- NCERT Chapter

Question 1

# 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}

(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}

Answer

**(i)** *A* = {1, 2, 3 … 13, 14}

R = {(*x*, *y*): 3*x* − *y* = 0}

**∴ **R = {(1, 3), (2, 6), (3, 9), (4, 12)}

R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.

Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]

Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.

[3(1) − 9 ≠ 0]

Hence, R is neither reflexive, nor symmetric, nor transitive.

**(ii)** R = {(*x*, *y*): *y* = *x* + 5 and *x* < 4} = {(1, 6), (2, 7), (3, 8)}

It is seen that (1, 1) ∉ R.

**∴ **R is not reflexive.

(1, 6) ∈R

But,

(1, 6) ∉ R.

**∴** R is not symmetric.

Now, since there is no pair in R such that (*x*, *y*) and (*y*, *z*) ∈R, then (*x*, *z*) cannot belong to R.

**∴** R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

**(iii)** *A* = {1, 2, 3, 4, 5, 6}

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

We know that any number (*x)* is divisible by itself.

(*x*, *x*) ∈R

**∴** R is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

(4, 2) ∉ R. [as 2 is not divisible by 4]

**∴ **R is not symmetric.

Let (*x*, *y*), (*y*, *z*) ∈ R. Then, *y* is divisible by *x* and *z* is divisible by *y*.

**∴ ***z* is divisible by *x*.

⇒ (*x*, *z*) ∈R

**∴ **R is transitive.

Hence, R is reflexive and transitive but not symmetric.

**(iv)** R = {(*x*, *y*): *x* − *y* is an integer}

Now, for every *x* ∈ **Z**, (*x*, *x*) ∈R as *x* − *x* = 0 is an integer.

**∴ **R is reflexive.

Now, for every *x*, *y* ∈ **Z** if (*x*, *y*) ∈ R, then *x* − *y* is an integer.

⇒ −(*x* − *y*) is also an integer.

⇒ (*y* − *x*) is an integer.

**∴** (*y*, *x*) ∈ R

**∴ **R is symmetric.

Now,

Let (*x*, *y*) and (*y*, *z*) ∈R, where *x*, *y*, *z* ∈ **Z**.

⇒ (*x* − *y*) and (*y* − *z*) are integers.

⇒ *x *− *z* = (*x* − *y*) + (*y* − *z*) is an integer.

**∴** (*x*, *z*) ∈R

**∴** R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

(*x*, *x*) ∈ R

**∴** R is reflexive.

If (*x*, *y*) ∈ R, then *x* and *y* work at the same place.

⇒ *y* and *x* work at the same place.

⇒ (*y*, *x*) ∈ R.

**∴ **R is symmetric.

Now, let (*x*, *y*), (*y*, *z*) ∈ R

⇒ *x* and *y* work at the same place and *y* and *z* work at the same place.

⇒ *x* and *z* work at the same place.

⇒ (*x*, *z*) ∈R

**∴** R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

Clearly (*x*, *x*) ∈ R as *x* and *x* is the same human being.

**∴** R is reflexive.

If (*x*, *y*) ∈R, then *x* and *y* live in the same locality.

⇒ *y* and *x* live in the same locality.

⇒ (*y*, *x*) ∈ R

**∴ **R is symmetric.

Now, let (*x*, *y*) ∈ R and (*y*, *z*) ∈ R.

⇒ *x* and *y* live in the same locality and *y* and *z* live in the same locality.

⇒ *x* and *z* live in the same locality.

⇒ (*x,* *z*) ∈ R

**∴ **R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

Now, (*x*, *x*) ∉ R

Since human being *x *cannot be taller than himself.

**∴ **R is not reflexive.

Now, let (*x*, *y*) ∈R.

⇒ *x* is exactly 7 cm taller than *y*.

Then, *y* is not taller than *x*.

**∴** (*y*, *x*) ∉R

Indeed if *x* is exactly 7 cm taller than *y*, then *y* is exactly 7 cm shorter than *x*.

