@Hugh Jones Most RUclipsrs have scripts, even if they write the video themselves. But in this case, it's a "well-known fact" that Liliana de Castro is an actor and that someone else writes the scripts for these videos. You can even look in the video description: "Teaching Assistant: Liliana de Castro Written & Directed by Michael Harrison Produced by Kimberly Hatch Harrison"
@@MuffinsAPlenty Thanks for sharing our credits! It's kind of funny how few people read the video descriptions - it's all in there! Lots of important info! 💜🦉
@@Socratica Indeed, IMO another important info is "Milne, Algebra Course Notes (available free online)": I've looked at the notes e.g. about Galois theory and at first sight this looks quite well explained, thank you also for it and of course for this superb video about integral domain!
I think this lecture is very well grounded; it’s got a lot of roots! In all seriousness, this is another phenomenal video, and I do plan on binge-watching the rest of the abstract algebra playlist today.
They were talking about zero divisors in my linear algebra course and it makes so much more sense now after this video I'd love to have a socratica video series about dual numbers
I was reading a book on abstract algebra that explained integral domains and it completely left out the part about the usage of modular arithmetic. Totally didn't understand. This makes it very clear, thank you.
I always revisit the series before any of my abstract algebra papers for basic definition clarity! Would love to see more being uploaded. Kudos to the author!
Liliana, these videos are so good, compressing down the succinct core knowledge on a topic into a few minutes, just what's needed to dip in and update as required.
Good explanation !, The book my university gave me just says "It's a ring in which a.b=0 implies that either a or b is zero" which is true but misses the whole point of defining a structure smh Also wanted to add that in some books "The integers mod n" are noted as Zn (The n being a subscript)
(5:42) Why doesn't the factoring and cancelling method require division? Isn't cancelling a number in both sides of an equation the product of doing multiplication of the original number by it's multiplicative inverse (and isn't this the essence of division in the first place)? Edit: Never mind, the last part of the video solved my doubts. I didn't realize that, being a common property of all rings, distribution could be used to cancel a common factor from both sides of any equation.
Although you figured it out, your intuition isn't bad. From any integral domain R, you can construct its field of fractions F, which is a field containing R as a subring and the inverses of every nonzero element of R (this works because R has no (nonzero) zero-divisors). If you have an equation xy = xz in R (with x not equal to 0), then it must still be valid in F (since R is a subring of F). Hence, you can multiply by x^-1 in F to get the equation y = z. But this is an equation consisting of elements in R, so it is also valid in R (since R is a subring of F). In general, given any commutative ring R in which x is a non-zero-divisor, if you know xy = xz, then you can conclude y = z, just from the properties of non-zero-divisors. But, since x is a non-zero-divisor, you can also find a ring S containing R as a subring in which x is a unit. Hence, you can use a "division" argument like you suggest to get the same result. So there is an interesting connection there anyway.
loving the videos here -- are there any notable groups/rings/etc with zero divisors other than groups of modular arithmetic? it seems that most of the caveat examples in this series' videos up to now come from modular arithmetic
Consider the (noncommutative) ring R of nxn matrices whose entries are real numbers. Every matrix with determinant 0 is a zero divisor in the ring. If you happen to have an idempotent element in your ring (an element e so that e^2 = e) which is not 0 or 1, then that element is guaranteed to be a zero divisor (e*(1-e) = 0). This is true in Boolean rings, for example, since the algebraic structure resembles that of Boolean logic. A lot of other examples come from performing operations on rings. For example, if you take the direct product of two rings, elements of the form (a,0) and (0,b) are zero divisors. Or if you take any ring R and form the polynomial ring R[x], then if R has zero divisors, it is possible that R[x] could have zero divisors too. Or if you have a ring R with zero divisors, then taking the ring of nxn matrices whose entries in R will have zero divisors as well (all matrices whose determinants are zero divisors in R). Or taking a quotient ring of a ring if you're modding out by an ideal that isn't prime (but this is like modular arithmetic).
We generally also require an integral domain's 1 element to be distinct from 0. In other words, we don't want the 0 ring to be classified as an integral domain, since we want an ideal I to be prime if and only if R/I is an integral domain.
Integral Domains were interesting to me because it's the minimum requirement for greater than or less than to be defined. Beneath that everything gets wibbly wobbly and is just there. Like colors, there's no way to define which one is greater or less than, but once you define color by it's frequency of light, you get the ROYGBIV we're familiar with red being the lower end and violet being the upper.
