The approach you took at the beginning of the explanation is what made this video different from all others. Sometimes persons start with the abstract and then lead to the examples by which time the viewer is completely confused. However, you took the difference approach and i understood perfectly. Thank you.
Wonderful.... Video helped me a lot in understanding the topic.... Thank you sir... For...providing Sach a easy way to understood these tough topics of math's ....
Excellent series. Thank you. And thank you especially for noting that many authors and mathematicians are sloppy with their notion when they fail to include the binary operator in the group name or definition. That confused me for so long.
THANK YOU VERY MUCH. But i think IS A GROUP. At times the inverse of an element can be the element itself mostly from the cayley's table But if am wrong you can help me too
I have a question on 11:00 you said it's not a group because zero is part of the set but why is the first one considered a group if zero is also part of the set and zero doesn't have an inverse, so doesn't that mean it shoudn't be a group?
Does this mean that when we consider a group of the set of rational numbers we consider subsets as well? In the example rational numbers under addition: if we add (1/2 + 1/2) we get an integer. Is it correct to say that the set of integers is a subset of the rational numbers therefore it does not violate closure under addition?
bizarre? think about a civilization that doesn't know abstract algebra. Would throw us back technologically into the early twentieth century, I would guess
@@kenmeyer100 : I am not talking about regular algebra as in algebra 1 and 2 or college algebra or precalculus, etc. I am talking about a course called abstract algebra usually taught in graduate school.
