Check whether the relation R in R defined by R = {(a, b) : a ≤ b^3} is reflexive, symmetric or transitive.

Class 12, NCERT Chapter 1, Exercise 1.1, Q5

It can be observed that (1/3, 1/3) ∉ R as 1/3 $\nleq$ 1/3²therefore, R is not reflexive.
Now, (1,3) ϵ R (as 1 < 3³ = 27)
But, (3,1) ∉ R (as 3³ > 1)
Therefore,given relation R is not symmetric.
now if we have (3, 3/2), (3/2, 6/5) ∈ R
as 3 < (3/2)³ and (3/2) < (6/5)³
but (3, 6/5) ∉ R as 3 > (6/5)³
therefore, given relation R is not transitive
Therefore, R is neither reflexive, nor symmetric, nor transitive.

Comments

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1 , L2 ) : L1 is perpendicular to L2 }. Show that R is symmetric but neither reflexive nor transitive.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1, Example 3 Solution Figure R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1 , L1 ) ∉ R. R is symmetric as (L1 , L2 ) ∈ R ⇒ L1 is perpendicular to L2 ⇒ L2 is perpendicular to L1 ⇒ (L2 , L1 ) ∈ R. R is not transitive. Indeed, if L1 is perpendicular to L2 and L2 is perpendicular to L3 , then L1 can never be perpendicular to L3 . In fact, L1 is parallel to L3 , i.e., (L1 , L2 ) ∈ R, (L2 , L3 ) ∈ R but (L1 , L3 ) ∉ R. Example1 Example2 Example3 ←you are here

Show that the function f : N → N, given by f(x) = 2x, is one-one but not onto

Class 12, NCERT Chapter 1, Example8 The function f is one-one, for f ( x 1 ) = f ( x 2 ) ⇒ 2 x 1 = 2 x 2 ⇒ x 1 = x 2 Further, f is not onto, as for 1 ∈ N , there does not exist any x in N , such that f(x) = 2x = 1.

Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2)/(x-3) Is f one-one and onto? Justify your answer.

Class 12, NCERT Chapter 1, Exercise 1.2, Q10 A = R - {3}, B = R - {1} f : A → B is defined as Let x,y∈A such that f(x)=f(y). ⇒ (x-2)/(x-3)=(y-2)/(y-3) ⇒ (x-2)(y-3)=(y-2)(x-3) ⇒ xy-3x-2y+6=xy-3y-2x+6 ⇒ -3x-2y=-3y-2x ⇒ 3x-2x=3y-2y ⇒ x=y ∴ f is one-one. Let y ∈B = R - {1}. Then, y ≠ 1. The function f is onto if there exists x ∈A such that f ( x ) = y. Now, f(x)=y ⇒(x-2)(x-3)=y ⇒x-2=xy-3y ⇒x(1-y)=-3y+2 ⇒ x=2-3y/1-y ∈A Thus, for any y ∈ B, there exists x=2-3y/1-y ∈A such that Hence, function f is one-one and onto.

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