At 4:46, do you mean "1" represents the multiplicative identity and "0" represents the additive identity? Because the members of the ring may not be numbers at all.
1. change the question to "how many unique solutions". And we do not have to worry about the "zero divisors" because it's quadratic. zero divisors like accidental solutions. So it gives 12C1 + 12C2 = 78 there should be 78 sets of different solutions, which gives 78 unique equations. 2. 12 = 2x2x3 2x6, 3x4, 3x8, 4x6, 4x9, 6x8, 6x10
mod 11 is also a field. It has multiplicative inverses which allows cancellation. Is there an example where no-zero-devisors alone makes cancellation possible?
I'm not sure if I'm understanding your question. Are you asking for an example of an integral domain which is not a field? If so, one example is the ring of integers, Z. The only units are 1 and -1, so it's not a field. But Z is an integral domain. In the ring of integers, if you have 2a = 0 for some element a in Z, it must be the case that a = 0, even though 2 is not invertible. You cannot find a finite example, though. In a finite commutative ring, every nonzero element must either be a unit or a zero-divisor. Therefore, all finite integral domains must be fields.
never studied abstract algebra, but I think this can be easily solved by setting the right hand side of the equation to 12n where n is an integer. so we can solve 2 equations, first when n=0 and then when n=1 so we get multiple solutions to the modular equation.
Finally, if a is divisible by 6, then the equation reduces to 6x^2 + bx + c ≡ 0 (mod 12), which has the same number of solutions as 3x^2 + bx + c ≡ 0 (mod 6), which we have already counted. Therefore, the total number of distinct quadratic equations modulo 12 is 144 + 864 + 72 = 1,080.
Should we make the distinction between trivial zero divisors, and non-trivial zero divisors? Since 0*0=0, or 0*2=0, etc., , 0 divides 0, and we had nonzero things times 0 giving 0, so 0 is a zero divisor. Also, any nonzero number like 1 satisfies 1*0=0, so 1 divides 0. However wikipedia says 1 needs to be multiplied by something nonzero and equaling 0 to be called a zero divisor, so 1 is not a zero divisor. So, 0 is a trivial zero divisor, and any nonzero element a dividing 0 with something nonzero (ie, ax=0 with x != 0) is a non-trivial zero divisor. When you say "zero divisor" it makes sense to mean "non-trivial zero divisor".
Different textbooks have different conventions on what "zero divisor" means. In areas of study where the focus is on integral domains (such as number theory), they like to exclude 0 from the definition of zero divisor because it makes it easier to define an integral domain (an integral domain is a commutative unital ring with 0 ≠ 1 and having no zero divisors). However, if you are working in a branch of mathematics which allows rings which are not integral domains, the typical definition is what Wikipedia says, and 0 is considered a zero divisor in nonzero rings. (The theory involving zero divisors is much easier to state if one considers 0 to be a zero divisor in such settings.)
Morning coffee, in the warm sun of May, lost nowhere in a village of french hilly countryside, surrounded by birds singing and soft wind in the trees, I was watching randomly your video, listening from one ear only. At 0:37 then you say "in abstract algebra, this isn't always true". I instantly stopped breathing for a second, then my heart went fast, starting to feel dizzy, sweating, thrills... I almost fell off my coffee mug on the ground. Then I said to myself, no, it's wrong, she must be wrong, how possible ? Then I watched the rest of the video, and now my whole world collapsed. I don't know what to do with this now. I feel anger cause I wasn't prepared for this. Shouldn't you have made a warning sign ? Hopefully I was sitting. I kind of understand how Pythagorician could feel at the time Hippase de Metaponte came to demonstrate the irrationality of square root of 2 xD Then hopefully the little stand-up comedy funny end was making all of my sadness disappear :D And now the birds sings sounds different, the wind seems way more mysterious and the sun tans my face with a new kind of light. You definitely change the world, at least mine :) Thanks so much for your videos Keep on !
