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Relations and Functions Relations and Functions

2. Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x²

(ii) f: Z → Z given by f(x) = x²

(iii) f: R → R given by f(x) = x²

(iv) f: N → N given by f(x) = x³

(v) f: Z → Z given by f(x) = x³

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Vishal Baghel

Contributor-Level 10

(i) $f : N \to N$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = x_{2}^{} \notin N$

So, $f$ is one-one/ injective

For $x \in N$ , i.e., $x = 1, 2, 3 . . . .$

Range of $f (x) = {1_{}^{2}, 2_{}^{2}, 3_{}^{2} . . .} = {1, 4, 9 . . .} \neq N$

i.e., co-domain of $N$

So, $f$ is not onto/ subjective

(ii) $f : Z \to Z$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in Z$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{} \notin Z$

i.e., $x_{1} = x_{2}$ and $x_{1} = - x_{2}$

So, $f$ is not one-one/ injective

For $x \in Z$ , $x = 0, \pm 1, \pm 2, \pm 3 . . . .$

Range of $f (x) = {0_{}^{2}, {(\pm 1)}_{}^{2}, {(\pm 2)}_{}^{2}, {(\pm 3)}_{}^{2} . . .}$

${0, 1, 4, 9 . . . .} \neq$ co-domain $Z$

So, $f$ is not onto/ subjective

(iii) $f : R \to R$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in R$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{}$

So, $f$ is not injective

For $x \in R$

Range of $f (x) = {x_{}^{2}, x \in R}$ gives a set of all positive real numbers

Hence, range of $f (x \neq)$ co-domain of $R$

So, $f$ is not subjective

(iv) $f : N \to N$ given by&n

...more

(i) $f : N \to N$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = x_{2}^{} \notin N$

So, $f$ is one-one/ injective

For $x \in N$ , i.e., $x = 1, 2, 3 . . . .$

Range of $f (x) = {1_{}^{2}, 2_{}^{2}, 3_{}^{2} . . .} = {1, 4, 9 . . .} \neq N$

i.e., co-domain of $N$

So, $f$ is not onto/ subjective

(ii) $f : Z \to Z$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in Z$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{} \notin Z$

i.e., $x_{1} = x_{2}$ and $x_{1} = - x_{2}$

So, $f$ is not one-one/ injective

For $x \in Z$ , $x = 0, \pm 1, \pm 2, \pm 3 . . . .$

Range of $f (x) = {0_{}^{2}, {(\pm 1)}_{}^{2}, {(\pm 2)}_{}^{2}, {(\pm 3)}_{}^{2} . . .}$

${0, 1, 4, 9 . . . .} \neq$ co-domain $Z$

So, $f$ is not onto/ subjective

(iii) $f : R \to R$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in R$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{}$

So, $f$ is not injective

For $x \in R$

Range of $f (x) = {x_{}^{2}, x \in R}$ gives a set of all positive real numbers

Hence, range of $f (x \neq)$ co-domain of $R$

So, $f$ is not subjective

(iv) $f : N \to N$ given by $f (x) = x_{}^{3}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{3} = x_{2}^{3}$

$\Rightarrow x_{1}^{} = x_{2}^{} \in N$

So, $f$ is injective

For $x \in N,$

Range of $f (x) = {1_{}^{3}, 2_{}^{3}, 3_{}^{3} . . . .}$

${1_{}^{}, 8, 2 7_{}^{} . . . .} \neq$ co-domain of $N$

So, $f$ is not subjective

(v) $f : R \to R$ given by $f (x) = x_{}^{3}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{3} = x_{2}^{3}$

$\Rightarrow x_{1}^{} = x_{2}^{} \in N$

So, $f$ is injective

For $x \in R,$

Range of $f (x) = {1_{}^{3}, 2_{}^{3}, 3_{}^{3} . . . .}$

${1_{}^{}, 8, 2 7_{}^{} . . . .} \neq$ co-domain of R

So, $f$ is not subjective

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4 months ago

Relations and Functions Relations and Functions

Relation and Function Exercise 1.2 Solutions

17. Show that the function f: R* → R* defined by f(x) = 1/x is one-one and onto, where: R* is the set of all non-zero real numbers. Is the result true, if the domain: R* is replaced by N with the co-domain being same as: R*?

