Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Harshit Chaturvedi 6 years, 8 months ago
- 1 answers
Deepak Gour 6 years, 8 months ago
Posted by Amit Mishra 6 years, 8 months ago
- 0 answers
Posted by Priyanshu Aman 6 years, 8 months ago
- 1 answers
K@Łp@N@ $Øđh!?? 6 years, 8 months ago
Posted by Prince Yadav 6 years, 8 months ago
- 0 answers
Posted by Balambika Thanulingeswaran 6 years, 8 months ago
- 2 answers
Posted by Vivek Shukla 6 years, 8 months ago
- 2 answers
Amol Gùpta 6 years, 8 months ago
Posted by Shivam Khatri 6 years, 8 months ago
- 0 answers
Posted by Dheeraj Jat 6 years, 8 months ago
- 2 answers
Premesh Tyagi 6 years, 8 months ago
Sakshi Rana 6 years, 8 months ago
Posted by Dheeraj Jat 6 years, 8 months ago
- 0 answers
Posted by Achyuth M H 6 years, 8 months ago
- 1 answers
Avantika Tayal 6 years, 8 months ago
Posted by Satyam Tiwari 6 years, 8 months ago
- 1 answers
Posted by Rashmi Tomar 6 years, 8 months ago
- 7 answers
Deepak Gour 6 years, 8 months ago
Posted by Deepak Gour 6 years, 8 months ago
- 1 answers
Bismi John 6 years, 8 months ago
Posted by Rachna Jain 6 years, 8 months ago
- 4 answers
Krishna Yadav 6 years, 8 months ago
Bismi John 6 years, 8 months ago
Priya Dharshini 6 years, 8 months ago
Posted by Pushpa Kadian 6 years, 8 months ago
- 1 answers
Posted by Amol Gùpta 6 years, 8 months ago
- 7 answers
Premesh Tyagi 6 years, 8 months ago
Achyuth M H 6 years, 8 months ago
Firefire Live With Funky Fames 6 years, 8 months ago
Ɽøⱨł₮ ₲Ʉ₱₮₳ 6 years, 8 months ago
Posted by Harjot Singh 6 years, 8 months ago
- 1 answers
Posted by Sahil Gupta 6 years, 8 months ago
- 0 answers
Posted by Rashmi Shukla 6 years, 8 months ago
- 1 answers
Neeti Singla 6 years, 7 months ago
Posted by Sejal Janghel 6 years, 8 months ago
- 1 answers
Posted by Rohit Rajput 6 years, 8 months ago
- 1 answers
Posted by Ishan Singh 6 years, 8 months ago
- 1 answers
Ɽøⱨł₮ ₲Ʉ₱₮₳ 6 years, 8 months ago
Posted by Ekta Gill 6 years, 8 months ago
- 0 answers
Posted by Gora Sekhon 6 years, 8 months ago
- 0 answers
Posted by Tajo Laa 6 years, 8 months ago
- 2 answers
Shivam Singh 6 years, 8 months ago
Posted by Surjit Singh 6 years, 8 months ago
- 0 answers
Posted by Rajnish Mishra 6 years, 8 months ago
- 3 answers
Purvesh Meena 6 years, 8 months ago
Posted by Rajeshbv Gupta 6 years, 8 months ago
- 1 answers
Sia ? 6 years, 8 months ago
We have, a relation R on N {tex}\times{/tex} N defined by {tex}(a, b)R(c, d){/tex}, if {tex}ad(b + c) = bc(a + d){/tex}
Reflexive :
Let (a, b) {tex}\in{/tex} N {tex}\times{/tex} N be any arbitrary element.
We have to show {tex}(a, b) R (a, b){/tex}, i.e. to show {tex}ab(b + a) = ba(a + b){/tex} which is trivially true as natural numbers are commutative under usual multiplication and addition.
Since, (a, b) {tex}\in{/tex} N {tex}\times{/tex} N was arbitrary, therefore R is reflexive.
Symmetric:
Let (a, b), (c, d) {tex}\in{/tex} N {tex}\times{/tex} N such that{tex} (a, b) R (c, d){/tex}, i.e. {tex}ad(b + c) = bc(a + d){/tex} ... (i)
To show, (c, d) R (a, b), i.e. to show {tex}cb(d + a) = da(c + b){/tex}
From Eq.(i), we have {tex}ad(b + c) = bc(a + d) da(c + b) = cb(d + a){/tex} [{tex}\because{/tex} natural numbers are commutative under usual addition and multiplication]
{tex}\Rightarrow{/tex} {tex}cb(d + a) = da(c + b){/tex}
{tex}\Rightarrow{/tex} {tex}(c, d) R (a, b){/tex}
Thus, R is symmetric.
Transitive:
Let (a, b), (c, d) and (e, f) {tex}\in{/tex} N {tex}\times{/tex} N such that (a, b) R (c, d) and (c, d R (e, f)
Now, (a, b) R (c, d) {tex}\Rightarrow{/tex}ad(b + c) = bc(a + d)
{tex}\Rightarrow \frac { b + c } { b c } = \frac { a + d } { a d }{/tex}{tex}\Rightarrow \frac { 1 } { b } + \frac { 1 } { c } = \frac { 1 } { a } + \frac { 1 } { d }{/tex}.......(ii)
and (c, d) R (e, f) {tex}\Rightarrow{/tex}cf(d + e) = de(c + f)
{tex}\Rightarrow \quad \frac { d + e } { d e } = \frac { c + f } { c f }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { d } + \frac { 1 } { e } = \frac { 1 } { c } + \frac { 1 } { f }{/tex}......(iii)
{tex}\Rightarrow \quad \frac { 1 } { b } + \frac { 1 } { e } = \frac { 1 } { a } + \frac { 1 } { f }{/tex}{tex}\Rightarrow \frac { e + b } { b e } = \frac { f + a } { a f }{/tex}
{tex}\Rightarrow {/tex}{tex}af(e + b) = be(f + a){/tex}
{tex}\Rightarrow {/tex}{tex}af(b+e)=be(a+f){/tex}
{tex}\Rightarrow{/tex} {tex}(a, b) R (e, f){/tex}
{tex}\Rightarrow{/tex}R is transitive
Thus, R is reflexive, symmetric and transitive, hence R is an equivalence relation.
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Johanna Eapen 6 years, 7 months ago
0Thank You