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11 months ago

Relations and Functions Maths

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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11 months ago

Information Technology Computers & IT (Non-Engg)

Can I get some details about the top BCA colleges in India?

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Muskan Chugh

Contributor-Level 10

Sure! The table below provides some details and information about the top BCA colleges in India.

Particulars	Parameters
No. of Colleges	120+ colleges
Fee	Less than INR 1 Lakh: 10+ colleges INR 1-2 Lakh: 15+ colleges INR 2-3 Lakh: 20 colleges INR 3-5 Lakh: 15+ colleges More than INR 5 Lakh: 5 colleges
Top BCA Colleges in India	Christ University, Symbiosis Institute of Computer Studies and Research, Kristu Jayanti College, Amity University, etc.
Accepted Entrance Exam	CUET, IPU CET, etc.
Top Specializations	Cyber Security, Digital Marketing, and Visual Communication

Disclaimer: This information is taken from the official website and may differ.

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11 months ago

Relations and Functions Maths

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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11 months ago

Relations and Functions Maths

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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11 months ago

Relations and Functions Maths

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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11 months ago

B.A Fashion Design Design

Which documents are required during BA in Fashion Design application process?

0 Follower 3 Views

Rashmi Srivastava

Contributor-Level 9

Here are some important documents you must carry during the application process of BA Fashion Design for a seamless process:

Class 10 and 12 mark sheets
Passport-sized photographs
A valid ID proof (Adhaar, Passport, PAN Card, etc.)
Migration Certificate, etc.

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11 months ago

Humanities & Social Sciences B.A

How many seats are reserved under the TFWS for BA course at Gokul Global University?

0 Follower 3 Views

Anushka Bidhi

Contributor-Level 10

According to the Gokul Global University, 5% of the total available seats in the BA course are reserved and allocated to the students who are enrolled under the Tuition Fee Waiver Scheme. Therefore, around 4-5 seats are reserved in the BA course as per TFWS. This information is taken from official source and is subject to revision.

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11 months ago

Relations and Functions Maths

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 28 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

...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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11 months ago

Kirit P. Mehta School of Law, NMIMS, Mumbai Law

What is the application fee for LLM course at Kirit P. Mehta School of Law?

0 Follower 4 Views

Indrani Kumar

Contributor-Level 10

After filling out the application form for the LLM course at Kirit P. Mehta School of Law, candidates are required to pay a registration fee, which is non-refundable. The amount of the registration fee is INR 2,500. The fee amount mentioned is sourced from the official website/sanctioning body and is subject to change. Therefore, it is indicative

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11 months ago

Relations and Functions Maths

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}.

0 Follower 39 Views

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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11 months ago

Design B.A Fashion Design

What are the benefits of pursuing a BA in Fashion Design?

0 Follower 5 Views

Akansha Bisht

Contributor-Level 10

Getting a degree in BA Fashion Design will benefit you in many ways, such as:

You'll get your hands on some important technical skills like hand sewing, using sewing machines, and designing illustrations using CAD.
You'll gain a deep knowledge of aesthetic and visualization principles, and the design process from sketching to outfit making.
Pursuing this course will teach you about how the industry works.
It will also help you develop communication skills necessary to grow the network and get a job.

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11 months ago

Information Technology MCA

Which colleges can I get for MCA through CET?

0 Follower 5 Views

Lakshita Saini

Beginner-Level 5

Some common options include:

Government engineering colleges in your state.
University departments offering MCA.
Top private institutes affiliated to state universities.
Regional colleges in cities like Mumbai, Pune, Bangalore, or Delhi (depending on your CET).
Self-financed colleges with good placement records.

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11 months ago

Relations and Functions Maths

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 6 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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11 months ago

Relations and Functions Maths

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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11 months ago

Relations and Functions Maths

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 4 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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11 months ago

Relations and Functions Maths

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

0 Follower 12 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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11 months ago

Colleges in India Colleges in Pune

I want to know about CDAC course offered at Sunbeam College Pune.

0 Follower 5 Views

New Question

11 months ago

Relations and Functions Maths

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

0 Follower 9 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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11 months ago

Business Management Studies MBA/PGDM

Does PTSNS University provide any financial aid to MBA candidates?

0 Follower 6 Views

Virajita Mehra

Contributor-Level 10

Yes, Pandit SN Shukla University provide scholarships through the MP scholarship board. MBA course aspirants can apply for these scholaships. In order to avail of a scheme, the aspirants must fulfil the eligibility requirement for the scheme. The important documents must also be submitted for verification purposes. Mentioned below are the available schemes: