- NCERT Chapter

Question 10

# Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Answer

**(i)** Let *A* = {5, 6, 7}.

Define a relation R on* A* as R = {(5, 6), (6, 5)}.

Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.

Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.

=> (5, 6), (6, 5) ∈ R, but (5, 5) ∉ R

∴R is not transitive.

Hence, relation R is symmetric but not reflexive or transitive.

**(ii) **Consider a relation R in **R **defined as:

R = {(*a*, *b*): *a* < *b*}

For any *a *∈ R, we have (*a*, *a*) ∉ R since *a* cannot be strictly less than *a* itself. In fact, *a* = *a*.

**∴ **R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2)

But, 2 is not less than 1.

**∴ **(2, 1) ∉ R

**∴ **R is not symmetric.

Now, let (*a*, *b*), (*b*, *c*) ∈ R.

⇒ *a* < *b* and *b* < *c*

⇒ *a* < *c*

⇒ (*a*, *c*) ∈ R

**∴ **R is transitive.

Hence, relation R is transitive but not reflexive and symmetric.

**(iii)** Let *A* = {4, 6, 8}.

Define a relation R on A as:

*A* = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

Relation R is reflexive since for every *a* ∈ *A*, (*a*, *a*) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.

Relation R is symmetric since (*a*, *b*) ∈ R ⇒ (*b*, *a*) ∈ R for all *a*, *b* ∈ R.

Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.

Hence, relation R is reflexive and symmetric but not transitive.

**(iv)** Define a relation R in **R** as:

R = {*a*, *b*): *a*^{3} ≥ *b*^{3}}

Clearly (*a*, *a*) ∈ R as *a*^{3} = *a*^{3}.

**∴ **R is reflexive.

Now,

(2, 1) ∈ R (as 2^{3} ≥ 1^{3})

But,

(1, 2) ∉ R (as 1^{3} < 2^{3})

**∴** R is not symmetric.

Now,

Let (*a*, *b*), (*b*, *c*) ∈ R.

⇒ *a*^{3} ≥ *b*^{3} and *b*^{3} ≥ *c*^{3}

⇒ *a*^{3} ≥ *c*^{3}

⇒ (*a*, *c*) ∈ R

**∴ **R is transitive.

Hence, relation R is reflexive and transitive but not symmetric.

**(v) ** Let *A* = {−5, −6}.

Define a relation R on *A* as:

R = {(−5, −6), (−6, −5), (−5, −5)}

Relation R is not reflexive as (−6, −6) ∉ R.

Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R.

It is seen that (−5, −6), (−6, −5) ∈ R. Also, (−5, −5) ∈ R.

**∴ **The relation R is transitive.

Hence, relation R is symmetric and transitive but not reflexive.

- Q:- Show that each of the relation R in the set A = { x ∈Z: 0≤x≤12}, A={x} given by

(i) R = { (a,b) : |a - b| is a multiple of 4}

(ii) R = {(a,b):a = b} is an equivalence relation.

Find the set of all elements related to 1 in each case. - Q:- Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation R in the set A = {1, 2, 3,13, 14} defined as

R = {(x, y): 3x − y = 0}

(ii) Relation R in the set N of natural numbers defined as

R = {(x, y): y = x + 5 and x < 4}

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as

R = {(x, y): y is divisible by x}

(iv) Relation R in the set Z of all integers defined as

R = {(x, y): x − y is as integer}

(v) Relation R in the set A of human beings in a town at a particular time given by

(a) R = {(x, y): x and y work at the same place}

(b) R = {(x, y): x and y live in the same locality}

(c) R = {(x, y): x is exactly 7 cm taller than y}

(d) R = {(x, y): x is wife of y}

(e) R = {(x, y): x is father of y} - Q:-
Check the injectivity and surjectivity of the following functions:

(i)

*f*:**N → N**given by*f(x*) = x^{2}(ii)

*f*:**Z → Z**given by*f(x)*= x^{2}(iii)

*f*:**R → R**given by*f(x)*= x^{2}(iv)

*f*:**N → N**given by*f(x)*= x^{3}(v)

*f*:**Z → Z**given by*f(x)*= x^{3 } - Q:- . Is f one-one and onto? Justify your answer.

">

Let A = R – {3} and B = R – {1}. Consider the function *f* : A → B defined by

. Is f one-one and onto? Justify your answer.

Prove that the Greatest Integer Function* f* : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Show that the Modulus Function *f* : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.

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.

- Q:-
Show that the Modulus Function

*f*: R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative. - Q:-
Maximise Z = 3

*x*+ 4*y*Subject to the constraints:

*x*+*y*≤ 4,*x*≥ 0,*y*≥ 0 - Q:- Let R be the relation in the set N given by R = {(a, b): a = b − 2, b > 6}. Choose the correct answer.

(A) (2, 4) ∈ R

(B) (3, 8) ∈R

(C) (6, 8) ∈R

(D) (8, 7) ∈ R - Q:- .">
Consider

*f*: R_{+}→ [– 5, ∞) given by*f(x)*= 9x^{2}+ 6x – 5. Show that*f*is invertible

with**.** - Q:- Show that the relation R defined in the set A of all triangles as R = {(T1, T2): T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related?
- Q:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
- Q:-
is neither one-one nor onto

">**.**Show that the Signum Function

*f*: R → R, given byis neither one-one nor onto

**.** - Q:- Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
- Q:-
Prove that the Greatest Integer Function

*f*: R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x. - Q:-
Let A and B be sets. Show that

*f*: A × B → B × A such that*f*(*a, b*) = (*b, a*) is bijective function.

callboy
2019-08-23 02:10:20

in part v set is trans. then (-6,-5) & (-5,-6) both are in relation

angshika
2019-08-21 09:28:45

Thanks for the help

Kajol
2018-12-23 12:01:37

In v. If -6,-6 belongs to R then it will be reflexive (a,a) belongs to R therefore v answer is correct

Sunny
2018-07-15 21:04:10

Try to improve much more

Sachin
2015-04-17 13:33:47

I think, it is correct because (-6,-6) does not belongs to relation set R. Properties of Relation is A realtion R on set A is reflexive if aRa for all a belongs to A i.e. is (a,a) belongs to R for all a belongs to R => each element a of A is related to itself. Ex: Let A = {a,b} and R = {(a,a),(a,b),(b,a)} then R is reflexive as aRa belongs to R but it is not reflexive for pair (b,b) does not belongs to R.

imer
2015-04-16 11:55:15

plz check part v it does not seems correct as -6,-6 doesnot belongs to R