Questions: Relations and Functions

Q1. Define a relation from set A to set B and state its domain and range.

Answer (English): A relation from A to B is any subset of the Cartesian product A x B, so it is a set of ordered pairs in which the first entry comes from A and the second from B. The domain is the set of all first entries that occur in the relation, and the range is the set of all second entries that occur. For example, A = {1, 2, 3} and B = {4, 5} with the rule b = a + 3 gives the relation {(1, 4), (2, 5)}; its domain is {1, 2} and its range is {4, 5}.

Hinglish: Relation A x B ka subset hota hai, yaani ordered pairs ka set. Domain pehle entries ka set aur range doosri entries ka set hota hai. Example me (1, 4) aur (2, 5) bane, to domain {1, 2} aur range {4, 5}.

Q2. When is a relation on a set called an equivalence relation?

Answer (English): A relation R on a set A is an equivalence relation when it is reflexive, symmetric and transitive together. Reflexive means (a, a) belongs to R for every a in A. Symmetric means (a, b) in R implies (b, a) in R. Transitive means (a, b) in R and (b, c) in R together imply (a, c) in R. Congruence of triangles is the standard example: every triangle is congruent to itself, congruence is mutual, and congruence carries through a chain of two other triangles.

Hinglish: Equivalence relation tab hota hai jab reflexive, symmetric aur transitive teeno ho. Teen definitions alag alag likho, ek hi line me fit karne ki koshish mat karo. Triangles ki congruence iska simple example hai.

Q3. For integers, link two numbers when their difference is a multiple of 2. Prove that this rule gives an equivalence relation.

Answer (English): Write the rule as a R b exactly when 2 divides a - b. Reflexivity: putting b = a gives a - a = 0, and 2 divides 0, so diagonal pairs qualify. Symmetry: if 2 divides a - b, then b - a is the negative of that number, and 2 divides it too. Transitivity: if 2 divides a - b and also b - c, then adding the two differences gives a - c, so 2 divides a - c. All three properties hold. The classes are the even integers and the odd integers.

Hinglish: Rule ko 2 divides (a - b) ki tarah likho. Reflexive me a - a = 0 aur 2 divides 0 hota hai. Symmetric me (a - b) ka negative bhi 2 se divide hota hai. Transitive me do differences ka sum a - c bhi 2 se divide hota hai. Isliye classes even aur odd integers ki banti hain.

Q4. How many relations are possible from a set with 3 elements to a set with 2 elements?

Answer (English): The Cartesian product has 3 x 2 = 6 ordered pairs. Every subset of this product is a relation, so the number of relations is 2 to the power 6, which equals 64. The number of functions between the same sets is 2 to the power 3, which equals 8, and none of those functions is one-one because two inputs cannot have two different outputs in a codomain of size 2.

Hinglish: 3 x 2 = 6 pairs bante hain, aur har subset relation hai, isliye 2^6 = 64 relations. Functions 2^3 = 8 hote hain, aur one-one koi nahi kyunki codomain chhota hai.

Q5. Show that the map f(x) = 2x + 5 on the real line is both injective and surjective, and write its inverse.

Answer (English): Take f(a) = f(b), so 2a + 5 = 2b + 5. Subtract 5 from both sides and divide by 2; the result is a = b, which proves injectivity. Given any real y, choose c = (y - 5)/2, a real number, and its image is 2((y - 5)/2) + 5 = y, which proves surjectivity. The equation y = 2x + 5 rearranges to x = (y - 5)/2, so the inverse map sends y to (y - 5)/2.

Hinglish: Pehle f(a) = f(b) maan kar a = b tak pahuncho, yeh injective proof hai. Phir kisi bhi y ke liye c = (y - 5)/2 chuno, image y aayega, isliye surjective. Inverse ke liye y = 2x + 5 ko x ke liye solve karo; jawab (y - 5)/2 milega.

Q6. Explain the difference between a relation and a function, with one example of each on the same sets.

Answer (English): Both are subsets of a Cartesian product. A relation only links elements of the two sets, and one element of the first set may be linked to several elements of the second set. A function is a relation with the extra condition that each element of the first set is used exactly once. On A = {1, 2, 3} and B = {4, 5, 6}, the set {(1, 4), (1, 5), (2, 5)} is a relation but not a function because 1 has two images. The set {(1, 4), (2, 5), (3, 6)} is a function.

Hinglish: Relation aur function dono A x B ke subset hote hain. Farq sirf itna hai ki function me pehle set ka har element exactly ek baar use hota hai. (1, 4), (1, 5) ek saath ho to relation hai, function nahi.

Q7. State the conditions for a function to be invertible and give the relation between the graphs of f and f⁻¹.

Answer (English): A function f : X → Y is invertible when it is one-one and onto, that is, bijective. Then there is a unique function f⁻¹ : Y → X with f⁻¹(f(x)) = x for every x in X and f(f⁻¹(y)) = y for every y in Y. The graph of f⁻¹ is the reflection of the graph of f in the line y = x, because the inverse swaps the roles of input and output and therefore exchanges the coordinates of every point.

Hinglish: Invertible function one-one aur onto hona chahiye, yaani bijective. Tab f⁻¹ milta hai. Graphs y = x line ke around mirror images hote hain kyunki coordinates swap ho jaate hain.

Q8. For f(x) = x² and g(x) = 2x - 1, work out both orders of composition and state whether they agree.

Answer (English): With f first and g afterwards, the input x gets squared and then doubled and reduced by 1, giving 2x² - 1. With g first and f afterwards, the input becomes 2x - 1 and is then squared, giving (2x - 1)² = 4x² - 4x + 1. At x = 2 the first order gives 7 while the second gives 9, so the two orders do not agree. Composition is not commutative.

Hinglish: Pehle wala order 2x² - 1 deta hai, ulta order (2x - 1)² deta hai. x = 2 par values 7 aur 9 alag aati hain, isliye order badalne se answer badal jaata hai.

Q9. Let A = {1, 2, 3, 4}. Find the number of equivalence relations on A for which 1 is related only to itself.

Answer (English): An equivalence relation with 1 alone in its class means the remaining elements 2, 3, 4 form classes among themselves. Equivalence relations on a three element set correspond to partitions, and the number of partitions of 3 elements is 5: three singleton classes, three ways of choosing a pair plus one singleton, and one class containing all three. So there are 5 equivalence relations on A that keep 1 in a class by itself.

Hinglish: 1 akela class me ho to baaki three elements 2, 3, 4 par partitions ginno. Three elements ke 5 partitions hote hain, isliye answer 5 hai. Partitions aur equivalence relations one to one hote hain.

Q10. On the reals define x R y when x² + y² = 1. Test R for reflexivity, symmetry and transitivity.

Answer (English): The pair (0, 0) fails the test because 0 + 0 is not 1, so R is not reflexive. Swapping x and y leaves the equation unchanged, so R is symmetric. For transitivity, both (1, 0) and (0, 1) satisfy the equation, but (1, 1) gives 1 + 1 = 2, so transitivity breaks down. R is symmetric and neither reflexive nor transitive.

Hinglish: (0, 0) fail hota hai kyunki 0 + 0 = 1 nahi, isliye reflexive nahi. Equation swap se nahi badalti, isliye symmetric haan. (1, 0) aur (0, 1) hain lekin (1, 1) nahi, isliye transitive nahi.

Quiz below: five multiple choice questions on relations, equivalence, one-one and onto maps, composition and inverse functions. Attempt them after revising the formula sheet and the worked examples.