∴R is not symmetric.

Now,

Let (*x*, *y*), (*y*, *z*) ∈ R.

⇒ *x* is exactly 7 cm taller than* y *and *y* is exactly 7 cm taller than z.

⇒ *x* is exactly 14 cm taller than *z *.

**∴ **(*x*, *z*) ∉R

**∴ **R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

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

Now, (*x*, *x*) ∉ R

Since *x* cannot be the wife of herself.

∴R is not reflexive.

Now, let (*x*, *y*) ∈ R

⇒ *x* is the wife of *y.*

Clearly *y* is not the wife of *x*.

**∴** (*y*, *x*) ∉ R

Indeed if *x* is the wife of *y*, then *y* is the husband of *x*.

**∴ **R is not transitive.

Let (*x*, *y*), (*y*, *z*) ∈ R

⇒ *x* is the wife of *y* and *y* is the wife of *z*.

This case is not possible. Also, this does not imply that *x* is the wife of *z*.

**∴** (*x*, *z*) ∉ R

**∴ **R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

**(e)** R = {(*x*, *y*): *x* is the father of *y*}

Now (*x*, *x*) ∉ R

As *x* cannot be the father of himself.

**∴**R is not reflexive.

Now, let (*x*, *y*) ∈R.

⇒ *x* is the father of *y.*

⇒ *y* cannot be the father of *y.*

Indeed, *y* is the son or the daughter of *y.*

**∴ **(*y*, *x*) ∉ R

**∴** R is not symmetric.

Now, let (*x*, *y*) ∈ R and (*y*, *z*) ∈ R.

⇒ *x* is the father of *y* and *y* is the father of *z*.

⇒ *x* is not the father of *z*.

Indeed *x* is the grandfather of *z*.

∴ (*x*, *z*) ∉ R

**∴**R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

- Q:- 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. - 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:-
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:- 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:- . 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:-
State with reason whether following functions have inverse

(i)

*f*: {1, 2, 3, 4} → {10} with*f*= {(1, 10), (2, 10), (3, 10), (4, 10)}(ii)

*g*: {5, 6, 7, 8} → {1, 2, 3, 4} with*g*= {(5, 4), (6, 3), (7, 4), (8, 2)}(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with

*h*= {(2, 7), (3, 9), (4, 11), (5, 13)} - Q:- Check whether the relation R in R defined as R = {(a, b): a ≤ b
^{3}} is reflexive, symmetric or transitive. - Q:-
Let

*f*: X → Y be an invertible function. Show that*f*has unique inverse.(Hint: suppose g1 and g2 are two inverses of f. Then for all

*y ∈ Y, fog1(y) = 1Y(y) = fog2(y)*. Use one-one ness of*f*). - Q:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
- Q:- , where R
_{+}is the set of all non-negative real numbers.">Consider

*f*: R_{+}→ [4, ∞) given by f(x) = x^{2}+ 4. Show that*f*is invertible with the inverse*f*^{–1}of f given by_{}, where R_{+}is the set of all non-negative real numbers. - Q:- is one-one. Find the inverse of the function
*f*: [–1, 1] → Range*f*.**(Hint: For***y*∈ Range*f*,*y*=, for some*x*in [ - 1, 1], i.e.,)Show that

*f*: [–1, 1] → R, given by is one-one. Find the inverse of the function*f*: [–1, 1] → Range*f*.**(Hint: For***y*∈ Range*f*,*y*=, for some*x*in [ - 1, 1], i.e.,) - Q:- Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation. - Q:- If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
- 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.

If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

Sant Kumar Hooda
2019-05-10 18:42:42

Thanks , Sant Kumar Hooda

2018-09-12 16:36:21

1 - ii is not correct

Manish Kumar
2018-05-01 09:10:13

2 is not correct

pankti naik
2017-09-01 15:03:11

solution of A={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15} S={(x,y)/y=5x ; xâA, yâA} find S is equivalence relation or not?

Hitesh
2017-03-26 20:54:15

Very helpful