Q.If P={x:x<3, x∈N}, Q={x:x≤2, x∈W}. Find (P∪Q)×(P∩Q), where W is the set of whole numbers.
Concept understanding — Cartesian Product
Cartesian Product: From Intuition to Definition
Imagine you're ordering a pizza. You have two choices to make: the size (Small, Medium, Large) and the topping (Cheese, Pepperoni, Veggie). How many different pizzas can you order?
You can pair each size with each topping:
- Small + Cheese, Small + Pepperoni, Small + Veggie
- Medium + Cheese, Medium + Pepperoni, Medium + Veggie
- Large + Cheese, Large + Pepperoni, Large + Veggie
That's 3×3=9 possible pizzas. What you just did — systematically pairing every element of one set with every element of another — is the Cartesian product in action.
The Intuition
The Cartesian product is a way to combine two sets to create a new set of ordered pairs. The order matters: (Small, Cheese) is different from (Cheese, Small) — one is a pizza order, the other is nonsense.
Think of it like a multiplication table for sets. If set A has m items and set B has n items, their Cartesian product has m×n items.
The name comes from René Descartes, who used this idea to create the coordinate plane — every point (x,y) on a graph is an element of the Cartesian product of the x-axis and y-axis.
The Precise Definition
Let A and B be two sets. The Cartesian product of A and B, written A×B, is the set of all ordered pairs (a,b) where a is from A and b is from B.
A×B={(a,b)∣a∈A and b∈B}
The vertical bar means "such that." So read it as: "The set of all ordered pairs (a, b) such that a belongs to A and b belongs to B."
Key Properties to Remember
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Order matters: A×B is generally not the same as B×A. For example, if A={1,2} and B={x,y}:
- A×B={(1,x),(1,y),(2,x),(2,y)}
- B×A={(x,1),(x,2),(y,1),(y,2)}
These are different sets because the pairs are ordered differently.
-
Size formula: If ∣A∣=m and ∣B∣=n, then ∣A×B∣=m×n. This holds even if one set is empty — then the product is empty.
-
Empty set: A×∅=∅ and ∅×B=∅. You can't form any pairs if one set has nothing to contribute.
A common mistake: thinking A×B contains all possible combinations of elements from A and B without caring about order. But (a,b) and (b,a) are different pairs unless a=b. Always treat ordered pairs as distinct based on position.
Examples to Cement the Idea
Example 1: A={1,2}, B={3,4}
A×B={(1,3),(1,4),(2,3),(2,4)}
Four pairs, as expected (2×2=4).
Example 2: A={a}, B={1,2,3}
A×B={(a,1),(a,2),(a,3)}
Three pairs — every element of B gets paired with the single element of A.
Example 3: A={0,1}, B={0,1}
A×B={(0,0),(0,1),(1,0),(1,1)}
This is the set of all possible 2-bit binary strings — a foundation for computer science.
Why This Matters
The Cartesian product is the mathematical backbone of:
- Coordinate geometry: Every point (x,y) in the plane is from R×R.
- Database tables: A table's rows are elements of the Cartesian product of its column domains.
- Probability: All possible outcomes of two independent events form a Cartesian product.
- Functions: A function from A to B is a subset of A×B with special properties.
The Cartesian product is not commutative (A×B=B×A in general), but it is associative: (A×B)×C can be thought of as A×B×C, the set of ordered triples. This extends naturally to any number of sets.
Quick Check for Yourself
If A={1,2} and B={2,3}, what is A×B? What is B×A? Are they the same?
Answer: A×B={(1,2),(1,3),(2,2),(2,3)}; B×A={(2,1),(2,2),(3,1),(3,2)}. They share only (2,2) — the rest are different because the order of coordinates is swapped.
The Cartesian product of two sets is introduced at the very start of the NCERT Class 11 Mathematics chapter on Relations and Functions, and "Cartesian product of sets definition and examples" is a commonly searched foundational topic for CBSE board and JEE Main preparation. This concept also underlies coordinate geometry and the formal definition of a function, both of which are frequently tested in "relations and functions important questions".
