Olympiad Exams Mastering this concept helps students quickly construct ordered pairs and analyze relations between sets.
State whether each of the following statements are true or false. If the statement
is false, rewrite the given statement correctly.
(i) If \(P = \{m,n\}\) and \(Q = \{n,m\}\), then \(P \times Q = \{(m,n),(n,m)\}\).
(ii) If \(A\) and \(B\) are non-empty sets, then \(A \times B\) is a non-empty set of ordered pairs \((x,y)\) such that \(x\in A\) and \(y\in B\).
(iii) If \(A=\{1,2\}\), \(B=\{3,4\}\), then \(A \times (B\cap \varnothing)=\varnothing\).
Concept Theory: Properties of Cartesian Product
The Cartesian product of two sets \(A\) and \(B\) is defined as \[ A\times B=\{(a,b)\mid a\in A, b\in B\} \] Important properties include:
- The number of elements in \(A\times B\) equals \(n(A)\times n(B)\).
- The order of elements in ordered pairs matters.
- If either set is empty, the Cartesian product is empty.
- If both sets are non-empty, the Cartesian product must also be non-empty.
Solution Roadmap
- Recall the definition of Cartesian product.
- List all possible ordered pairs where required.
- Check whether the given statement matches the definition.
- If incorrect, rewrite the correct statement.
Solution
(i) We are given \[ P=\{m,n\}, \quad Q=\{n,m\} \] The Cartesian product \(P\times Q\) contains all ordered pairs \((x,y)\) such that \(x\in P\) and \(y\in Q\). \[ P\times Q \] \[ =\{(m,n),(m,m),(n,n),(n,m)\} \] The given statement lists only two ordered pairs. Therefore, the statement is false. The correct statement is \[ P\times Q=\{(m,n),(m,m),(n,n),(n,m)\} \]
(ii) If \(A\) and \(B\) are non-empty sets, then there exists at least one element \(x\in A\) and at least one element \(y\in B\). Thus the ordered pair \((x,y)\) belongs to \(A\times B\). Hence, \[ A\times B\ne\varnothing \] Therefore, the given statement is true.
(iii) We are given \[ A=\{1,2\}, \quad B=\{3,4\} \] First find the intersection: \[ B\cap\varnothing=\varnothing \] Now, \[ A\times(B\cap\varnothing)=A\times\varnothing \] The Cartesian product of any set with the empty set is the empty set. \[ A\times\varnothing=\varnothing \] Thus, the given statement is true.
Why This Question is Important
This question tests fundamental logical properties of Cartesian products, which are essential for understanding:
- Relations between sets
- Functions and mappings
- Ordered pair structures
- Discrete mathematics
Such conceptual true/false questions frequently appear in • CBSE Board Exams • JEE Main • CUET • Olympiad Exams A strong understanding of these properties helps students quickly identify incorrect statements in objective questions.
If \(A=\{-1,1\}\), find \(A\times A\times A\).
Concept Theory: Cartesian Product of Three Sets
The Cartesian product of sets forms ordered tuples. For two sets: \[ A\times B=\{(a,b)\mid a\in A, b\in B\} \] For three sets: \[ A\times B\times C=\{(a,b,c)\mid a\in A, b\in B, c\in C\} \] If a set \(A\) has \(n\) elements, then \[ n(A\times A\times A)=n(A)^3 \] because each position in the ordered triple can take any element from the set.
Solution Roadmap
- Write the elements of set \(A\).
- First compute \(A\times A\).
- Then take the Cartesian product of the result with \(A\).
- List all ordered triples.
Solution
We are given \[ A=\{-1,1\} \]
First compute the Cartesian product \(A\times A\): \[ A\times A \] \[ =\{-1,1\}\times\{-1,1\} \] \[ =\{(-1,-1),(-1,1),(1,-1),(1,1)\} \]
Now take the Cartesian product with \(A\): \[ A\times A\times A \] \[ =\{(-1,-1),(-1,1),(1,-1),(1,1)\}\times\{-1,1\} \]
\[ A\times A\times A = \] \[ \{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1), \] \[ (1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)\} \]
Thus, the Cartesian product \(A\times A\times A\) contains \[ 8 \text{ ordered triples} \]
Why This Question is Important
This problem demonstrates how Cartesian products extend to higher dimensions and how ordered triples are formed. Understanding this concept is important for topics such as:
- Relations and Functions
- 3-dimensional coordinate systems
- Binary and ternary relations
- Discrete mathematics and computer science
Questions involving Cartesian products and counting ordered tuples frequently appear in • CBSE Board Exams • JEE Main • CUET • Olympiad Exams A quick method is to remember the formula: \[ n(A\times A\times A)=n(A)^3 \]
If \(A \times B=\{(a,x),(a,y),(b,x),(b,y)\}\), find the sets \(A\) and \(B\).
Concept Theory: Recovering Sets from a Cartesian Product
The Cartesian product of two sets \(A\) and \(B\) is defined as \[ A\times B=\{(a,b)\mid a\in A,\; b\in B\} \] Each ordered pair contains: • the first component from set \(A\) • the second component from set \(B\) Therefore: - the set of all first elements of the ordered pairs forms \(A\) - the set of all second elements forms \(B\)
Solution Roadmap
- Observe the first components of each ordered pair.