There are 11*12*12 different quadratics, if you mean just different coefficients and not the same functions x^2 + 2x + 4 is an example of a quadratic with only 2 solutions mod 12 (2 and 8)
What they did is this: A quadratic will be of the form ax^2 + bx + c, and we require a to be non-zero so that it is a quadratic. Now, this means that for a there are 11 choices *mod 12* (1,2,3,4,5,6,7,8,9,10,11). b and c can be zero if we want, so there are 12 choices for them mod 12 (0,1,2,3,4,5,6,7,8,9,10,11). Now, 11 choices for a, 12 for b, 12 for c gives a total of 11*12*12 possible quadratic equations mod 12. Now I am not sure if this is the correct answer, I wouldn't be able to come up with anything else. I suppose that we might consider two equations different mod 12 if there is an input (x value) which gives a different output for each equation.
I asked the question on Mathematics Stack Exchange and here's the answer: ( it's (12^3)/2 ) math.stackexchange.com/questions/3185444/number-of-different-quadratic-functions-mod-12
Fields (and integral domains) have one (and only one) zero-divisor: zero. The video is good but misleading as says that domains (like Z_{11}) have no zero divisors, which is false.
There are different conventions for handling the trivial case of 0. For example, Serge Lang defines them as "Elements x, y of A are said to be zero divisors if x!=0, y!=0, and xy = 0." Richard Dean at Caltech used the same convention. So you'll frequently hear mathematicians say certain rings do not have zero divisors. Thanks for watching!
Socratica, do you know if there is a reason why Lang and Dean prefer that 0 not be considered a zero-divisor? I understand why in commutative algebra we consider 0 to be a zero-divisor (except in the 0 ring): the set of zero divisors is the union of the associated prime ideals of the ring. If you don't want to consider 0 as a zero-divisor, then you have to remove 0 from this union. So it makes things slightly easier to consider 0 as a zero-divisor in nonzero rings if you care about associated prime ideals. Is there something similar for excluding 0 from the set of zero-divisors, or is it just to make the definition of an integral domain simpler?
@@MuffinsAPlenty I think it ties to making the definition for cancellation more clear. You don't want x to be zero and neither y, so that when xy=0 then you can definitely say whether you have zero divisors. If you include zero, then it becomes trivial and muddles it up a bit. If you count zero, then you lose some detail in the definitions. I mean, I'm here trying to figure out how to do my homework though LOL! So I may be completely wrong, but that's how I interpreted it.... take my comment with a grain of salt :P
@@Taricus Very delayed response to you, but I think you're pretty much right! In number theory, most of the rings they care about are integral domains. So, if you exclude 0 from the classification of zero-divisor, it makes it easier to define integral domains, which is pretty much all they care about anyway. In some sense, they are like, "Hey, in arbitrary ring theory, this bad thing will happen. But we won't consider any such situation." However, more general ring theory (such as commutative algebra) really does want to consider rings with zero-divisors. The theory of zero-divisors is cleaner when 0 is considered to be zero-divisor (except in the 0 ring). So in this context, it makes sense to consider 0 to be a zero-divisor. It's sort of a matter of convenience. If you want your study to include rings with zero-divisors, it's easier to consider 0 to be a zero-divisor (except in the 0 ring); but if your study only includes integral domains, then the only reason you care about zero-divisors is to say "bad thing doesn't happen", and it gives a cleaner definition of integral domain.
I really liked the video, but it would be much better if it was little bit slower :) I had to stop n play again lots of times cuz it was too fast for my brain lol
I already watched this before our class in Abstract Algebra started but it turns out that her explanation and examples are similar to what is given here so I left the meet and watched this video again. Srlsly, I don't understand my prof's explanation.
Nope! There are not infinitely many quadratic polynomials modulo 12. A quadratic polynomial is of the form ax^2 + bx + c. All polynomials are determined by their coefficients. So choosing the values of a, b, and c determine the polynomial. Usually, we have infinitely many values of a, b, and c to choose from. How many values can we choose for a, b, and c if we're dealing with integers modulo 12?
Ok.. i get it now. Y'know since the 2nd question ask about solutions, i thought the 1st answer being infinitely many but just that not all of which has solution, kek
It seems to me that the cancellation property should hold for non-integral domains. It doesn't require commutation to work, just the lack of zero-divisors... or maybe I'm missing something.
