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Maths Relations and Functions

8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a,b) : |a-b| is even} is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

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Vishal Baghel

Contributor-Level 10

We have,

R= ${(a, b) ∴ | a - b |$ is even $}$ is a relation in set A= ${1, 2, 3, 4, 5}$

For all $a \in A$ , $| a - a | = 0$ is even.

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

For $a, b \in A$ and $(a, b) \in R$

$| a - b |$ is even

$| - b + a |$ is even $|$

$| - (b - a) |$ is even

$| b - a |$ is even

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

Hence, R is symmetric.

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

We have $| a - b |$ is even

and $| b - c |$ is even

then, $| a - b | + | b - c |$ is even as even + even=even

$| a - b + b - c |$ is even

$| a - c |$ is even

$∴$ $(a, c) \in R$

So, R is transitive.

$∴$ R is an equivalence relation

All elements of [1,3,5] are odd positive numbers and its subset are odd and their difference given an even number. Hence, they are related to each other.

Similarly,

...more

We have,

R= ${(a, b) ∴ | a - b |$ is even $}$ is a relation in set A= ${1, 2, 3, 4, 5}$

For all $a \in A$ , $| a - a | = 0$ is even.

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

For $a, b \in A$ and $(a, b) \in R$

$| a - b |$ is even

$| - b + a |$ is even $|$

$| - (b - a) |$ is even

$| b - a |$ is even

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

Hence, R is symmetric.

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

We have $| a - b |$ is even

and $| b - c |$ is even

then, $| a - b | + | b - c |$ is even as even + even=even

$| a - b + b - c |$ is even

$| a - c |$ is even

$∴$ $(a, c) \in R$

So, R is transitive.

$∴$ R is an equivalence relation

All elements of [1,3,5] are odd positive numbers and its subset are odd and their difference given an even number. Hence, they are related to each other.

Similarly, all elements of [2,4] are even positive numbers and its subset are even and their difference gives an even number. Hence, they are related to each other.

However, elements of [1,3,5] are related to elements of [2,4] since the difference of the two subsets is never even.

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New answer posted

a year ago

Maths Relations and Functions

3. If G = {7, 8} and H = {5, 4, 2}, find G * H and H * G.

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Payal Gupta

Contributor-Level 10

3. Given, G = {7, 8} and H = {5, 4, 2}

By the definition of the Cartesian product,

G *H = { (x, y): x∈G and y = ∈ H}

= { (7, 5), (7, 4), (7, 2), (8, 5), (8,4), (8,2)}

H* G = { (x, y): x∈ H and y ∈G}

= { (5, 7), (5, 8), (4,7), (4, 8), (2, 7), (2,8)}

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New answer posted

a year ago

Maths Relations and Functions

7. Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y): x and y have same number of pages} is an equivalence relation.

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

We have,

R= ${(x, y) : x & y$ have same number of pages $}$ is a relation in set of A of all books in

For $(x, y) \in R & x, y \in A$

As x=y=same no. of pages

Then, $(x, x) \in R$

Hence, R is reflexive.

For $(x, y) \in R$ and $x, y \in A$

Also, $(y, x) \in R$ , $∴ x = y$

Hence, R is symmetric.

For $x, y, z \in A$ and $(x, y) \in R$ and $(y, z) \in R$

x=y and y=z

x=z

i.e., $(x, z) \in R$

hence, R is also transitive

$∴$ R is an equivalence relation.

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New answer posted

a year ago

Maths Relations and Functions

6. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

We have,

R= ${(1, 2), (2, 1)}$ is a relation in set ${1, 2, 3}$

Then, as $(1, 1) \notin R$ and $(2, 2) \notin R$

So, R is not reflective

As $(1, 2) \in R$ and $(2, 1) \in R$

So, R is symmetric

And as $(1, 2) \in R, (2, 1) \in R$ but $(1, 1) \notin R$

So, R is not transitive.

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New answer posted

a year ago

Maths Relations and Functions

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

0 Follower 40 Views

Vishal Baghel

Contributor-Level 10

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.

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New answer posted

a year ago

Maths Relations and Functions

4. Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.

0 Follower 13 Views

Vishal Baghel

Contributor-Level 10

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

For, $a \in R$ ,

$a \leq b$ but $b \leq a$ is not possible i.e., $(b, a) \notin R$

Hence, R is not symmetric.

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

$a \leq b$ and $b \leq c$

So, $a \leq c$

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

$∴$ R is transitive.