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Vishal Baghel

Contributor-Level 10

The $f x$ n is $f (x) = \frac{1}{x}$ , which is a $f : R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_{1}, x_{2} \in R_{*}, f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}}$

$\Rightarrow x_{1} = x_{2}$ So, $f$ is one-one

For, $y \in R_{*},$ $x = \frac{1}{f (x)} = \frac{1}{y}$ such that

So, $f (x) = y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f : N \to R_{*}$ such that $f (x) = \frac{1}{x}$

For, $x_{1}, x_{2} \in N,$ $f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}}$

$\Rightarrow x_{1} = x_{2}$ So, $f$ is one-one

For, $y \in R_{*}$ and $f (x) = y$ we have $x = \frac{1}{y} \notin N$

Eg., $3 \in R_{*}$ so $x = \frac{1}{3} \notin N$

So, $f$ is not onto

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4 months ago

Relations and Functions Relations and Functions

16. 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

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Vishal Baghel

Contributor-Level 10

The given relation in set N defined by

$R = {(a, b) : a = b - 2, b > 6}$

For (2,4), 4>6 is not true

For (3,8), 8>6 but 3= 8-2 ⇒3=6 is not true

For (6,8), 8>6 and 6= 8-2 ⇒6=6 is true

And for (8,7), 7>6 but 8= 7-2 ⇒8=5 is not true

Hence, option (C) is correct

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New answer posted

4 months ago

Relations and Functions Relations and Functions

15. 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.

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Vishal Baghel

Contributor-Level 10

The set in $A = {1, 2, 3, 4}$

The relation in this set $A$ is given by

$R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}$

$R$ is reflexive as $(1, 1), (2, 2), (3, 3), (4, 4) \in R$

As, $(1, 2) \in R$ but $(2, 1) \notin R$

$R$ is not symmetric

For $(1, 2) \in R$ and $(2, 2) \in R; (1, 2) \in R$

And for $(1, 3) \in R$ and $(3, 2) \in R; (1, 3) \in R$

∴ $R$ is transitive

Hence, option (B) is correct

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4 months ago

Relations and Functions Relations and Functions

14. 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.

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Vishal Baghel

Contributor-Level 10

The given relation in the set $L =$ all lines in $X Y -$ plane is defined as

$R = {(L_{1}, L_{2}) : L_{1}$ is parallel to $L_{2}}$

Let $L_{1} \in A$ then as $L_{1}$ is parallel to $L_{1}$ ,

$(L_{1}, L_{1}) \in R$

So, $R$ is reflexive

Let $L_{1}, L_{2} \in A$ and $(L_{1}, L_{1}) \in R$

Then, $L_{1}$ is parallel to $L_{2}$

$L_{2}$ is parallel to $L_{1}$

So, $(L_{2}, L_{1}) \in R$

i.e., $R$ is symmetric

Let $L_{1}, L_{2}, L_{3} \in A$ and $(L_{1}, L_{2})$ and $(L_{2}, L_{3}) \in R$

Then, $L_{1} ? L_{2}$ and $L_{2} ? L_{3}$

So, $L_{1} ? L_{3}$

i.e., $(L_{1}, L_{2}) \in R$

So, $R$ is transitive

Hence, $R$ is an equivalence relation

The set of lines related to $y = 2 x + 4$ is given by the equation $y = 2 x + C$ where $C$ is some constant.

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Relations and Functions Relations and Functions

13. Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂): P_{1 and} P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right-angle triangle T with sides 3, 4 and 5?