You are correct! You can get cancellation pretty generally provided that the thing you're trying to cancel is not a zero-divisor of any kind. Commutativity is not required. For example, if you have a non-commutative ring and know xy = xz, where x is not a left zero-divisor (meaning that xw = 0 implies w = 0), then you can conclude that y = z. Similarly, if you know yx = zx and x is not a right zero-divisor (meaning wx = 0 implies w = 0), then you can conclude y = z. Of course, without commutativity, you can no longer conclude that xy = zx implies that y = z, even if x is not a zero-divisor (left, right, or two-sided).
I think it depends on the definition of each author. Sometimes adding some properties simplify proofs in the whole theory. That's because removing some properties could lead to not interesting cases that doest not matter for the theory purposes and just complicate the proofs.
Thank you! Could anyone be kind to tell me what's the solution to the 2 puzzles? My answers correct(?: *FALSE*): there exist (12^2-12)/2 + 12 = 78 quadratic equations; the number of those who have only 2 solutions is 78 - 4*12 = 30, where 12 eqns out of 30 are in the form of (x-a)^2 = 0 (that is a = b), while for the other 18 eqns, (x-a)*(x-b) = 0 (where a!=b). Could anyone tell me the correct answer? (I'm *WRONG*)
@@gundamlh Could you explain how you found this please? I've been trying to practice with smaller values such as mod 3, mod 4, etc, but I have yet to find a good way to calculate everything short of just manually checking every possible quadratic equation.
Wow, very interesting ! Bet this is at the core of some cryptographic/decryption methods. Regardless, no way to really understand w/o trying a bunch of related basic problems in these topics, bet it won't take long to get totally stuck/ confused ;-)
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ples madm you sport me
Whoever wrote the scripts for this playlist did an absolutely outstanding job.
@Hugh Jones Most RUclipsrs have scripts, even if they write the video themselves.
But in this case, it's a "well-known fact" that Liliana de Castro is an actor and that someone else writes the scripts for these videos. You can even look in the video description:
"Teaching Assistant: Liliana de Castro
Written & Directed by Michael Harrison
Produced by Kimberly Hatch Harrison"
@@MuffinsAPlenty Thanks for sharing our credits! It's kind of funny how few people read the video descriptions - it's all in there! Lots of important info! 💜🦉
@@Socratica Indeed, IMO another important info is "Milne, Algebra Course Notes (available free online)": I've looked at the notes e.g. about Galois theory and at first sight this looks quite well explained, thank you also for it and of course for this superb video about integral domain!
I think this lecture is very well grounded; it’s got a lot of roots! In all seriousness, this is another phenomenal video, and I do plan on binge-watching the rest of the abstract algebra playlist today.
They were talking about zero divisors in my linear algebra course and it makes so much more sense now after this video
I'd love to have a socratica video series about dual numbers
I was reading a book on abstract algebra that explained integral domains and it completely left out the part about the usage of modular arithmetic. Totally didn't understand. This makes it very clear, thank you.
Why does abstract algebra seem so much simpler this way? Love this. Maybe you could add one on the definition of an algebra.
These are all wonderful videos, brilliantly narrated, keep up the good work
Your channel deserves a lot more attention.
I always revisit the series before any of my abstract algebra papers for basic definition clarity! Would love to see more being uploaded. Kudos to the author!
Just great! I just discovered your videos, and it is fun to watch them!
We're so glad you've found us! Thanks for watching! :)
Liliana, these videos are so good, compressing down the succinct core knowledge on a topic into a few minutes, just what's needed to dip in and update as required.
Thanks for adding a new video about Abstract Algebra...Plz upload more videos on Pure Mathematics...
Your explanation technique is great...
Good explanation !, The book my university gave me just says "It's a ring in which a.b=0 implies that either a or b is zero" which is true but misses the whole point of defining a structure smh
Also wanted to add that in some books "The integers mod n" are noted as Zn (The n being a subscript)
You are my all time favourite Teacher.
The video is good throughout, but the end part alone is worth a thumbs up.
Really best video for abstract algebra: the instructor explains clearly in reasoning
(5:42) Why doesn't the factoring and cancelling method require division? Isn't cancelling a number in both sides of an equation the product of doing multiplication of the original number by it's multiplicative inverse (and isn't this the essence of division in the first place)? Edit: Never mind, the last part of the video solved my doubts. I didn't realize that, being a common property of all rings, distribution could be used to cancel a common factor from both sides of any equation.
