To click on "Thombs Up" in order to indicate your acknowledgement of your satisfaction about the lecture in this and other videos of Dr. Salomone is like giving a drop for the ocean of satisfaction that he deserves. I wish there was no limit of clicking on that "Thombs Up" icon so that I could click on it 10 ^ 23, Avogadro's number of times. Thank you Matthew.
06:48 So let me get this straight, and please correct me if I'm wrong: When we _don't_ have the identity element in our set, there also _cannot_ be any inverses in our set, right? Because if there was at least _one_ element that _has_ an inverse, we could take those two and combine together with our operation, and it would (by definition of inverses) produce the identity element, and this element _must_ be included in our set too, because otherwise we wouldn't have _closure_ for our operation. Right? But now let's suppose we _have_ the identity element in our set. This, in turn, _requires_ that there is at least _one_ element that has inverse, right? Because this identity element would be produced by our operation taking elements from _outside_ of our set as input, which also is wrong - it must take them from our set. So whatever it takes to produce the identity element, it must be in our set: both the element and its inverse. So having the identity in our set requires at least _one_ inverse to exist. There's no way that we have the identity but no inverses. Merely by the fact that the identity itself is its own inverse, always. So at lest _this_ element is guaranteed to be invertible. Others - not necessarily so. So now the only questions that remain are: whether we have any _other_ inverses besides the identity element, and whether _every_ element is invertible or just a few of them. For the set to be a group, we need _all_ of them to be invertible, not just _some_ of them. If only a few are invertible, or just the identity alone, we have a Monoid. But it's perfectly fine to have the identity and yet not all elements to be invertible. However, there's no way to have inverses and not have the identity, because that would break the closure. Correct? Then there are those zero divisors. From what I see, if all elements have inverses, there is no way to also have any zero divisors, because there's no one left in the set to play that role. But what if not every element has an inverse? Does it imply that there must be at least one zero-divisor? Or is it perfectly fine for those non-invertible elements to _not_ be zero divisors either? In other words: does the existence of non-invertible elements imply the existence of zero-divisors, or not?
By ring definition, + and x have their own substructures within a ring. Also, x structure makes no claims of requiring inverse as part of the x operation. So while + identity is an element of the set R, the monoid (R, x) has no requirement that the + identity element to have an multiplicative inverse. IE division by zero is not part of being a ring structure or substructure monoid, let alone any other multiplicative inverse.
This is almost certainly the only video on youtube that combines Beyoncé and abstract algebra. Thank you!
I just now found Matt's channel 11 years later, and sometimes it feels like I'm doing a bit of archeology. This is awesome!
To click on "Thombs Up" in order to indicate your acknowledgement of your satisfaction about the lecture in this and other videos of Dr. Salomone is like giving a drop for the ocean of satisfaction that he deserves. I wish there was no limit of clicking on that "Thombs Up" icon so that I could click on it 10 ^ 23, Avogadro's number of times. Thank you Matthew.
Holy mol-e that's a lot of times!
0:33 🤣This channel is solid gold!💍
06:48 So let me get this straight, and please correct me if I'm wrong:
When we _don't_ have the identity element in our set, there also _cannot_ be any inverses in our set, right? Because if there was at least _one_ element that _has_ an inverse, we could take those two and combine together with our operation, and it would (by definition of inverses) produce the identity element, and this element _must_ be included in our set too, because otherwise we wouldn't have _closure_ for our operation. Right?
But now let's suppose we _have_ the identity element in our set. This, in turn, _requires_ that there is at least _one_ element that has inverse, right? Because this identity element would be produced by our operation taking elements from _outside_ of our set as input, which also is wrong - it must take them from our set. So whatever it takes to produce the identity element, it must be in our set: both the element and its inverse. So having the identity in our set requires at least _one_ inverse to exist. There's no way that we have the identity but no inverses. Merely by the fact that the identity itself is its own inverse, always. So at lest _this_ element is guaranteed to be invertible. Others - not necessarily so.
So now the only questions that remain are: whether we have any _other_ inverses besides the identity element, and whether _every_ element is invertible or just a few of them. For the set to be a group, we need _all_ of them to be invertible, not just _some_ of them. If only a few are invertible, or just the identity alone, we have a Monoid. But it's perfectly fine to have the identity and yet not all elements to be invertible. However, there's no way to have inverses and not have the identity, because that would break the closure. Correct?
Then there are those zero divisors. From what I see, if all elements have inverses, there is no way to also have any zero divisors, because there's no one left in the set to play that role. But what if not every element has an inverse? Does it imply that there must be at least one zero-divisor? Or is it perfectly fine for those non-invertible elements to _not_ be zero divisors either? In other words: does the existence of non-invertible elements imply the existence of zero-divisors, or not?
By ring definition, + and x have their own substructures within a ring. Also, x structure makes no claims of requiring inverse as part of the x operation. So while + identity is an element of the set R, the monoid (R, x) has no requirement that the + identity element to have an multiplicative inverse. IE division by zero is not part of being a ring structure or substructure monoid, let alone any other multiplicative inverse.
If my other answer is insufficient, consider 42.
I nearly swallowed my tongue when Prof Matt dances Beyoncé! LOL
NOTE: Creator is not responsible for any joke-induced injuries. :)
Sir, I find your videos extraordinary
Can I get information about the song which was played at beginning of the video, please?
Put a ring on it (Beyoncé)
Great explanations!!
thanks for clarification
Awesome
please never do that again xD
just kidding. Glad youre having fun and learning