-3 first of all is not in the set of z+, of course it satisfies closurity property, but the reason that it is not a group is because It does'nt have an identity element and inverse just as (N, +).
Sir your videos are really intelligible and I liked them very much .Could u please upload the videos in a faster rate? I am so eager to learn the whole subject
yeah bro thanks according to me the question you gave at last is z(binary operation), +) is it a group soln (according to me i think it is right) CAIN - closure - yes a,b (belongs to )P a*b (belongs to )P yes associative subtraction is not associative so i stopped here it self so it is not an group that's all !
Hw answer Z+ --> set of positive integers Identity -->0 ( identity elmt in addition) Consider inverse property.. Since z+ consists of all positive integers, obviously sum of any two elements won't give identity elmt 0. So each element in z+ does not have a corresponding inverse ---> violation of CAIN ---> hence { Z+ , +} not a group
Answer to H.W : Z* = {... -3, -2, -1, 1, 2, 3 ...} under Addition. Closure => Any Two Numbers : -1, 1 ∈ Z* but -1 + 1 = 0 ∉ Z* Since, one of the property out of CAIN is not satisfied. Hence, (Z*, +) is not Group. Right ?
It's a group Because 0 is the identity element for addition we are not talking about whether it is belongs to integers or not. And same as for 1. Yes when it comes to multiplication then it's not a group.
@@michaellee7933 it fails all except associativity. a+(-a)=0, but 0 is not in the group and hence can't act as an identity, meaning that there is nowhere for an inverse to map to. This means it's not closed, it's got no identity and from having no identity can have no inverses.
Home work question given at 10:13 is not a group as 3 + (-3) = 0 and 0 is not the Z'. So its not closed with addition operation.
-3 first of all is not in the set of z+, of course it satisfies closurity property, but the reason that it is not a group is because It does'nt have an identity element and inverse just as (N, +).
@@tesfayeyisahak
@@tesfayeyisahak the set used in the HW ques is Z* and not Z+ .
Beautifully explained with such a direct approach. Loved it.
Beautiful explanation ...In Closure property ,5-2=3
Thanks a lot for this explanation. It was all going over my head while learning vector space. Now I'll be able to understand it all.
then teach us vector spaces too @nishith sir please :)
@@shakirashaikh Now, you've seen the same video and hence you can understand vector spaces yourself ma'am.
insanely good explanation
You did a fantastic job with this topic. Thank you.
Really smooth explanation. Understood with just one watch. 🥇🏆🏅
For the 1st time I understood this topic very well
Thankyouu so much sir for that cain short cut ❤ the whole video is amazing
Very clearly explained, thanks so much!
At 6:14, 5-2 must be +3
Yes you are right it should be +3
6:09 it should be 3 not -3. But it is good and comprehensible presentation 👏👏
Sir your videos are really intelligible and I liked them very much .Could u please upload the videos in a faster rate? I am so eager to learn the whole subject
very intuitive explaination
Excellently explained ❤️
❤❤❤ your explanation is excellent
very well explained. thank you for all the help.
and i want more videos from you with fields odered fields because those are the concepts i am stuck with
This is great... thankyou
Nice explanation sir
Wonderful explanation 😊
THANK YOU SIR❤
That was helpful...do one for binary operations
very clear and useful!! many thanks; ❤️
thank you or this lovely explanation
Excellent explained
thank you so much it was an amazing explanation
Sir the result of associative and commutave must be their in group or just need to get RHS=LHS
we need to get just RHS=LHS
Excellent sir very useful
Sir in closure property a=5 ,b=-2 then it is 3 not -3 sir...
Thanks a lot 🎉
homework question , initially , + is not binary operation on Z*
i love this. Thank you so much
Sir, 5+(-2) = 3 ...
Thank u so so much sir ♥️♥️♥️♥️
Sir...N is not a group under addition and multiplication ...is it correct
Love you very clear uni explanation terrible yours good
damn, too good presentation & explanation man, please keep up the good work.
i feel like teacher is not same as in c programming and preposition logic
Thank you so much sir!!!☺️
yeah bro thanks according to me the question you gave at last is z(binary operation), +) is it a group
soln (according to me i think it is right)
CAIN - closure - yes a,b (belongs to )P a*b (belongs to )P yes
associative subtraction is not associative so i stopped here it self so it is not an group that's all !
how is 5+(-2)=-3??? {at 6:02}
Hw answer
Z+ --> set of positive integers
Identity -->0 ( identity elmt in addition)
Consider inverse property..
Since z+ consists of all positive integers, obviously sum of any two elements won't give identity elmt 0. So each element in z+ does not have a corresponding inverse ---> violation of CAIN ---> hence { Z+ , +} not a group
Thanks sir
(Z*, +) is not a group as it does not satisfy inverse.
it was just excellent
(z*,+) is not a group because no 0 exists, eliminating the identity criteria.
Hi
Z+ means positive integers know bro
I hope
Yes
😊
Thankyou ❤️
Please put videos for subgroup and normal subgroup
Thanku sir
Answer to H.W :
Z* = {... -3, -2, -1, 1, 2, 3 ...} under Addition.
Closure => Any Two Numbers : -1, 1 ∈ Z* but -1 + 1 = 0 ∉ Z*
Since, one of the property out of CAIN is not satisfied. Hence, (Z*, +) is not Group. Right ?
Z is denoted as an integer and an integer is a set of positive and negative numbers including 0 hence Z,+ is satisfied and it is an abelian group
@@Anonyymous818 Careful, that Z* is all integets but 0. Doesn't satisfy.
yup it's not a group but a semi group cuz it satisfies closure and associative property.
Thanku a lot sir
Thankyou
Thank u....
Where did 5 in closure came from??
time?
Sir, can you please solve this problem Prove that G is a abelian if b^{-1}a^{-1}ba=e for all a, b€G
Do you know the answer
O thank you!
(Z*,+) is not a group
It's a ring
Yes because inverse element is not exist in z+
@@A-ONE991doesn't*
@@omkar6649😂
It's a group
Because
0 is the identity element for addition we are not talking about whether it is belongs to integers or not.
And same as for 1.
Yes when it comes to multiplication then it's not a group.
Sir always we take identity element is zero
No, it is not a group since closure property not satisfied
Sorry lads but I haven’t got the patience!
Y’all gotta find someone else
(Z* +) is not a group sir
How
@@NoobROHIT20 i believe it failed the inverse and identity element
@@michaellee7933 it fails all except associativity.
a+(-a)=0, but 0 is not in the group and hence can't act as an identity, meaning that there is nowhere for an inverse to map to.
This means it's not closed, it's got no identity and from having no identity can have no inverses.
@@xinpingdonohoe3978 how 0 is not in Z
@@comradenikolatesla1998
The set Z* is defined as Z\{0}
❤❤❤
it is not a group since there is no identity element which is zero in that set
Eduphile??
verryyyy nice
calculation for closure is wrong not -5
is it me or does he sound like chris griffin ?
💥💥💥💥💥💥💥💥💥💥
"L."
Gulo n'yo naman. Edi hindi group yan?
No inverse property unsatisfied.
Very good explanation
❤❤❤
(Z+,+) is not a group