Although you figured it out, your intuition isn't bad. From any integral domain R, you can construct its field of fractions F, which is a field containing R as a subring and the inverses of every nonzero element of R (this works because R has no (nonzero) zero-divisors). If you have an equation xy = xz in R (with x not equal to 0), then it must still be valid in F (since R is a subring of F). Hence, you can multiply by x^-1 in F to get the equation y = z. But this is an equation consisting of elements in R, so it is also valid in R (since R is a subring of F).
In general, given any commutative ring R in which x is a non-zero-divisor, if you know xy = xz, then you can conclude y = z, just from the properties of non-zero-divisors. But, since x is a non-zero-divisor, you can also find a ring S containing R as a subring in which x is a unit. Hence, you can use a "division" argument like you suggest to get the same result.
So there is an interesting connection there anyway.
this is like the 5th of your vids so far, thanks for helping me pass my algebra exam :D
loving the videos here -- are there any notable groups/rings/etc with zero divisors other than groups of modular arithmetic? it seems that most of the caveat examples in this series' videos up to now come from modular arithmetic
Consider the (noncommutative) ring R of nxn matrices whose entries are real numbers. Every matrix with determinant 0 is a zero divisor in the ring.
If you happen to have an idempotent element in your ring (an element e so that e^2 = e) which is not 0 or 1, then that element is guaranteed to be a zero divisor (e*(1-e) = 0). This is true in Boolean rings, for example, since the algebraic structure resembles that of Boolean logic.
A lot of other examples come from performing operations on rings. For example, if you take the direct product of two rings, elements of the form (a,0) and (0,b) are zero divisors. Or if you take any ring R and form the polynomial ring R[x], then if R has zero divisors, it is possible that R[x] could have zero divisors too. Or if you have a ring R with zero divisors, then taking the ring of nxn matrices whose entries in R will have zero divisors as well (all matrices whose determinants are zero divisors in R). Or taking a quotient ring of a ring if you're modding out by an ideal that isn't prime (but this is like modular arithmetic).
@@MuffinsAPlenty wow this is awesome, thank you!
Love your way of explanation from India Ma'am . 👍🏻
We generally also require an integral domain's 1 element to be distinct from 0. In other words, we don't want the 0 ring to be classified as an integral domain, since we want an ideal I to be prime if and only if R/I is an integral domain.
I was taught about integral domains in our college, but this really made me understand the idea behind these.... Great Video
Good videos. Can you upload Topology videos?
Love your channel! Glad to know you are uploading new videos. Could you please talk more about commutative rings?
This is quite clear and easy to understand.
THANK YOU .GOT TO LEARN ALOT OF NEW THINGS
Innovative lectures I liked it so much
nice explanation mam.....i have studied before but never understood......this example makes me clear all doubt about integral domain ......thnxx mam
😂😂, I love this woman, you are superb. God bless you very much for helping some of us
Integral Domains were interesting to me because it's the minimum requirement for greater than or less than to be defined. Beneath that everything gets wibbly wobbly and is just there. Like colors, there's no way to define which one is greater or less than, but once you define color by it's frequency of light, you get the ROYGBIV we're familiar with red being the lower end and violet being the upper.
Your videos are very well done and helpful. Keep it up!
Always I am interested when I am watching videos u present!
I like it.
At 4:46, do you mean "1" represents the multiplicative identity and "0" represents the additive identity? Because the members of the ring may not be numbers at all.
these videos are very helpful,May I get more videos in ring theory by socratica,these are pretty particular abt topic..i need them
thanks for yet another video Socratica
I use these videos as motivation before I read my text THE SECOND TIME 😂 ❤ 😅
1. change the question to "how many unique solutions". And we do not have to worry about the "zero divisors" because it's quadratic. zero divisors like accidental solutions. So it gives 12C1 + 12C2 = 78
there should be 78 sets of different solutions, which gives 78 unique equations.
2. 12 = 2x2x3
2x6, 3x4, 3x8, 4x6, 4x9, 6x8, 6x10
Very informative and easy to follow. 9/10.
Bryan Villegas what is the deduction of 1 for?
mod 11 is also a field. It has multiplicative inverses which allows cancellation. Is there an example where no-zero-devisors alone makes cancellation possible?
I'm not sure if I'm understanding your question. Are you asking for an example of an integral domain which is not a field?
If so, one example is the ring of integers, Z. The only units are 1 and -1, so it's not a field. But Z is an integral domain. In the ring of integers, if you have 2a = 0 for some element a in Z, it must be the case that a = 0, even though 2 is not invertible.