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New answer posted

a year ago

Maths Relations and Functions

3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

We have,

R= ${(a, b) : b = a + 1}$ is a relation in set ${1, 2, 3, 4, 5, 6}$

So, R= ${(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}$

As, $(1, 1) \notin R$ , R is not reflexive

As, $(1, 2) \in R$ but $(2, 1) \notin R$ , R is not symmetric

And as $(1, 2)$ & $(2, 3) \in R$ but $(1, 3) \notin R$

Hence, R is not transitive.

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New answer posted

a year ago

Maths Relations and Functions

2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A*B).

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

2. Given, n (A) = 3

n (B) = 3 or B = {3,4,5}

So, number of elements in A* B = n (A* B) = n (A)* n (B) = 3 *3 = 9.

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New answer posted

a year ago

Maths Relations and Functions

2. Show that the relation R in the set R of real numbers, defined as

R = {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive.

0 Follower 26 Views

Vishal Baghel

Contributor-Level 10

We have,

R= ${(a, b) : a \leq b^{2}}$ is a relation in R.

For $a \in R$ then is $b = a, a \leq a^{2}$ is not true for all real number less than 1.

Hence, R is not reflexive.

Let $(a, b) \in R$ and a=1 and b=2

Then, $a \leq b^{2}$ = $1 \leq 2^{2}$ = $1 \leq 4$ so, $(1, 2) \in R$

But $(b, a) = (2, 1)$

i.e., $2 \leq 1^{2}$ = $2 \leq 1$ is not true

so, $(2, 1) \notin R$

hence, R is not symmetric.

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

We have, $a = 10 \leq 4^{2} = b^{2}$ => $10 \leq 16$ is true

So, $(10, 4) \in R$

And $4 \leq 2^{2}$ => $4 \leq 4$ So, $(4, 2) \in R$

But $10 \leq 2^{2}$ => $10 \leq 4$ is not true.

So, $(10, 2) \notin R$

Hence, R is not transitive.

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New answer posted

a year ago

Maths Relations and Functions

Relation and Function Exercise 1.1 Solutions

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)

...more

0 Follower 75 Views

Vishal Baghel

Contributor-Level 10

(i) We have, $R = {(x, y) : 3 x - y = 0}$ a relation in set A= ${1, 2, 3 . . . . . . . . . . 14}$

For $x \in A, y = 3 x$ or $y \neq x$ i.e.,

$(x, x)$ does not exist in R

$∴$ R is not reflexive.

For $(x, y) \in R, y = 3 x$

Then $(y, x) x \neq 3 y$

So $(y, x) \notin R$

$∴$ R is not symmetric

For $(x, y) \in R$ and $(y, z) \in R$ . We have

$y = 3 x$ and $z = 3 y$

Then $z = 3 (3 x) = 9 x$

i.e., $(x, z) \notin R$

$∴$ R is not Transitive

(ii) We have,

R= ${(x, y) : y = x + 5 & x < 4}$ is a relation in N

= ${(1, 1 + 5), (2, 2 + 5), (3, 3 + 5)}$

= ${(1, 6), (2, 7), (3, 8)}$

Clearly, R is not reflexive as $(x, x) \notin R$ and $x < 4 & x \in N$

Also, R is not symmetric as $(1, 6) \in R$ but $(6, 1) \notin R$

And for $(x, y) \in R (y, z) \notin R$ . Hence, R is not Transitive.

(iii) R= ${(x, y); y$ is divisible by x $}$ is a relation in set

A= ${1, 2, 3, 4, 5, 6}$

So, R= ${(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)}$

Hence, R is reflexive because $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) \in R$ i.e., $(x, x) \in R$

R is not sy

...more

(i) We have, $R = {(x, y) : 3 x - y = 0}$ a relation in set A= ${1, 2, 3 . . . . . . . . . . 14}$

For $x \in A, y = 3 x$ or $y \neq x$ i.e.,

$(x, x)$ does not exist in R

$∴$ R is not reflexive.

For $(x, y) \in R, y = 3 x$

Then $(y, x) x \neq 3 y$

So $(y, x) \notin R$

$∴$ R is not symmetric

For $(x, y) \in R$ and $(y, z) \in R$ . We have

$y = 3 x$ and $z = 3 y$

Then $z = 3 (3 x) = 9 x$

i.e., $(x, z) \notin R$

$∴$ R is not Transitive

(ii) We have,

R= ${(x, y) : y = x + 5 & x < 4}$ is a relation in N

= ${(1, 1 + 5), (2, 2 + 5), (3, 3 + 5)}$

= ${(1, 6), (2, 7), (3, 8)}$

Clearly, R is not reflexive as $(x, x) \notin R$ and $x < 4 & x \in N$

Also, R is not symmetric as $(1, 6) \in R$ but $(6, 1) \notin R$

And for $(x, y) \in R (y, z) \notin R$ . Hence, R is not Transitive.

