In physics, the k = 8 eightfold way is an organizational scheme for a class of subatomic particles known as hadrons that led to the development of the quark model. Working alone, both the American physicist Murray Gell-Mann and the Israeli physicist Yuval Ne'eman proposed the idea in 1961. Wikipedia
+Matthew Salomone I didn't understand why adding two elements: i and sqrt(2) to the rationals is simpler than just adding one element: zeta(8). As you say, you're generating isomorphic fields in both cases.
What specifically did you need? Have you tried watching the videos in Matt's "Exploring Abstract Algebra II" playlist in order? ua-cam.com/play/PLL0ATV5XYF8DTGAPKRPtYa4E8rOLcw88y.html&feature=shared
Something isn't clicking for me in this video. I don't see how adjoining a root of unity to Q is any "simpler" or adds "less" non-real numbers than just adjoining i to Q. I get that powers of the roots of unity just spiral around the unit circle and don't grow/shrink w/o bound, but since we are in a field we also have any rational multiple of that root of unity, and powers of that would be able to grow/shrink w/o bound. And whether we've adjoined (1+i), (1+i)/2, or (1+i)/sqrt(2), we still get that i itself must be in the extended field so I don't see how we've added "the bare minimum" non-reals to Q. In fact, Q(1+i) = Q( (1+i)/2 ) = Q(i), whereas Q( 1+i)/sqrt(2) ) = Q(i, sqrt(2)), so if anything this extension is less "simple" than just adjoining i to Q, and we still have both (1+i) and (1+i)/2 in Q( (1+i)/sqrt(2) ). What am I missing here??
The goal was never to merely add i; it was to add a specific root. In the example "a=1+i", imagine "a" is some root, then by definition of the field every multiplication of "a" needs to be part of the field as well. That's the spiraling part as you seem to understand; these numbers never modulate to the residue classes as the 8th root of unity does. This is presented as an observation: if you want to adjoin new numbers you can minimalize the set required by making sure they "loop around". I think part of your confusion simply lies in the fact you didn't really understand the nature of the example.
I'm not exactly sure what you mean, but that sequence indeed gets as close as you want it to zero (because its modulus is smaller than 1). If you mean that there is a lower bound (0)...yeak ok but that was what he meant
You're right - I was mistaken. Despite reading (1+i/2)^n and typing it that way, I somehow managed to equate it in my mind to 1+(i/2)^n or something similar that does not have a limit.
In physics, the k = 8 eightfold way is an organizational scheme for a class of subatomic particles known as hadrons that led to the development of the quark model. Working alone, both the American physicist Murray Gell-Mann and the Israeli physicist Yuval Ne'eman proposed the idea in 1961. Wikipedia
Excellent explanation. I don't understand how there are only 140 thumbs up for such a brillant math video !
Because the prerequisite knowledge needed to understand this involves 2 years of college level math.
64 people over the past year (myself included) have given it another thumbs up!
Very much appreciate your videos. Your explanation of abstract algebra concepts have been the best I've found.
Glad to hear it, thanks!
a Real colorful spectacle!
Convergent (spiralling inwards) is dual to divergent (spiralling outwards).
It is not often I have to slow down my videos because the instructor is speaking too quickly. Brilliant!
Fantastic explanation, thank you so much!
You Sir have already helped me tremendously. Thank you
+Matthew Salomone I didn't understand why adding two elements: i and sqrt(2) to the rationals is simpler than just adding one element: zeta(8). As you say, you're generating isomorphic fields in both cases.
Great Video. Thanks you so much.
brilliant.
thanks!
once again sir , thank you for your efforts
Cyclotomic fields? More like "Cool videos, and hilarious puns that bring the feels!" 🤣👍
You have so many lectures, I can't find what I need.
What specifically did you need? Have you tried watching the videos in Matt's "Exploring Abstract Algebra II" playlist in order?
ua-cam.com/play/PLL0ATV5XYF8DTGAPKRPtYa4E8rOLcw88y.html&feature=shared
@@PunmasterSTP 3d curve derivative
Beautiful
I want to see why these numbers are closed fields under addition and multiplication. Show the lattice please. I've had some trouble with this.
sir , you are a very good teacher ..... i seriously appreciate your efforts.
Something isn't clicking for me in this video. I don't see how adjoining a root of unity to Q is any "simpler" or adds "less" non-real numbers than just adjoining i to Q. I get that powers of the roots of unity just spiral around the unit circle and don't grow/shrink w/o bound, but since we are in a field we also have any rational multiple of that root of unity, and powers of that would be able to grow/shrink w/o bound. And whether we've adjoined (1+i), (1+i)/2, or (1+i)/sqrt(2), we still get that i itself must be in the extended field so I don't see how we've added "the bare minimum" non-reals to Q.
In fact, Q(1+i) = Q( (1+i)/2 ) = Q(i), whereas Q( 1+i)/sqrt(2) ) = Q(i, sqrt(2)), so if anything this extension is less "simple" than just adjoining i to Q, and we still have both (1+i) and (1+i)/2 in Q( (1+i)/sqrt(2) ).
What am I missing here??
The goal was never to merely add i; it was to add a specific root. In the example "a=1+i", imagine "a" is some root, then by definition of the field every multiplication of "a" needs to be part of the field as well. That's the spiraling part as you seem to understand; these numbers never modulate to the residue classes as the 8th root of unity does.
This is presented as an observation: if you want to adjoin new numbers you can minimalize the set required by making sure they "loop around". I think part of your confusion simply lies in the fact you didn't really understand the nature of the example.
can we plot the nth roots on Latex by the value of n ?
I don't think that [(1+i)/2 ]^n shrinks "without bound".
I'm not exactly sure what you mean, but that sequence indeed gets as close as you want it to zero (because its modulus is smaller than 1). If you mean that there is a lower bound (0)...yeak ok but that was what he meant
You're right - I was mistaken. Despite reading (1+i/2)^n and typing it that way, I somehow managed to equate it in my mind to 1+(i/2)^n or something similar that does not have a limit.
Bu that still has a limit xd. It will go to 1 for n to infinity (because the second term goes to zero). It doesn't go to zero if you mean that