I previously had a video on this, but I used a needlessly complicated definition which made the lesson a few minutes longer than it needed to be. So I redid it.
Thanks for the videos; they are immensely helpful!!! Question about the example presented at 4:40: The way I look at it is: Cycle 1: Start with 1: 1 --> 6 (1) 6 --> 4 (2) 4 -> 2 (3) 2 -> 1 (4) I started with 1 and ended with 1 so to get back to 1, I had to make 4 “jumps” so 4 is a possible answer for our order. Cycle 2: Skip all the numbers covered in Cycle 1 so I start with 3. 3 --> 3 (0) Since 3 maps to itself, I do not consider a “jump”. 0 would be a possible answer but since it is non-positive, I can eliminate it as a possibility. Cycle 3: 5 --> 5 (0) Again, not a “jump” because 5 maps to itself. Again, 0 can be eliminated as a possibility since it is non-positive. So that is why the order is 4. Is this way of thinking correct or is there something important I’m missing? Thank you!!
If e is the identity element of G, and e is also a power of some a ∈ G, then can we say ord(e) = 0, or is it ord(e) = 1 because the order of an element is strictly for non-zero positive integers? Edit: Nevermind I watched the next video
I was wondering why not 0 for the order, and the reason it is not 0 is because every element to the 0th power is the identity element, so it has to be a positive integer to be non trivial
Hi W.O.M Would love some videos on ramsey theory! Could be a great fit in your graph theory playlist, or just within a general combinatorics theme. There is definitely room for a more intuitive explanation on UA-cam. Love the Abstract algebra vids.
am i the only one who couldn't understand this 😭😭? I'm feeling very dumb, i watched it so many times and still can't get it well idk why🥹!! ( I'm a math major)
I previously had a video on this, but I used a needlessly complicated definition which made the lesson a few minutes longer than it needed to be. So I redid it.
Thanks for the videos; they are immensely helpful!!!
Question about the example presented at 4:40:
The way I look at it is:
Cycle 1:
Start with 1:
1 --> 6 (1)
6 --> 4 (2)
4 -> 2 (3)
2 -> 1 (4)
I started with 1 and ended with 1 so to get back to 1, I had to make 4 “jumps” so 4 is a possible answer for our order.
Cycle 2:
Skip all the numbers covered in Cycle 1 so I start with 3.
3 --> 3 (0)
Since 3 maps to itself, I do not consider a “jump”.
0 would be a possible answer but since it is non-positive, I can eliminate it as a possibility.
Cycle 3:
5 --> 5 (0)
Again, not a “jump” because 5 maps to itself.
Again, 0 can be eliminated as a possibility since it is non-positive.
So that is why the order is 4.
Is this way of thinking correct or is there something important I’m missing?
Thank you!!
If e is the identity element of G, and e is also a power of some a ∈ G, then can we say ord(e) = 0, or is it ord(e) = 1 because the order of an element is strictly for non-zero positive integers?
Edit: Nevermind I watched the next video
I was wondering why not 0 for the order, and the reason it is not 0 is because every element to the 0th power is the identity element, so it has to be a positive integer to be non trivial
Thank you.
Glad to help!
Good explanation! Keep going, You do good job.
Thanks, will do!
So helpful thanks
Thanks for watching!
Thank u so much 🖤
You're welcome!
very interesting
I think so too!
You can never go wrong with Wrath of Math!
Such is the order of things!
The frog is the icing on the cake.
It always is!
i love you
Hi W.O.M
Would love some videos on ramsey theory!
Could be a great fit in your graph theory playlist, or just within a general combinatorics theme.
There is definitely room for a more intuitive explanation on UA-cam.
Love the Abstract algebra vids.
am i the only one who couldn't understand this 😭😭? I'm feeling very dumb, i watched it so many times and still can't get it well idk why🥹!! ( I'm a math major)