-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
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
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 !
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 ?
@@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.
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.
Beautifully explained with such a direct approach. Loved it.
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+ .
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.
Beautiful explanation ...In Closure property ,5-2=3
Thankyouu so much sir for that cain short cut ❤ the whole video is amazing
For the 1st time I understood this topic very well
insanely good explanation
Really smooth explanation. Understood with just one watch. 🥇🏆🏅
You did a fantastic job with this topic. Thank you.
Very clearly explained, thanks so much!
6:09 it should be 3 not -3. But it is good and comprehensible presentation 👏👏
At 6:14, 5-2 must be +3
Yes you are right it should be +3
Excellently explained ❤️
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
❤❤❤ 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
very intuitive explaination
very clear and useful!! many thanks; ❤️
damn, too good presentation & explanation man, please keep up the good work.
Wonderful explanation 😊
Nice explanation sir
THANK YOU SIR❤
This is great... thankyou
That was helpful...do one for binary operations
Sir in closure property a=5 ,b=-2 then it is 3 not -3 sir...
Wawooo this explaination is crealy
homework question , initially , + is not binary operation on Z*
thank you so much it was an amazing explanation
Thank u so so much sir ♥️♥️♥️♥️
Excellent sir very useful
thank you or this lovely explanation
Excellent explained
i love this. Thank you so much
Sir, 5+(-2) = 3 ...
Thank you so much sir!!!☺️
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 a lot 🎉
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
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 !
Love you very clear uni explanation terrible yours good
(Z*, +) is not a group as it does not satisfy inverse.
Sir...N is not a group under addition and multiplication ...is it correct
how is 5+(-2)=-3??? {at 6:02}
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.
i feel like teacher is not same as in c programming and preposition logic
(z*,+) is not a group because no 0 exists, eliminating the identity criteria.
it was just excellent
Please put videos for subgroup and normal subgroup
Sir always we take identity element is zero
Thankyou ❤️
Thanks sir
Hi
Z+ means positive integers know bro
I hope
Yes
😊
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
Thanku a lot sir
Where did 5 in closure came from??
time?
Thanku sir
No, it is not a group since closure property not satisfied
Thank u....
Thankyou
(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}
(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.
it is not a group since there is no identity element which is zero in that set
verryyyy nice
O thank you!
❤❤❤
calculation for closure is wrong not -5
Sorry lads but I haven’t got the patience!
Y’all gotta find someone else
is it me or does he sound like chris griffin ?
Eduphile??
💥💥💥💥💥💥💥💥💥💥
Gulo n'yo naman. Edi hindi group yan?
"L."
No inverse property unsatisfied.
Very good explanation
Thanks
❤❤❤
(Z+,+) is not a group