(iii) R= ${(x, y); y$ is divisible by x $}$ is a relation in set

A= ${1, 2, 3, 4, 5, 6}$

So, R= ${(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)}$

Hence, R is reflexive because $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) \in R$ i.e., $(x, x) \in R$

R is not symmetric as $(1, 2) \in R$ but $(2, 1) \notin R$

And for $x, y, z \in A$ and $(x, y) \in R & (y, z) \in R$

$\frac{y}{z} = n$ and $\frac{z}{y} = m$ where $n & m \in N$

Then, $z = m y = m (n x) = m n . x$

$\frac{z}{x} = m . n, m, m n \in N$

Hence, $(x, z) \in R$

$∴$ R is transitive

(iv) R= ${(x, y) : x - y$ is an integer $}$ is a relation in set Z

For $x \in Z$

$x - x = 0$ is an integer

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

For $x, y \in Z$ and $(x, y) \in R$

$x - y$ is an integer

$- (y - x)$ is an integer

$(y - x)$ is an integer

So, $(y, x) \in R$ i.e., R is symmetric

For $x, y, z \in R$ and $(x, y) \in R & (y, z) \in R$ We have,

$x - y$ is an integer

$y - z$ is an integer

So, $(x - y) + (y - z)$ is also an integer

$x - z$ is an integer

So, $(x, z) \in R$ i.e., R is transitive.

(v) (a) R= ${(x, y) : x$ and $y$ work at same place $}$ in set A of human being.

For $x \in A$ we get

$x & x$ work at same place

$(x, x) \in R$ So, R is reflexive.

For $x, y \in A$ and $(x, y) \in R$ We get,

$x & y$ work at same place

$y & x$ work at same place

$(y, x) \in R$ . So, R is symmetric.

For $x, y, z \in A$ and $(x, y)$ and $(y, z) \in R$ . We have,

$x & y$ work at same place

$y & z$ work at same place

$x & z$ work at same place

i.e., $(x, z) \in R$ . So, R is Transitive.

(b) R= ${(x, y) : x & y$ live in same locality $}$

For $x \in A$ , we get,

$x & x$ live in same locality

$(x, x) \in R$ So, R is reflexive.

For $x, y \in A$ and $(x, y) \in R$ we get,

$x & y$ live in same locality

$y & x$ live in same locality

i.e., $(y, x) \in R$ . So, R is symmetric.

For $x, y, z \in A$ and $(x, y)$ & $(y, z) \in R$ . Then

$x, y$ live in same locality

$y, z$ live in same locality

$x, z$ live in same locality

So, $(x, z) \in R$ i.e., R is transitive.

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

For $x \in A$ ,

Height $(x) \neq 7 +$ height of $(x)$

So, $(x, x) \notin R$ . i.e., R is not reflexive.

For, $x, y \in A$ and $(x, y) \in R$ we have,

Height $(x)$ $= 7 + h e i g h t (y)$

But $h e i g h t (y) \neq 7 + h e i g h t (x)$

So, $(y, x) \in R$ i.e., R is not symmetric.

For $x, y, z \in A$ and $(x, y) \in R$ and $(y, z) \in R$ we have,

Height $(x) = 7 + h e i g h t (y)$

And $h e i g h t (y) = 7 + h e i g h t (z)$

So, $h e i g h t (x) = 7 + 7 + h e i g h t (z) = 14 + h e i g h t (z)$

i.e., $(x, z) \notin R$

So, R is not transitive.

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

For $x \in A,$

X is not wife of x.

So, $(x, x) \notin R$ i.e., R is not reflexive.

For $x, y \in A$ and $(x, y) \in R$ ,

X is wife of y but y is not wife of x

So, $(y, x) \notin R$ . i.e., R is not symmetric.

For $x, y, z \in A$ and $(x, y)$ and $(y, z) \in R$

X is wife of y

Y is wife of z

But y can never be husband & wife simultaneously

So, R is not transitive.

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

For $x \in A$ ,

X is not a father of x

So, $(x, x) \notin R$ i.e., R is not reflexive.

For $x, y \in A$ and $(x, y) \in R$

X is father of y

Y is father of z

But x is not father of z

So, $(x, z) \notin R$ i.e., R is not transitive.

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