Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b^2} is neither reflexive nor symmetric nor transitive

Class 12, NCERT Chapter 1, Exercise 1.1, Q2

Solution
We have R = {(a,b): $a\leqslant b^2$ }

It can be observed that

(1 4, 1 4) \notin R

Bcoz,

1 4 > (1 4) 2

R is not Reflexive.

Now (1,4)∈R as 1<16

But (4,1)∉R as

4 ≮ 12

Therefore R is not symmetric.

Now (3,2),(2,1.5)∈R [As 3 < 4 and 2 < 2.25 ]

But 3 > 2.25

Therefore ( 3, 1.5 )∉R

Therefore R is not Transitive.

Hence R is neither reflexive nor symmetric nor transitive.

Comments

. Let f : R → R be defined as f(x) = 3x. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto.

Class 12, NCERT Chapter 1, Exercise 1.2, Q12 f : R → R defined as f ( x ) = 3 x . Let x , y ∈ R such that f ( x ) = f ( y ). ⇒ 3 x = 3 y ⇒ x = y ∴ f is one-one. Also, for any real number ( y) in co-domain R , there exists in R such that . ∴ f is onto. Hence, function f is one-one and onto. The correct answer is A.

Let f : N → N be defined by f (n) ={ (n+1)/2, if n is odd and (n-1)/2, if n is even, for all n ∈ N.State whether the function f is bijective. Justify your answer

Class 12, NCERT Chapter 1, Exercise 1.2, Q9 Thus it is bijective.

Show that a one-one function f : {1, 2, 3} → {1, 2, 3} must be onto.

Class 12, NCERT Chapter 1, Example14 Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f. Hence, f has to be onto .

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