The Cartesian product A × A has 16 elements among which are (0, 2) and (1, 3). Which of the following statements is/are correct? 1. It is possible to determine set A. 2. A × A contains the element (3, 2). Select the correct answer using the code given below:
Both 1 and 2
Understanding the Cartesian Product
The question asks about a Cartesian product A × A. The Cartesian product of two sets A and B, denoted by A × B, is the set of all ordered pairs (a, b) where 'a' is an element of A, and 'b' is an element of B. In this case, the product is A × A, meaning it's the set of all ordered pairs (a, b) where both 'a' and 'b' are elements of the set A.
The number of elements in the Cartesian product A × A is given as 16. The cardinality (number of elements) of a Cartesian product is the product of the cardinalities of the individual sets. So, |A × A| = |A| × |A| = |A|2.
Given that |A × A| = 16, we have |A|2 = 16.
To find the cardinality of set A, we take the square root of 16. Since the cardinality must be a non-negative integer, |A| = $\sqrt{16} = 4$. So, set A has exactly 4 elements.
Analyzing Statement 1: Possibility to Determine Set A
Statement 1 says it is possible to determine set A. We know |A| = 4. We are also given that the elements (0, 2) and (1, 3) are present in A × A.
According to the definition of the Cartesian product A × A, if an ordered pair (x, y) is in A × A, then 'x' must be an element of A, and 'y' must be an element of A.
- Since (0, 2) $\in$ A × A, it means that 0 $\in$ A and 2 $\in$ A.
- Since (1, 3) $\in$ A × A, it means that 1 $\in$ A and 3 $\in$ A.
From these two given elements, we know that the numbers 0, 1, 2, and 3 must all be elements of set A. So, the set {0, 1, 2, 3} is a subset of A ({0, 1, 2, 3} $\subseteq$ A).
We previously determined that the cardinality of A is 4 (|A| = 4). We have found 4 distinct elements (0, 1, 2, and 3) that must be in A. Since A contains exactly 4 elements and we have identified 4 elements that it must contain, these must be the only elements in A.
Therefore, set A must be {0, 1, 2, 3}. We have successfully determined set A.
Statement 1 is correct.
Analyzing Statement 2: A × A Contains Element (3, 2)
Statement 2 asks if A × A contains the element (3, 2). Based on our analysis of Statement 1, we determined that the set A is {0, 1, 2, 3}.
The Cartesian product A × A is formed by taking all possible ordered pairs where the first element comes from A and the second element comes from A.
A × A = {(a, b) | a $\in$ A and b $\in$ A}
Using A = {0, 1, 2, 3}, the elements of A × A are:
| First Element (from A) | Second Element (from A) | Ordered Pair |
|---|---|---|
| 0 | 0, 1, 2, 3 | (0, 0), (0, 1), (0, 2), (0, 3) |
| 1 | 0, 1, 2, 3 | (1, 0), (1, 1), (1, 2), (1, 3) |
| 2 | 0, 1, 2, 3 | (2, 0), (2, 1), (2, 2), (2, 3) |
| 3 | 0, 1, 2, 3 | (3, 0), (3, 1), (3, 2), (3, 3) |
To check if (3, 2) is in A × A, we need to see if the first element, 3, is in A and if the second element, 2, is in A.
From A = {0, 1, 2, 3}, we see that:
- 3 $\in$ A (True)
- 2 $\in$ A (True)
Since both elements 3 and 2 are in set A, the ordered pair (3, 2) is indeed an element of A × A.
Statement 2 is correct.
Conclusion on Statements
Both Statement 1 and Statement 2 are correct.
- Statement 1 is correct because knowing the cardinality of A × A and two specific elements allowed us to deduce the unique set A.
- Statement 2 is correct because once we determined A = {0, 1, 2, 3}, we could confirm that (3, 2) is an element of A × A as both 3 and 2 belong to A.
Revision Table: Key Concepts
| Concept | Explanation | Application in Problem |
|---|---|---|
| Cartesian Product (A × B) | Set of all ordered pairs (a, b) where a $\in$ A, b $\in$ B. | Used to define A × A and its elements. |
| Cardinality (|S|) | Number of elements in a set S. | $|$A × A$|$ = $|$A$|^2$; Used to find $|$A$|$. |
| Elements of A × A | If (x, y) $\in$ A × A, then x $\in$ A and y $\in$ A. | Used to deduce elements of A from (0, 2) and (1, 3). |
| Determining Set A | Knowing $|$A$|$ and identifying all its elements. | $|$A$|$=4 and {0, 1, 2, 3} $\subseteq$ A $\implies$ A = {0, 1, 2, 3}. |
Additional Information: Properties of Cartesian Products
- The order of elements in an ordered pair matters, i.e., (a, b) is generally not equal to (b, a) unless a = b.
- If A or B is an empty set ($\emptyset$), then A × B is also an empty set ($\emptyset$).
- If A and B are finite sets, then $|$A × B$|$ = $|$A$|$ $\times$ $|$B$|$.
- The Cartesian product is not commutative, i.e., A × B $\neq$ B × A unless A = B or A or B is empty. In our case, A × A = A × A is trivially true because it's the same product.
- The Cartesian product can be extended to more than two sets, e.g., A × B × C would be a set of ordered triples (a, b, c).
Important Questions from Sets
Consider the proper subsets of {1, 2, 3, 4}. How many of these proper subsets are a superset of the set {3}?
In a class of $200$ students numbered $1$ to $200$, those whose number is divisible by $2$ opted for Literature, those whose number is divisible by $3$ opted for History, and those whose number is divisible by $7$ opted for Philosophy. Then the number of students who did not opt for any of the three courses is:
Let $f(x) = |x - 2| + |x - 8|$; $x \in R$. Then the set of all values of $x$, at which the function, $g(x) = f(f(x))$ is not differentiable, is:
If A = {x : x is a multiple of 7},
B = {x : x is a multiple of 5} and
C = {x : x is a multiple of 35}
Then which of the following is null set?
Consider the following statements in respect of two non-empty sets A and B :
1. A ∪ B = A ∩ B if A = B
2. A Δ B = ϕ if A = B
Which of the above statements is/are correct ?