You cannot find a finite example, though. In a finite commutative ring, every nonzero element must either be a unit or a zero-divisor. Therefore, all finite integral domains must be fields.
@@MuffinsAPlenty huh, make sense. Thanks!
never studied abstract algebra, but I think this can be easily solved by setting the right hand side of the equation to 12n where n is an integer. so we can solve 2 equations, first when n=0 and then when n=1 so we get multiple solutions to the modular equation.
Fantastic. If only my University could make videos like this
You are a wonderful teacher...
Thank you very much
Beautifully explained and laid out. I wish I had access to this material when I studied abstract algebra!
product of 2 terms is 0 so one must at least be 0...ya had to do that in LinAl...saw how beautiful Math can be....you are a wonderful teacher.
So what's the answer to the first puzzle? I only could come up with an upper bound of 11*12*12.
Here's the answer: math.stackexchange.com/questions/3185444/number-of-different-quadratic-functions-mod-12
@@Macieks300 the page doesn't open
@@gayecelebi8975 Copy the link manually.
Finally, if a is divisible by 6, then the equation reduces to 6x^2 + bx + c ≡ 0 (mod 12), which has the same number of solutions as 3x^2 + bx + c ≡ 0 (mod 6), which we have already counted. Therefore, the total number of distinct quadratic equations modulo 12 is 144 + 864 + 72 = 1,080.
You've made this interesting. Thanks.
I enjoyed that parody of "stand-up comedy" part....
Great series 👍. Thank you.
Should we make the distinction between trivial zero divisors, and non-trivial zero divisors? Since 0*0=0, or 0*2=0, etc., , 0 divides 0, and we had nonzero things times 0 giving 0, so 0 is a zero divisor. Also, any nonzero number like 1 satisfies 1*0=0, so 1 divides 0. However wikipedia says 1 needs to be multiplied by something nonzero and equaling 0 to be called a zero divisor, so 1 is not a zero divisor. So, 0 is a trivial zero divisor, and any nonzero element a dividing 0 with something nonzero (ie, ax=0 with x != 0) is a non-trivial zero divisor. When you say "zero divisor" it makes sense to mean "non-trivial zero divisor".
Different textbooks have different conventions on what "zero divisor" means. In areas of study where the focus is on integral domains (such as number theory), they like to exclude 0 from the definition of zero divisor because it makes it easier to define an integral domain (an integral domain is a commutative unital ring with 0 ≠ 1 and having no zero divisors).
However, if you are working in a branch of mathematics which allows rings which are not integral domains, the typical definition is what Wikipedia says, and 0 is considered a zero divisor in nonzero rings. (The theory involving zero divisors is much easier to state if one considers 0 to be a zero divisor in such settings.)
Mam please share class about unique factorisation domain
Best stand up Maths I ever understood! :-)
Un excelente video, me gustaría que hablaran un poco más de los diferentes tipos de Dominios como los de factorización única o los euclídeanos.
Please make a video on unique factorisation domain too
Great Video 👍👍
what a nice explanation ma'am. Lucid one ..
Morning coffee, in the warm sun of May, lost nowhere in a village of french hilly countryside, surrounded by birds singing and soft wind in the trees, I was watching randomly your video, listening from one ear only.
At 0:37 then you say "in abstract algebra, this isn't always true".
I instantly stopped breathing for a second, then my heart went fast, starting to feel dizzy, sweating, thrills... I almost fell off my coffee mug on the ground.
Then I said to myself, no, it's wrong, she must be wrong, how possible ? Then I watched the rest of the video, and now my whole world collapsed. I don't know what to do with this now. I feel anger cause I wasn't prepared for this. Shouldn't you have made a warning sign ? Hopefully I was sitting. I kind of understand how Pythagorician could feel at the time Hippase de Metaponte came to demonstrate the irrationality of square root of 2 xD
Then hopefully the little stand-up comedy funny end was making all of my sadness disappear :D
And now the birds sings sounds different, the wind seems way more mysterious and the sun tans my face with a new kind of light.
You definitely change the world, at least mine :)
Thanks so much for your videos
Keep on !
The world is more beautiful because you are in it, Socratica Friend! 💜🦉
If anybody wants a trick to do modular arithmetic fast, try using the Chinese Remainder Theorem
Theses videos are so much better. Indians are just obsessed with Definitions.
