Class 12 Maths, Worked in Full Past papersGet updates
CBSE Sample Paper 2020-21 · Part A: Section I · 1 mark

Q3: Is {(a,b):a=b}\{(a, b) : \sqrt a = b\} a function? Or: union and intersection of equivalence classes

The question

(a) A relation R in the set of real numbers R defined as R={(a,b):a=b}R = \{(a, b) : \sqrt{a} = b\} is a function or not. Justify

OR

(b) An equivalence relation R in A divides it into equivalence classes A1,A2,A3A_1, A_2, A_3⁠. What is the value of A1∪A2∪A3A_1 \cup A_2 \cup A_3 and A1∩A2∩A3A_1 \cap A_2 \cap A_3

Exam answer

(a)

Since a\sqrt{a} is not defined (in R) for a∈(−∞,0)a \in (-\infty, 0)⁠, a negative real number aa has no image bb⁠.

Why?

A relation in R is a function only when every real number aa has exactly one image bb⁠.

The square of a real number is never negative, so a\sqrt{a} is a real number only when a≥0a \ge 0⁠. For example, no real bb satisfies b=−4b = \sqrt{-4}⁠, so −4-4 has no image.

∴\therefore a=b\sqrt{a} = b⁠, that is R, is not a function.

(b)

The equivalence classes are disjoint and together they make up the whole set A.

Why?

Every element aa of A lies in its own class, because (a,a)∈R(a, a) \in R (R is reflexive). So the classes together cover all of A.

Two different equivalence classes never share an element. So no element is in all three of A1,A2,A3A_1, A_2, A_3⁠, and their intersection is the empty set ϕ\phi⁠.

Example: on {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}⁠, "a−ba - b is divisible by 3" gives the classes {1,4}\{1, 4\}⁠, {2,5}\{2, 5\}⁠, {3,6}\{3, 6\}⁠.

A1∪A2∪A3=AA_1 \cup A_2 \cup A_3 = A and A1∩A2∩A3=ϕA_1 \cap A_2 \cap A_3 = \phi⁠.

Where this comes from

Chapter 1, Relations and Functions: Functions as relations; equivalence classes.

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →