Group Theory Introduction Part 1
36:33
Introduction to Modern Algebra
32:24
3 жыл бұрын
Ring Homomorphism : Part 1
20:25
3 жыл бұрын
Euclidean Ring
9:59
3 жыл бұрын
Ideals of a Ring
21:13
3 жыл бұрын
Subring of a ring
9:23
3 жыл бұрын
Characteristic of a Ring
12:00
3 жыл бұрын
Permutation Groups  :Part 2
25:51
3 жыл бұрын
Subgroups   - Group Theory : Part 09
31:41
Пікірлер
@ramsharma580
@ramsharma580 21 күн бұрын
Sir can i get your all recorded lecture ❤
@mrsimphiwebazzano877
@mrsimphiwebazzano877 4 ай бұрын
Well explained, thank you for doing it in English
@user-kj2iu8xp1s
@user-kj2iu8xp1s 4 ай бұрын
Thankyou sir
@nowardchaselenkana8596
@nowardchaselenkana8596 5 ай бұрын
Perfect explanation sir in class I didn't get many concepts
@crtlallaw
@crtlallaw 5 ай бұрын
great job sir🎉
@H3XED_OwO
@H3XED_OwO 5 ай бұрын
this is so clear, thanks so much!
@visionbestapproach
@visionbestapproach 6 ай бұрын
Mind blowing concept on entire KZbin on Riemann Integral
@visionbestapproach
@visionbestapproach 7 ай бұрын
Mind blowing approach Sir. Salute to you Sir.
@ndamonatuyoleni3614
@ndamonatuyoleni3614 10 ай бұрын
His videos are waking up Mr West ❤🙏🏿🔥
@ndamonatuyoleni3614
@ndamonatuyoleni3614 10 ай бұрын
His videos are waking up Mr West ❤🙏🏿🔥
@aldtzyml2060
@aldtzyml2060 10 ай бұрын
Very understandable, i have always search for English lecture and finally found u,🥺thank u much SIR.
@shubham8192
@shubham8192 11 ай бұрын
One of the best, clear cut explaination, each step is crystal clear Thank you sir 🎉
@markarok
@markarok 11 ай бұрын
It gives me immense pleasure when I hear that someone got benefitted by watching my video. Thank you for your compliment.
@sikaripro4443
@sikaripro4443 11 ай бұрын
Very very great explanation sir U r just a unique personality on KZbin
@sulojan665
@sulojan665 Жыл бұрын
Wow.....Sir you clear my doubt fully.I am struggling with this question for so many time.Thanks a lot SIr
@markarok
@markarok Жыл бұрын
It gives me immense pleasure when I hear that someone got benefitted by watching my video. Thank you for your compliment.
@kpkrkr4038
@kpkrkr4038 Жыл бұрын
Thank you sir so much you help me a lot .
@VijayMishra-uz5dc
@VijayMishra-uz5dc Жыл бұрын
Sir why do we take limit of upper darboux sum to find infimum of it
@markarok
@markarok Жыл бұрын
Thank you for watching my videos. The answer to your question is explained in the video (link provided). kzbin.info/www/bejne/oqjQcn-PqJqLj9k (Concept of Integrability of a bounded function given by Riemann)
@VijayMishra-uz5dc
@VijayMishra-uz5dc Жыл бұрын
@@markarok thank you so much sir 🙏
@shailendarprasad6234
@shailendarprasad6234 Жыл бұрын
Excellent
@jim7090
@jim7090 Жыл бұрын
thank you sir
@MuhammadMunaf-tb8ig
@MuhammadMunaf-tb8ig Жыл бұрын
Assalamualaikum Sir ap thoda Urdu main bhi is ki vezahat Karin shukriya main aapko Islam qabool karna ki bhi dawat deta hon taqa ap is dunya aur aakherat main kamyab hon shukriya
@krisaty166
@krisaty166 Жыл бұрын
Hat's off 🌟
@markarok
@markarok Жыл бұрын
Thank you for your compliment.
@josephatangaagana8993
@josephatangaagana8993 Жыл бұрын
educative Thanks Sir
@markarok
@markarok Жыл бұрын
Thank you for your compliment
@s.velmurugangac9756
@s.velmurugangac9756 Жыл бұрын
Very nice super sir
@markarok
@markarok Жыл бұрын
Thank you sir for your compliment
@pradeeptabehera157
@pradeeptabehera157 Жыл бұрын
thanx😃
@markarok
@markarok Жыл бұрын
Thank you for your compliment
@PavanKumar-ko7ch
@PavanKumar-ko7ch Жыл бұрын
Sir
@PavanKumar-ko7ch
@PavanKumar-ko7ch Жыл бұрын
What does it mean <,> ?
@PardeepVats2002
@PardeepVats2002 Жыл бұрын
great sir G😁😁😁🙏
@markarok
@markarok Жыл бұрын
Thank you for your compliment.
@rahulthakur9039
@rahulthakur9039 Жыл бұрын
Very Very Thankyou sir 🫂
@Naveenbabuborugadda
@Naveenbabuborugadda Жыл бұрын
At 11: 40 . Is ε very small real number here?
@markarok
@markarok Жыл бұрын
yes. ε is positive but very very small as we please.
@Naveenbabuborugadda
@Naveenbabuborugadda Жыл бұрын
Is upper bound doesn't need to be in the set? Like if I consider S={1,2,3.....60} . Here 60 is the upper bound. So any number greater than 60 is also an upper bound as per the point. Is it right? Also you said number at 9:47. Is it real number or rational number or something else like complex number?