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Vishal Baghel

Contributor-Level 10

The given relation in set A of all polygons is defined as

R= ${(P_{1}, P_{2}) : P_{1}$ and $P_{2}$ have same number of sides $}$

Let $P_{1} \in A$ ,

As number of sides $(P_{1})$ = number of sides $(P_{1})$

$(P_{1}, P_{1}) \in R$

So, R is reflexive.

Let $P_{1}, P_{2} \in A$ and $(P_{1}, P_{2}) \in R$

Then, number of sides of $P_{1}$ = number of sides of $P_{2}$

Number of sides of $P_{2}$ = number of sides of $P_{1}$

i.e., $(P_{2}, P_{1}) \in R$

so, R is symmetric.

Let $P_{1}, P_{2}, P_{3} \in A$ and $(P_{1}, P_{2})$ and $(P_{2}, P_{3}) \in R$

Then, number of sides $(P_{1})$ = number of sides $(P_{2})$

Number of sides $(P_{2})$ = number of sides $(P_{3})$

So, number of sides $(P_{1})$ = number of sides $(P_{3})$

I.e., $(P_{1}, P_{3}) \in R$

So, R is transitive.

Hence, R is an equivalence relation.

...more

The given relation in set A of all polygons is defined as

R= ${(P_{1}, P_{2}) : P_{1}$ and $P_{2}$ have same number of sides $}$

Let $P_{1} \in A$ ,

As number of sides $(P_{1})$ = number of sides $(P_{1})$

$(P_{1}, P_{1}) \in R$

So, R is reflexive.

Let $P_{1}, P_{2} \in A$ and $(P_{1}, P_{2}) \in R$

Then, number of sides of $P_{1}$ = number of sides of $P_{2}$

Number of sides of $P_{2}$ = number of sides of $P_{1}$

i.e., $(P_{2}, P_{1}) \in R$

so, R is symmetric.

Let $P_{1}, P_{2}, P_{3} \in A$ and $(P_{1}, P_{2})$ and $(P_{2}, P_{3}) \in R$

Then, number of sides $(P_{1})$ = number of sides $(P_{2})$

Number of sides $(P_{2})$ = number of sides $(P_{3})$

So, number of sides $(P_{1})$ = number of sides $(P_{3})$

I.e., $(P_{1}, P_{3}) \in R$

So, R is transitive.

Hence, R is an equivalence relation.

The set of all triangles will be the elements in A related to the right-angle triangle T with sides 3,4 and 5 as they all have three number of sides.

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Relations and Functions Relations and Functions

12. 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?

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Vishal Baghel

Contributor-Level 10

The given relation to set A of all triangles is defined as

R= ${(T_{1}, T_{2}) : T_{1}$ is similar to $T_{2}}$

For $T_{1} \in A$ ,

$T_{1}$ is always similar to $T_{1}$

So, $(T_{1}, T_{1}) \in R$ . Hence R is reflexive.

For $T_{1,} T_{2} \in A$ and $(T_{1}, T_{2}) \in R$ we have

$T_{1} \sim T_{2} (s i m i l a r)$

$T_{2} \sim T_{1}$ i.e., $(T_{2}, T_{1}) \in R$

so, R is symmetric.

for, $T_{1}, T_{2}, T_{3} \in A$ and $(T_{1}, T_{2}) \in R$ and $(T_{2}, T_{3}) \in R$

$T_{1} \sim T_{2}$ and $T_{2} \sim T_{3}$

i.e., $T_{1} \sim T_{3}$ $\to (T_{1}, T_{3}) \in R$

so, R is transitive

$∴$ R is an equivalence relation.

Given, sides of $T_{1}$ are 3,4,5

Sides of $T_{2}$ are 5,12,13

Sides of $T_{3}$ are 6,8,10

As $\frac{3}{5} \neq \frac{4}{12} \neq \frac{5}{13}$ we conclude that $T_{1}$ is not similar to $T_{2}$

As $\frac{5}{6} \neq \frac{12}{8} \neq \frac{13}{10}$ we conclude that $T_{2}$ is not similar to $T_{3}$

But as $\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$ we conclude that

...more

The given relation to set A of all triangles is defined as

R= ${(T_{1}, T_{2}) : T_{1}$ is similar to $T_{2}}$

For $T_{1} \in A$ ,

$T_{1}$ is always similar to $T_{1}$

So, $(T_{1}, T_{1}) \in R$ . Hence R is reflexive.