Concept: Cartesian Product — the set of all ordered pairs where the first element comes from the first set and the second from the second set.
Step 1: List the elements of P and Q.
P={x:x<3, x∈N}={1,2} (natural numbers start at 1).
Q={x:x≤2, x∈W}={0,1,2} (whole numbers include 0).
Step 2: Find P∪Q and P∩Q.
P∪Q={0,1,2}
P∩Q={1,2}
Step 3: Form the Cartesian product.
(P∪Q)×(P∩Q)={0,1,2}×{1,2}
List all ordered pairs: (0,1),(0,2),(1,1),(1,2),(2,1),(2,2)
The set is {(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)}.
(P∪Q)×(P∩Q)={0,1,2}×{1,2}={(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)} — 6 ordered pairs.
List the sets. Using the Indian convention N={1,2,3,…} and W={0,1,2,3,…}:
- P={x:x<3, x∈N}={1,2}
- Q={x:x≤2, x∈W}={0,1,2}
Union and intersection:
P∪Q={0,1,2},P∩Q={1,2}
Cartesian product — first element from {0,1,2}, second from {1,2}:
(P∪Q)×(P∩Q)={(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)}
This has ∣P∪Q∣×∣P∩Q∣=3×2=6 pairs, as expected.
(P∪Q)×(P∩Q)={(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)}.
- TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQQ.If A={x∈R∣x2−8x+15∈R} and B={x∈R∣2x−5x−3<2x−11x−6}, then A∩B= (A) ϕ (B) (25,3]∪[5,211) (C) (25,421) (D) (25,211)
›Reveal solutionSolution
We find the domain of the square root (set A) and solve the rational inequality (set B), then intersect them. The intersection is (25,3]∪[5,211), which matches option (B).
Concept & Intuition
We have two sets defined by conditions on real numbers.
- Set A is simply the domain of the square root: the expression inside must be non-negative.
- Set B is the solution set of a rational inequality. The trick: bring all terms to one side, combine into a single fraction, and use a sign chart. The intersection is the set of numbers that satisfy both conditions. The answer choices are intervals, so we expect a union of intervals.
Step-by-step solution
- Find set A The condition is x2−8x+15∈R, so the radicand must be ≥0:
x2−8x+15≥0.
Factor: (x−3)(x−5)≥0.
The quadratic opens upward, so it is non-negative outside the roots:
x≤3orx≥5.
Hence
A=(−∞,3]∪[5,∞).
- Find set B Solve
2x−5x−3<2x−11x−6.
Bring all to one side:
2x−5x−3−2x−11x−6<0.
Common denominator (2x−5)(2x−11):
(2x−5)(2x−11)(x−3)(2x−11)−(x−6)(2x−5)<0.
Expand numerator:
(2x2−11x−6x+33)−(2x2−5x−12x+30)=(2x2−17x+33)−(2x2−17x+30)=3.
So the inequality reduces to
(2x−5)(2x−11)3<0.
Since 3>0, the sign is determined by the denominator:
(2x−5)(2x−11)<0.
This is a quadratic opening upward, negative between its roots.
Roots: 2x−5=0⇒x=25, and 2x−11=0⇒x=211.
Hence
B=(25,211).
Watch outDo not forget that the original fractions are undefined at x=25 and x=211. These points are excluded from B, which is already an open interval.
- Intersect A and B
A∩B=((−∞,3]∪[5,∞))∩(25,211).
- The part of A in (−∞,3] intersects (25,211) to give (25,3]. (Note: 3 is included because 3∈A and 3<211, so 3∈B as well — the inequality is strict, but 3 is not a problematic point for the rational expression.)
- The part of A in [5,∞) intersects (25,211) to give [5,211). (Note: 5 is included because 5∈A and 5>25, so 5∈B.)
Therefore
A∩B=(25,3]∪[5,211).
TipThe endpoints 25 and 211 are excluded because they make the denominator zero in the inequality. The endpoints 3 and 5 are included because they satisfy both conditions (the square root is defined and the inequality holds at those points).
✓Final answerThe correct option is (B).
ANSWER: B
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