- These elements form set \(A\).
- Observe the second components of the ordered pairs.
- These elements form set \(B\).
Solution
We are given \[ A\times B=\{(a,x),(a,y),(b,x),(b,y)\} \]
The first components of the ordered pairs are \[ a,\; b \] Hence, \[ A=\{a,b\} \]
The second components of the ordered pairs are \[ x,\; y \] Therefore, \[ B=\{x,y\} \]
Thus, the required sets are \[ A=\{a,b\}, \quad B=\{x,y\} \]
Why This Question is Important
This question helps students understand how sets can be reconstructed from a given Cartesian product. This skill is useful in topics such as:
- Relations between sets
- Functions and mappings
- Discrete mathematics
- Computer science data structures
Questions of this type often appear in • CBSE Board Exams • JEE Main • CUET • Olympiad exams Students should remember that the first elements of ordered pairs form set \(A\) and the second elements form set \(B\).
Let \(A=\{1,2\}\), \(B=\{1,2,3,4\}\), \(C=\{5,6\}\) and \(D=\{5,6,7,8\}\).
Verify that
(i) \(A\times(B\cap C)=(A\times B)\cap(A\times C)\)
(ii) \(A\times C \subseteq B\times D\)
Concept Theory: Cartesian Product and Set Operations
The Cartesian product of two sets \(A\) and \(B\) is defined as \[ A\times B=\{(a,b)\mid a\in A,\; b\in B\} \] Important properties involving set operations include:
- Cartesian product distributes over intersection.
- If \(A\subseteq B\) and \(C\subseteq D\), then \(A\times C\subseteq B\times D\).
- The Cartesian product forms ordered pairs combining elements from two sets.
Verification Roadmap
- Find the intersection \(B\cap C\).
- Compute \(A\times(B\cap C)\).
- Compute \(A\times B\) and \(A\times C\).
- Find their intersection and compare both sides.
- For part (ii), compute \(A\times C\) and verify whether all ordered pairs belong to \(B\times D\).
Solution
We are given the sets \[ A=\{1,2\} \] \[ B=\{1,2,3,4\} \] \[ C=\{5,6\} \] \[ D=\{5,6,7,8\} \]
(i) First find the intersection of sets \(B\) and \(C\). Since there are no common elements, \[ B\cap C=\varnothing \]
Thus, \[ A\times(B\cap C)=A\times\varnothing=\varnothing \]
Now compute \(A\times B\): \[ A\times B \] \[ =\{(1,1),(1,2),(1,3),(1,4), \] \[ (2,1),(2,2),(2,3),(2,4)\} \]
Next compute \(A\times C\): \[ A\times C \] \[ =\{(1,5),(1,6),(2,5),(2,6)\} \]
There is no ordered pair common in the two sets. Therefore, \[ (A\times B)\cap(A\times C)=\varnothing \]
Hence, \[ A\times(B\cap C)=(A\times B)\cap(A\times C) \] Thus, part (i) is verified.
(ii) First compute \(A\times C\): \[ A\times C \] \[ =\{(1,5),(1,6),(2,5),(2,6)\} \]
Now consider \[ B\times D \] \[ =\{1,2,3,4\}\times\{5,6,7,8\} \]
Each ordered pair of \(A\times C\) has • first element in \(B\) • second element in \(D\)
Therefore, \[ A\times C\subseteq B\times D \] Hence, part (ii) is verified.
Why This Question is Important
This question tests important properties of Cartesian products involving intersection and subset relationships. Understanding these properties is essential for studying:
- Relations and Functions
- Set theory identities
- Discrete mathematics
- Computer science logic structures
Such verification problems frequently appear in • CBSE Board Exams • JEE Main • CUET • Olympiad Exams Students should remember that Cartesian products often preserve subset relationships between sets.
Let \(A=\{1,2\}\) and \(B=\{3,4\}\). Write \(A\times B\). How many subsets will \(A\times B\) have? List them.
Concept Theory
The Cartesian product of two sets \(A\) and \(B\) is defined as \[ A\times B=\{(a,b)\mid a\in A,\; b\in B\} \] If \[ n(A)=m,\quad n(B)=n \] then \[ n(A\times B)=m\times n \] If a set contains \(k\) elements, then the total number of its subsets is \[ 2^k \] This set of all subsets is called the power set.
Solution Roadmap
- Write the Cartesian product \(A\times B\).
- Find the number of elements in \(A\times B\).
- Use the subset formula \(2^n\).
- List all subsets of the Cartesian product.
Solution
We are given \[ A=\{1,2\} \] \[ B=\{3,4\} \]
The Cartesian product is \[ A\times B \] \[ =\{1,2\}\times\{3,4\} \] \[ =\{(1,3),(1,4),(2,3),(2,4)\} \]
Number of elements: \[ n(A)=2,\quad n(B)=2 \] \[ n(A\times B)=2\times2=4 \]
Number of subsets: \[ 2^4=16 \] Hence \(A\times B\) has 16 subsets.