It ix humble request to u ....plx make vedio 1 hour ....even I can watch your your vedio 10 sec even 10 hours also....best of best explanation...
Good lecture
This is so much better!
Thank you. You helped me giving sense to such a kind of concept. I am a self learning wanna be mathematician.
Watch socratica instead of sleeping on professor's notes... Time saved
Big help, thank you!
There are 11*12*12 different quadratics, if you mean just different coefficients and not the same functions
x^2 + 2x + 4 is an example of a quadratic with only 2 solutions mod 12 (2 and 8)
hi, I'm confused, could you be kind to explain your solution a little bit?
What they did is this: A quadratic will be of the form ax^2 + bx + c, and we require a to be non-zero so that it is a quadratic. Now, this means that for a there are 11 choices *mod 12* (1,2,3,4,5,6,7,8,9,10,11). b and c can be zero if we want, so there are 12 choices for them mod 12 (0,1,2,3,4,5,6,7,8,9,10,11). Now, 11 choices for a, 12 for b, 12 for c gives a total of 11*12*12 possible quadratic equations mod 12. Now I am not sure if this is the correct answer, I wouldn't be able to come up with anything else. I suppose that we might consider two equations different mod 12 if there is an input (x value) which gives a different output for each equation.
11*12*12 : [939 (no solution), 144 (unique solution), 324 (just 2 solutions), 8 (have 3 solutions), 138, 0, 18, 0, 12, 0, 0, 0]
Yeah but some of those could be the same quadratic congruence.
I asked the question on Mathematics Stack Exchange and here's the answer: ( it's (12^3)/2 )
math.stackexchange.com/questions/3185444/number-of-different-quadratic-functions-mod-12
Thank you so much.
Thanks for amazing video
Plzzz ma'am tell me how to find zero divisor and unit in Qotient ring e.g R/i
Good teaching. Every one need example better
Fields (and integral domains) have one (and only one) zero-divisor: zero. The video is good but misleading as says that domains (like Z_{11}) have no zero divisors, which is false.
There are different conventions for handling the trivial case of 0. For example, Serge Lang defines them as "Elements x, y of A are said to be zero divisors if x!=0, y!=0, and xy = 0." Richard Dean at Caltech used the same convention. So you'll frequently hear mathematicians say certain rings do not have zero divisors. Thanks for watching!
Socratica, do you know if there is a reason why Lang and Dean prefer that 0 not be considered a zero-divisor?
I understand why in commutative algebra we consider 0 to be a zero-divisor (except in the 0 ring): the set of zero divisors is the union of the associated prime ideals of the ring. If you don't want to consider 0 as a zero-divisor, then you have to remove 0 from this union. So it makes things slightly easier to consider 0 as a zero-divisor in nonzero rings if you care about associated prime ideals.
Is there something similar for excluding 0 from the set of zero-divisors, or is it just to make the definition of an integral domain simpler?
@@MuffinsAPlenty I think it ties to making the definition for cancellation more clear. You don't want x to be zero and neither y, so that when xy=0 then you can definitely say whether you have zero divisors. If you include zero, then it becomes trivial and muddles it up a bit. If you count zero, then you lose some detail in the definitions.
I mean, I'm here trying to figure out how to do my homework though LOL! So I may be completely wrong, but that's how I interpreted it.... take my comment with a grain of salt :P
@@Taricus Very delayed response to you, but I think you're pretty much right!
In number theory, most of the rings they care about are integral domains. So, if you exclude 0 from the classification of zero-divisor, it makes it easier to define integral domains, which is pretty much all they care about anyway. In some sense, they are like, "Hey, in arbitrary ring theory, this bad thing will happen. But we won't consider any such situation."
However, more general ring theory (such as commutative algebra) really does want to consider rings with zero-divisors. The theory of zero-divisors is cleaner when 0 is considered to be zero-divisor (except in the 0 ring). So in this context, it makes sense to consider 0 to be a zero-divisor.
It's sort of a matter of convenience. If you want your study to include rings with zero-divisors, it's easier to consider 0 to be a zero-divisor (except in the 0 ring); but if your study only includes integral domains, then the only reason you care about zero-divisors is to say "bad thing doesn't happen", and it gives a cleaner definition of integral domain.
love this channel
Keep it up! Looking forward for more videos from you, don't stop!
