At 0:45 you have given an example the set G = {1,2,3} and binary operation + : GxG -> G (the ordinary addition). But (G, +) is not a group, since the identity element of addition 0 is not an element of G. Furthermore, for x=1, y=3 in G, x+y = 1 + 3 = 4 is not in G. Even the inverse of x = 1 in G , i.e. x⁻1 = -1 is not in G.
I would suggest another example : G={2*k : k integer } = {..., -4,-2.0,2,4,...} \subset ZZ .Then (G,+) is a group
And since it equals e then we know that (y^-1.x^-1) is the inverse of (x.y). So if we get the inverse of (x.y) ==> (x.y)^-1 then it is equal to (y^-1.x^-1)
and so (x.y)^-1=(y^-1.x^-1)
I hope that made sense if not then i will do a video for you
Define ''closure'' and ''operations''.
I believe the operation might be a function which actually may contain several operations in the ordinary arithmetic sense.
wlwak 1 year ago
At 0:45 you have given an example the set G = {1,2,3} and binary operation + : GxG -> G (the ordinary addition). But (G, +) is not a group, since the identity element of addition 0 is not an element of G. Furthermore, for x=1, y=3 in G, x+y = 1 + 3 = 4 is not in G. Even the inverse of x = 1 in G , i.e. x⁻1 = -1 is not in G.
I would suggest another example : G={2*k : k integer } = {..., -4,-2.0,2,4,...} \subset ZZ .Then (G,+) is a group
radocool1 2 years ago
Thanks, Dr.
The prove is in order.
Thanks for posting..
I will continue watching your video..
whotookmynickkokokai 2 years ago
how do you prove (x.y)-1= y-1.x-1?
whotookmynickkokokai 2 years ago
Sorry for late reply i was away for a week.
first let us prove something else:
(x.y).(y^-1.x^-1)=x.((y.y^-1).x^-1)
=x.(e.x^-1)
=x.x^-1
=e
You should be able to follow that
And since it equals e then we know that (y^-1.x^-1) is the inverse of (x.y). So if we get the inverse of (x.y) ==> (x.y)^-1 then it is equal to (y^-1.x^-1)
and so (x.y)^-1=(y^-1.x^-1)
I hope that made sense if not then i will do a video for you
burny1 2 years ago