5. Check whether the relation R in R defined as R = {(a, b): a ≤ b3} is reflexive, symmetric or transitive.

We have,R= {(a,b):a≤b3} is a relation in R.For, (a,b)∈R and a=12 we can writea≤a3 => 12≤(12)3 => 12≤18 which is not true.So, R is not reflexive.For (a,b)=(1,2)∈R we have,a≤b3 => 1≤23 => 1≤8 is true.So, (1,2)∈RBut 2≤13 => 2≤1 is not trueSo, (2,1)∉R and (b,a)∉RHence, R is not symmetric.For, (a,b)=(10,4) and (b,c)=(4,2)∈R10≤43 => 10≤64 is true=> (10,4)∈R4≤23 => 4≤8 is true=> (4,2)∈RBut 10≤23 => 10≤8 is not true=> (10,2)∉RHence, for (a,b),(b,c)∈R,(a,c)∉RSo, R is not transitive.

Ask & Answer Question

Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th Relations and Functions

5. Check whether the relation R in R defined as R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.

4 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

We have,

R= ${(a, b) : a \leq b^{3}}$ is a relation in R.

For, $(a, b) \in R$ and $a = \frac{1}{2}$ we can write

$a \leq a^{3}$ => $\frac{1}{2} \leq {(\frac{1}{2})}^{3}$ => $\frac{1}{2} \leq \frac{1}{8}$ which is not true.

So, R is not reflexive.

For $(a, b) = (1, 2) \in R$ we have,

$a \leq b^{3}$ => $1 \leq 2^{3}$ => $1 \leq 8$ is true.

So, $(1, 2) \in R$

But $2 \leq 1^{3}$ => $2 \leq 1$ is not true

So, $(2, 1) \notin R$ and $(b, a) \notin R$

Hence, R is not symmetric.

For, $(a, b) = (10, 4)$ and $(b, c) = (4, 2) \in R$

$10 \leq 4^{3}$ => $10 \leq 64$ is true=> $(10, 4) \in R$

$4 \leq 2^{3}$ => $4 \leq 8$ is true=> $(4, 2) \in R$

But $10 \leq 2^{3}$ => $10 \leq 8$ is not true=> $(10, 2) \notin R$

Hence, for $(a, b), (b, c) \in R, (a, c) \notin R$

So, R is not transitive.

0 Upvotes 0 Downvotes

Similar Questions for you

Let S = {1, 2, 3 ,....,20}

R₁ = {(a, b) : a divide b}

R₂ = {(a, b) : a is integral multiple of b} a, b Î s

n(R₁ – R₂) = ?

alok kumar singh

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R₁) = 66

R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

$f (x) = {\begin{matrix} e^{- x}, & x < 0 \\ l n x, & x > 0 \end{matrix}$

$g (x) = {\begin{matrix} e^{x}, & x < 0 \\ x, & x > 0 \end{matrix}$

The gof : A->R is

alok kumar singh

$g o f (x) = {\begin{matrix} f (x), & f (x) < 0 \\ f (x), & f (x) > 0 \end{matrix}$

$= {\begin{array}{l} e^{l n x} = x & (0, 1) & e^{- x} \\ (- \infty, 0) & l n x & (1, \infty) \end{array}$

If A = {1, 2, 3, … 100}, R = {(x, y) | 2x = 3y, x, y ∈ A} is symmetric relation on A and the number of elements in R is n, the smallest integer value of n is

alok kumar singh

$?$ R is symmetric relation

⇒ (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

∴ Value of n = 0

Let : f(0, R)->∞ be increasing function such that $\underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)} = 1$ then $\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1)$ is equal to

alok kumar singh

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

$\frac{f (x)}{f (x)} < \frac{f (5 x)}{f (x)} < \frac{f (7 x)}{f (x)}$

$\underset{x \to \infty}{l i m} \frac{f (x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)}$

-> $1 < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < 1$ $\underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} = 1$

$\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1) = 0$

Let A = ${1, 2, 3, . . . . ., 10} a n d f : A \to A$ be defined as

$f (k) = {\begin{matrix} k + 1 & i f k i s o d d \\ k & i f k i s e v e n \end{matrix}$

Then the number of possible functions g : A ® A such that gof = f is:

alok kumar singh

Given f (k) = ${\begin{cases} k + 1, k i s o d d \\ k, k i s e v e n \end{cases}$

$? g : A \to A$ such that g (f (x) = f (x)

Case I : If x is even then g (x) = x . (i)

Case II : If x is odd then g (x + 1) = x + 1 . (ii)

From (i) & (ii), g (x) = x, when x is even

So total no. of functions = 10⁵ × 1 = 10⁵

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
682k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

5. Check whether the relation R in R defined as R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question