These people are GREAT! Love the vids.
Thanks
I enjoyed the video! Keep it up!
I really liked the video, but it would be much better if it was little bit slower :) I had to stop n play again lots of times cuz it was too fast for my brain lol
wat is mod 12
I already watched this before our class in Abstract Algebra started but it turns out that her explanation and examples are similar to what is given here so I left the meet and watched this video again.
Srlsly, I don't understand my prof's explanation.
What is a mod 12?
I had one doubt, why we didn't take 1 and 6 and we took 9 10 when we got 4 answers
mam
Integral domain: If ab=0, then either a or b is 0 where a,b in R.
awesome video
When is mod used? It's a new concept to me.
It's used all throughout number theory. It also has applications in cryptography
Thanks a lot!!!
Q1. 11*12*12 assuming rhs is 0.
Q2. Yes x^2=0 has 0,6 has solutions mod 12.
Q1. Why not 12^3?
ax^2+bx+c = 0
s.t a,b,c belong to Z12.
12 possibilities for a, 12 for b, 12 for c. 12^3 combinations.
@@Yougottacryforthis because if a=0 it's not a quadratic equation
Mam excellent class
Thank you, kind Socratica Friend! 💜🦉
Very interesting
I am stuck with the puzzle she asked, there should be infinitely many quadratic equation right? right?
Nope! There are not infinitely many quadratic polynomials modulo 12.
A quadratic polynomial is of the form ax^2 + bx + c. All polynomials are determined by their coefficients. So choosing the values of a, b, and c determine the polynomial. Usually, we have infinitely many values of a, b, and c to choose from.
How many values can we choose for a, b, and c if we're dealing with integers modulo 12?
Ok.. i get it now. Y'know since the 2nd question ask about solutions, i thought the 1st answer being infinitely many but just that not all of which has solution, kek
It seems to me that the cancellation property should hold for non-integral domains. It doesn't require commutation to work, just the lack of zero-divisors... or maybe I'm missing something.
You are correct! You can get cancellation pretty generally provided that the thing you're trying to cancel is not a zero-divisor of any kind. Commutativity is not required.
For example, if you have a non-commutative ring and know xy = xz, where x is not a left zero-divisor (meaning that xw = 0 implies w = 0), then you can conclude that y = z. Similarly, if you know yx = zx and x is not a right zero-divisor (meaning wx = 0 implies w = 0), then you can conclude y = z.
Of course, without commutativity, you can no longer conclude that xy = zx implies that y = z, even if x is not a zero-divisor (left, right, or two-sided).
I think it depends on the definition of each author. Sometimes adding some properties simplify proofs in the whole theory. That's because removing some properties could lead to not interesting cases that doest not matter for the theory purposes and just complicate the proofs.
Brilliant work! Keep it up!
Nice! Keep it up!
I was lost in the first 10 seconds. Why factor and not just move the -4 to the other side? x^2 = 4?
I think thz example is to relate to the kind of algebra that only deal with + and *. If you introduce square root here, that is another story~
casually introducing a new operation is a whole pandora box
thank you madam...................
Fantastic video! Keep it up!
Thank you! Could anyone be kind to tell me what's the solution to the 2 puzzles? My answers correct(?: *FALSE*): there exist (12^2-12)/2 + 12 = 78 quadratic equations; the number of those who have only 2 solutions is 78 - 4*12 = 30, where 12 eqns out of 30 are in the form of (x-a)^2 = 0 (that is a = b), while for the other 18 eqns, (x-a)*(x-b) = 0 (where a!=b). Could anyone tell me the correct answer? (I'm *WRONG*)
11*12*12 : [939 (no solution), 144 (unique solution), 324 (just 2 solutions), 8 (have 3 solutions), 138, 0, 18, 0, 12, 0, 0, 0]
@@gundamlh Could you explain how you found this please? I've been trying to practice with smaller values such as mod 3, mod 4, etc, but I have yet to find a good way to calculate everything short of just manually checking every possible quadratic equation.
Its very very good
I'm impressed
Wow, very interesting ! Bet this is at the core of some cryptographic/decryption methods. Regardless, no way to really understand w/o trying a bunch of related basic problems in these topics, bet it won't take long to get totally stuck/ confused ;-)
omg the integral domain joke at the end LMAO