The subsets are \[ \varnothing \] \[ \{(1,3)\},\{(1,4)\},\{(2,3)\},\{(2,4)\} \] \[ \{(1,3),(1,4)\}, \{(1,3),(2,3)\}, \{(1,3),(2,4)\} \] \[ \{(1,4),(2,3)\}, \{(1,4),(2,4)\}, \{(2,3),(2,4)\} \] \[ \{(1,3),(1,4),(2,3)\}, \{(1,3),(1,4),(2,4)\} \] \[ \{(1,3),(2,3),(2,4)\}, \{(1,4),(2,3),(2,4)\} \] \[ \{(1,3),(1,4),(2,3),(2,4)\} \]
Why This Question is Important
This problem combines two important ideas of set theory:
- Cartesian products of sets
- Counting subsets using powers of 2
These concepts frequently appear in • CBSE Board Exams • JEE Main • CUET • Olympiad Exams Students should remember the key formula: \[ \text{Number of subsets of a set with } n \text{ elements } = 2^n \]
Let \(A\) and \(B\) be two sets such that \(n(A)=3\) and \(n(B)=2\). If \((x,1),(y,2),(z,1)\) are in \(A\times B\), find \(A\) and \(B\), where \(x,y,z\) are distinct elements.
Concept Theory: Elements of a Cartesian Product
The Cartesian product of two sets \(A\) and \(B\) is \[ A\times B=\{(a,b)\mid a\in A,\; b\in B\} \] Thus: - the first component of each ordered pair belongs to set \(A\) - the second component belongs to set \(B\) Therefore, by examining ordered pairs we can determine the elements of the original sets.
Solution Roadmap
- Observe the first components of the ordered pairs.
- These elements must belong to set \(A\).
- Observe the second components.
- These elements must belong to set \(B\).
- Use the given sizes \(n(A)=3\) and \(n(B)=2\).
Solution
We are given \[ n(A)=3,\quad n(B)=2 \] and the ordered pairs \[ (x,1),\;(y,2),\;(z,1) \] belong to \(A\times B\).
The first elements of the ordered pairs come from set \(A\). The first components are \[ x,\;y,\;z \] Since these are distinct and \(n(A)=3\), \[ A=\{x,y,z\} \]
The second elements of the ordered pairs come from set \(B\). The second components are \[ 1,\;2 \] Thus, \[ B=\{1,2\} \]
Therefore, the required sets are \[ A=\{x,y,z\},\qquad B=\{1,2\} \]
Why This Question is Important
This question develops the ability to reconstruct sets from a given Cartesian product. Such reasoning is essential for understanding:
- Relations between sets
- Functions and mappings
- Discrete mathematics
- Computer science set structures
Questions involving ordered pairs and Cartesian products frequently appear in • CBSE Board Exams • JEE Main • CUET • Olympiad Exams Students should remember that the first elements determine set \(A\) and the second elements determine set \(B\).
The Cartesian product \(A\times A\) has 9 elements among which are found \((-1,0)\) and \((0,1)\). Find the set \(A\) and the remaining elements of \(A\times A\).
Concept Theory: Cardinality of Cartesian Products
For two finite sets \(A\) and \(B\), \[ n(A\times B)=n(A)\times n(B) \] If both sets are the same, \[ A\times A \] then \[ n(A\times A)=n(A)^2 \] Thus, if we know the number of elements in \(A\times A\), we can determine the number of elements in \(A\). Also, if an ordered pair \((a,b)\) belongs to \(A\times A\), then both \(a\) and \(b\) must be elements of the set \(A\).
Solution Roadmap
- Use the formula \(n(A\times A)=n(A)^2\).
- Determine the number of elements in \(A\).
- Identify elements appearing in the ordered pairs.
- Write the complete Cartesian product \(A\times A\).
Solution
We are given \[ n(A\times A)=9 \] Using the Cartesian product formula, \[ n(A)^2=9 \] \[ n(A)=3 \] Thus, the set \(A\) contains three elements.
We are also told that \[ (-1,0)\in A\times A \] and \[ (0,1)\in A\times A \] This means that the elements \[ -1,\;0,\;1 \] must belong to set \(A\). Since \(A\) has exactly three elements, \[ A=\{-1,0,1\} \]
Now we write the complete Cartesian product: \[ A\times A \] \[ =\{-1,0,1\}\times\{-1,0,1\} \]
\[ A\times A= \] \[ \{(-1,-1),(-1,0),(-1,1), \] \[ (0,-1),(0,0),(0,1), \] \[ (1,-1),(1,0),(1,1)\} \]
Thus, \[ A=\{-1,0,1\} \] and the remaining elements of \(A\times A\) are the ordered pairs listed above.
Why This Question is Important
This question tests two fundamental ideas in set theory:
- Cardinality of Cartesian products
- Determining sets from ordered pairs
Such questions frequently appear in • CBSE Board Exams • JEE Main • CUET • Olympiad Exams Students should remember the important identity \[ |A\times A|=|A|^2 \] which allows quick determination of the number of elements in a set.