@markarok
@markarok Жыл бұрын
THANK YOU FOR WATCHING THIS VIDEO. The set S can be any subset of R, the set of real numbers. Here I have taken only natural Numbers for understanding. In S 60 is an upper bound and any real number greater than 60 is also an upper bound. Yes, an upper bound need not be in the set under consideration.
@Naveenbabuborugadda
@Naveenbabuborugadda Жыл бұрын
@@markarok Thanks so it's difficult to find out number of upper bounds to any subset of real numbers since infinite upper bounds exist for any given subset of real numbers but we can identify least upper bound and similarly greatest lower bound. In some cases like infinite sets may be it's difficult to find out or do not have also these least upper bound and greatest lower bound.
@markarok
@markarok Жыл бұрын
@@Naveenbabuborugadda . For types of sets (1) bounded below but not bounded above (2) bounded above but not bounded below (3) bounded above and bounded below ( bounded set) (4) neither bounded above nor bounded below. Examples are given in this Video. 🙏🙏🙏
@insignificant154
@insignificant154 Жыл бұрын
I don't understand that , because I want to know what is infimum and supremum of upper sum and lower sum respectively ? Also, I want to know what is darboux theorem in reiman integral ?
@markarok
@markarok Жыл бұрын
Thank you for your quarry. Please go through my videos on these topics .
@md.mehedihasan6782
@md.mehedihasan6782 Жыл бұрын
you are great sir
@markarok
@markarok Жыл бұрын
Thank you for your compliment
@bksmathsacademy3167
@bksmathsacademy3167 Жыл бұрын
Please continue uploading video sir
@markarok
@markarok Жыл бұрын
Sure. Thank you Sir
@Leo-kq5sw
@Leo-kq5sw Жыл бұрын
Thank you sir
@abhitiwari9865
@abhitiwari9865 Жыл бұрын
This was so helpful Thank you sir Love from bihar ❤️
@thrisha212
@thrisha212 2 жыл бұрын
Sir pls continue the classes on group theory........
@markarok
@markarok 2 жыл бұрын
Thank you for your compliments. Please subscribe my channel. I have made 36 videos on Abstract (Modern) Algebra.
@shubham8192
@shubham8192 2 ай бұрын
​@@markarok Sir where are you, why you discontinue teaching??
@markarok
@markarok 2 ай бұрын
Sure. Soon I will upload some more videos.
@thrisha212
@thrisha212 2 жыл бұрын
Tq so much sir......
@thrisha212
@thrisha212 2 жыл бұрын
Tq so much sir ........ helped me lot......
@architapatel454
@architapatel454 2 жыл бұрын
Excellent explanation sir finally i understood this topic
@AdityaSingh-qn6pi
@AdityaSingh-qn6pi 2 жыл бұрын
Thank you sir......
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
What are a and b in [ a, b ]
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
I think there are no monotonically increasing sequences are bounded but monotonically increasing and decreasing sequence are bounded.
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
Could you please provide examples for monotonically increasing sequence sequence and bounded other than <1, 1 , 1, 1, 1,1,1,1,1,.....> such examples
@markarok
@markarok 2 жыл бұрын
< n/(n+1)> is a bounded and an increasing sequence. 1/2<=n/(n+1)<1 . Infimum (Lower bound) is 1/2 and Supremum (upper bound) is 1.
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
Is there any theorem that shows the relation between the convergence of a sequence and series? Similarly, relation between the divergence of a series and divergence of a sequence. I think there is no such relation right because < 1/n > sequence converges and Σ 1/n diverges. But, in the video you said monotonically increasing sequence is bounded but <Sn> is not a sequence it is a series. * If it is true then for p= 1 the power series is bounded above but monotonically increasing and hence it will also converges.
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
When I take power series for p=1 and I added up to n=10. I got the summation as 2.8289682539682 and when I added up to n= 9, I got 2.8289682539682111111 * From the above observation, I increased the n values above 10 and found the summation. * I observed that as the n value increases for the power p=1, the summation cannot cross the 2.9 value and hence it is bounded above. * To represent the infinite series what we should take on the x-axis? Should we take partial sums i.e a1,a2, a3.... and on the y-axis we should take real numbers right. Please verify whether I am right or wrong
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
How u suffix n / u suffix n+1 is < l from where it is determined. The limit of u suffix n / u suffix n+1 is less than l right. I am asking how u suffix n / u suffix n+1 is < l.
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
Where is the 6th form
@markarok
@markarok 2 жыл бұрын
See the next Video
@markarok
@markarok 2 жыл бұрын
Sixth form in available in kzbin.info/www/bejne/eauviayNmcaNmsk&lc=Ugxx5TQxmwMXNbMG14F4AaABAg
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
At 11:5 How u suffix n converges without knowing the u suffix n value to use it comparison test.
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
For 7:12 *. if we compare 1 and 2 with 1000000000 ( large number) 1 and 2 looks like they are very very small but 1≠ 2 ( 1 is never equal to 2) * Similarly Ɛ is very very small and Ɛ/2 is also very very small positive real number but Ɛ≠Ɛ/2 ( Ɛ is not equal to Ɛ/2 for limit of a sequence) * Hence, for given Ɛ >0 | a suffix n+p -a suffix n | < 2Ɛ
@Naveenbabuborugadda
@Naveenbabuborugadda 2 жыл бұрын
Sir , what are a and b in open interval are they real numbers or integers