For $T_{1,} T_{2} \in A$ and $(T_{1}, T_{2}) \in R$ we have

$T_{1} \sim T_{2} (s i m i l a r)$

$T_{2} \sim T_{1}$ i.e., $(T_{2}, T_{1}) \in R$

so, R is symmetric.

for, $T_{1}, T_{2}, T_{3} \in A$ and $(T_{1}, T_{2}) \in R$ and $(T_{2}, T_{3}) \in R$

$T_{1} \sim T_{2}$ and $T_{2} \sim T_{3}$

i.e., $T_{1} \sim T_{3}$ $\to (T_{1}, T_{3}) \in R$

so, R is transitive

$∴$ R is an equivalence relation.

Given, sides of $T_{1}$ are 3,4,5

Sides of $T_{2}$ are 5,12,13

Sides of $T_{3}$ are 6,8,10

As $\frac{3}{5} \neq \frac{4}{12} \neq \frac{5}{13}$ we conclude that $T_{1}$ is not similar to $T_{2}$

As $\frac{5}{6} \neq \frac{12}{8} \neq \frac{13}{10}$ we conclude that $T_{2}$ is not similar to $T_{3}$

But as $\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$ we conclude that $T_{1} \sim T_{3}$

Hence, $T_{1}$ is related to $T_{3}$

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Relations and Functions Relations and Functions

11. Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

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Vishal Baghel

Contributor-Level 10

The given relation in set A of points in a plane is

R= ${(P, Q) :$ distance of point P from origin=distance of point Q from $o r i g i n}$

If O is the point of origin

R= ${(P, Q) : P O = Q O}$

Then, for $P \in A$ we have PO=PO

So, $(P, P) \in R$

i.e., P is reflexive

for, $P, Q \in A$ and $(P, Q) \in R$ we have

PO=QO

QO=PO i.e., $(Q, P) \in R$

i.e., R is symmetric

for $P, Q, S \in A$ and $(P, Q) & (Q, S) \in R$

PO=QO and QO=SO

PO=SO

i.e., $(P, S) \in R$

so, R is transitive

Hence, R is an equivalence relation

For a point $P \neq (o, o)$ the set of all points related to P i.e., distance from origin to the points are equal is a circle with center at origin (o, o) by the definition of circle

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New answer posted

4 months ago

Relations and Functions Relations and Functions

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.

0 Follower 19 Views

Vishal Baghel

Contributor-Level 10

Let A= ${a, b, c}$

(i) R= ${(a, b), (b, a)}$ is a relation in set A

So, $(a, b) \in R$ and $(b, a) \in R \to$ Symmetric

$(a, a) \notin R \to$ not reflexive

$(a, b) \in R, (b, a) \in R$ but $(a, a) \notin R$ $\to$ not transitive

(ii) R= ${(a, b), (b, c), (a, c)}$ is a relation in set A

So, $(a, a) \notin R \to$ not reflexive

$(a, b) \in R$ but $(b, a) \notin R \to$ not symmetric

$(a, b) \in R & (b, c) \in R$ and also $(a, c) \in R \to$ transitive

(iii) R= ${(a, a), (b, b), (c, c), (a, b), (b, a), (a, c), (c, a)}$

So, $(a, a), (b, b), (c, c) \in R \to$ Reflexive

$(a, b) \in R \to (b, a) \in R \to$ Symmetric

$(a, c) \in R \to (c, a) \in R$

$(b, a) \in R$ and $(a, c) \in R$

But $(b, c) \notin R \to$ not transitive

(iv) R= ${(a, a), (b, b), (c, c), (a, b), (b, c), (a, c)}$ is s relation in set A

