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In example 2^n, where n is an integer, don't you think, its not an algebraic structure? because if you put n=-1, the output is 0.5 which is not an interger. Please correct me if I'm wrong.
Thanks sir for this video and a request you can make video on coset, lattice ,Homomorphism, Isomorphism and Automorphism of groups. Please sir a humble request
if we do (5/0), as both 5 and 0 are integers the answer would be infinity which is not an integer number, so it does not follow closure property and so it is not an algebraic structure and hence it is not a semigroup.
Sir if we take -ve value of n in the (2^n) example then the answer we will get in -ve power of 2 which is not a integer That means it is not closer So it is not semi-group
The division will not be a semi-group with respect to the set of all integers Z, because it does not even follow closure property, i.e., it will even not be an algebraic structure.
in the last expression n is an integer so it can be negetive aswell consider (2^-3)*(2^2)=0.5 which does not belong to integer so closure is not met how is this semi group please clarify
(Z, ÷) does not follow closure property as well as associative property.
Kaise
6, 2
@Utsav so?... There is infinity
Because integer can not kept decimal number so it's not hold clouser and associative property for division
i'm a MCA student it's my first year i started study after years again ...
because of you i scored in my first and second MST sem exam is at door ...thank you sooo much sir ..
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S= find the member of semigroup S using breeding coset enumeration technique
you will never find a better teacher then him
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The last example which is given for 2^n is not closure if one take n= 1 and n = -2, then the ans will be a rational number.
Your explanation is always awesome.... Whenever I am stuck in any subject and that is available on your channel then i feel very relaxed that if i watch it before one day a paper , the paper will be awesome 😎....
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Semi group : an algebraic structure (S , *) is semi group if it follows associative property
Algebraic structure must follow closure prop.
Clear cut explanation sir🎉 thank you so much ❤ 😊
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Thankyou is much sir ! This helped a lot.. I was struggling with algebraic structure until I watched your videos... keep going 🌟😇🔆
(Z, ÷) => first of all its not an algebraic structure ,so no need to check whether it's semi group or not.
5÷ 2= 2.5 doesn't belong to Z.
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Sir what about monoid for integer numbers with multiplication?
Z under multiplication is a Monoid as it has a Multiplicative identity element as 1.
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In example 2^n, where n is an integer, don't you think, its not an algebraic structure? because if you put n=-1, the output is 0.5 which is not an interger. Please correct me if I'm wrong.
Yaa bro , you are right 👍
Exactly, i am thinking same ....and you are right its not follow closure property.
Right ✅
How bro
2^-1 * 2^-4 ... if base same power add 2^-5 which is integer
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Very nice explanation every std can understand thanks a lot sir plz upload more and more videos
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you are so great
Thanks sir for this video and a request you can make video on coset, lattice ,Homomorphism, Isomorphism and
Automorphism of groups. Please sir a humble request
if we do (5/0), as both 5 and 0 are integers the answer would be infinity which is not an integer number, so it does not follow closure property and so it is not an algebraic structure and hence it is not a semigroup.
Sir if we take -ve value of n in the (2^n) example then the answer we will get in -ve power of 2 which is not a integer
That means it is not closer
So it is not semi-group
Yes... I have same question also
n should be an integer..not 2^n according to question..check again..
@@anweshabiswas19-ve is also an integer
@@introverts_site yeah -ve whole numbers are Also integers..
@@anweshabiswas19 finally uhh got this 😁
Great explanation! Thank you sir
5:20 I think multiplication doesn't satisfying associative property
sir, could you please make one video for triggers in SQL . i found you haven't made video on triggers . please sir .
In last, example 2^n is not an algebraic structure because if we take 2^0*2^-1 =0.5 is not an integer
Given n is integer not 2^n
So 2^0.2^(-1) gives 2^(-1) only where -1 is integer so it's algebric structure
Sir division doesn't follow associativity ,so can't be semigroup .. correct me if I am wrong
It doesn't follow closer property...
Integer (Z) ke isme division check krne ki kya jarurt sir wo closure property hi nhi follow krti..
Right🤞🏻
Division is not a semigroup in case of Z
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Sir pls upload the full videos of this topic....
In 2^n if we take n=-1, then it is not an integer.
2^n is not following closure property , if i take n = -5 which is integer 2^-5 is 1/32 is not an integer hence , (2^n, . ) is not a SG
thinking same bro
Thank you sooo much sir its very helpful....
The division will not be a semi-group with respect to the set of all integers Z, because it does not even follow closure property, i.e., it will even not be an algebraic structure.
Product and Quotient of semi groups please make video on it
Excellent brother
Sir please upload video on K-map.
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Sir please continue to teach
Integers can be negative too then it will not satisfy closure property then it is not a semi group
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Nice explain sir
Sir devision ke method me natural number flow (semigraph) hoga
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What If we take negative values In 2^n /n is integer w.r.t. * {n=-1,-2
Plz explain
Sir please upload videos for group and abelian group..
Will 2^n follow associative property? As an integer we can have -ve power also.
yup at that case we can get decimal number which dont come in integer
last example mein n=-1 daal du toh? it should not be a SEMI GROUP, right? Please clarify this doubt
we are checking the value whether n is coming integer or not ,we are not checking the overall value of 2^n
@@bishwadeeprana6856 oh, thanks
Good but when binary operation is * over set then how come + is defined to add power
in the last expression n is an integer so it can be negetive aswell
consider (2^-3)*(2^2)=0.5 which does not belong to integer so closure is not met
how is this semi group please clarify
So, 2^n is not closer
Tysm sir!❤️
Division is not a semi group because first of all it doesn't follow closer property and not a algebric structure 🤘😅
(Z , /) can not be a algebraic structure yet , so it is not a semi group
Thank you
Thank sir.
For semi group first have to proof closure property
Division is not a semi group because it's doesn't follow associative property.
Sir please Solve the GATE 2021 paper topic wise
thank you sir
Division on Integers set does not follow associative property and that's why it is not a semi-group.
Thanks !!!
thank you
sir last question me n -ve value le to decimal me value ayega vo integer nahi hai to kaise closure property follow hui
(z,/)Does not follow closure property so it does not comes under algebraic structure so how it comes under Semi group .
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Thankyou thankyou
Sir if (z,.) Monoid group hai to kya semi group hoga?
Book discreate mathematics ka kn sa version hy
z is not a semi group as well as algebric structure bcoz it doesnt follow closure property as well as associative property .
Sir (N,.) is not shown closer property
Sir ic m truth table wly topics bhe cover krvain
Sir (5-3)-2=5-(3+2) na sir, not 5-(3-2) right? In the example u have given
At time 5.53
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hello, I have a question about semigroup, can you solve my question?
Integer is not semi group
(Z, /) does not follow closure property, so its not a semigroup
(Z,÷) nahi hoga kyu ki wo algebraic structure form nahi kar raha ha
Topp brother
2 ki power integers m h to minus m lo power to wo rational number bnjaega to wo follow hi ni krega
(z,/) does not even follow closure property ..so it's not a semi group
Z,/ does not a semi-group
they does not follows a closure property
and does not follows a associative property
Does not follow the closure property
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Hlo sir please ring k question v upload krdo ring with unity vale