So, $(a, a), (b, b), (c, c) \in R \to$ reflexive

$(a, b) & (b, c) \in R$ so, $(a, c) \in R \to$ transitive

$(a, b) \in R$ but $(b, a) \notin R \to$ not symmetric

(v) R= ${(a, a), (a, b), (b, a)}$

So, $(b, b) \notin R \to$ not reflexive

$(a, b) \in R$ and $(b, a) \in R \to$ symmetric

And $(a, b) \in R & (b, a) \in R$

and also $(a, a) \in R \to$ transitive

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Relations and Functions Relations and Functions

9. Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12} ,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.

0 Follower 13 Views

Vishal Baghel

Contributor-Level 10

We have,

A= ${x \in 2, 0 \leq x \leq 1 2}$

The relation in set A is defined by

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

For all $a \in A$ ,

$| a - a | = 0$ is a multiple of 4

So, $(a, a) \in R$ i.e., R is reflexive

For $a, b \in A & (a, b) \in R$ we have,

$| a - b |$ is multiple of 4

$| - (b - a) |$ is multiple of 4

$| b - a |$ is multiple of 4

So, $(b, a) \in R$

i.e., R is symmetric

for $a, b, c \in A$ $& (a, b) \in R & (b, c) \in R$

$| a - b |$ & $| b - c |$ is a multiple of 4

So $| a - b | + | b - c |$ is also a multiple of 4

$| a - b + b - c |$ is a multiple of 4

$| a - c |$ is a multiple of 4

So, $(a, c) \in R$

i.e., R is transitive

Hence, R is an equivalence relation.

Finding all set of elements related to 1

For $a \in A$

Then, $(a, 1) \in R$ i.e., $| a - 1 |$ is a multiple of 4

So, a can be 0 ≤ a ≤ 12

Only,

$| 1 - 1 | = 0$

$| 5 - 1 | = 4$ is a multiple of 4

...more

We have,

A= ${x \in 2, 0 \leq x \leq 1 2}$

The relation in set A is defined by

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

For all $a \in A$ ,

$| a - a | = 0$ is a multiple of 4

So, $(a, a) \in R$ i.e., R is reflexive

For $a, b \in A & (a, b) \in R$ we have,

$| a - b |$ is multiple of 4

$| - (b - a) |$ is multiple of 4

$| b - a |$ is multiple of 4

So, $(b, a) \in R$

i.e., R is symmetric

for $a, b, c \in A$ $& (a, b) \in R & (b, c) \in R$

$| a - b |$ & $| b - c |$ is a multiple of 4

So $| a - b | + | b - c |$ is also a multiple of 4

$| a - b + b - c |$ is a multiple of 4

$| a - c |$ is a multiple of 4

So, $(a, c) \in R$

i.e., R is transitive

Hence, R is an equivalence relation.

Finding all set of elements related to 1

For $a \in A$

Then, $(a, 1) \in R$ i.e., $| a - 1 |$ is a multiple of 4

So, a can be 0 ≤ a ≤ 12

Only,

$| 1 - 1 | = 0$

$| 5 - 1 | = 4$ is a multiple of 4

$| 9 - 1 | = 8$

Hence, sets of elements related to A= ${1, 5, 9}$

The relation in set A is defined as R= ${(a, b) : a = b}$

For all $a \in A$ ,

a=a is true

so, $(a, a) \in R$

i.e., R is reflexive

for $a, b \in A & (a, b) \in R$ we have,

a=b

b=a i.e., $(b, a) \in R$

so, R is symmetric

for $a, b, c \in A$ , $(a, b) \in R$ and $(b, c) \in R$ . We have,

a=b and b=c

a=c i.e., $(a, c) \in R$

$∴$ R is transitive

Hence, R is an equivalence relation

Find all sets of elements related to 1

For $a \in A$ , we need $(a, 1) \in R$

i.e., a=1

so, set of elements related to 1 in A is {1}.

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