Z/12Z is a very cool group in relation to music, since the octave is commonly divided into 12 equal steps. The only two units beside the obvious 1 and -1 are 5 and 7, which correspond to the perfect fifth and its inverse, the perfect fourth. Because the perfect fifth is a unit, you can use it to generate the entire scale. That's the "circle of fifths". You can't use any interval besides the fifth and the step (and their inverses) to organize every possible key signature.
@@ilguerrierodragone129 Basically, yeah! Instead of a "Circle of Fifths" you could call it a "Circle of Seven Half-step Jumps". The interval of seven half-steps is called a Fifth by musicians because it's the fifth note in the major scale. (and it doesn't help that each of the 12 equal steps in an octave are called "half-steps") It doesn't really make sense from a math perspective.
@@nomukun1138I'm sorry i didn't understand very well. I've read your comment again and I think it's not correct that -1 is in Z/12Z, also there is 11 which is a unit too in Z/12Z. Said that i want to understand better your concept, I don't know very well what is the cricle of fifths in music, could you explain it to me? I think that 7 is not the inverse of 5 in Z/12Z. What is the perfect fourth? And how i organize the signatures based on the circle of fifths?
@@ilguerrierodragone129 Music terminology is extremely confusing. -1 in the context of Z/12Z is the same as 11. I thought calling it -1 emphasized that it steps backward through the ring. It's the same as negative one. 7 is the additive inverse of 5 in Z/12Z. The word "inverse" usually means the "multiplicative inverse" but I meant "additive inverse". In music theory, the Perfect Fourth Interval (5 half-steps) is called the inverse of the Perfect Fifth Interval (7 half-steps), that's the reason I used the word "inverse" without considering that it would be confusing. Let me try to explain the circle of fifths... Every melody or piece of music (in this traditional European system of music) has a Key Signature describing which notes are used most frequently. A melody with the Key Signature of C would have the notes C D E F G A B, and a melody in the Key Signature of G would have the notes G A B C D E F#. The Key Signature of C# includes the notes C# D# E# F# G# A# B#, which has no overlap with the Key of C, even though the root notes are only one space apart. Playing a melody in C# at the same time as a melody in C sounds horrible because none of the notes match. However playing a melody in G at the same time as a melody in C will often sound very good, because many of the notes in the Key Signature are the same. (and because 7 generates Z/12Z, every possible key signature is some precise number of Fifths away from every other key signature, and this number of Fifths is a good measure of similarity of the key signatures) The Key of C is "closer" in sound to the Key of G even though the root notes are a Fifth apart. The Key of C is "distant" in sound from the Key of C# even though the root notes are only one half-step apart. We say that C and G are "neighbors on the Circle of Fifths". Changing a piece of music by seven half-steps, corresponding to a Fifth, sounds pleasant and changing it by one half-step sounds rough.
6:20 If you view the matrices as linear transformation, the determinant is the factor by which it scales the plane. If the matrix scales the plane by any other factor than 1 (or -1), say a, the inverse would have to scale back space by 1/a (a-inverse). Since 1 and -1 are the only two unit elements in the ring of whole numbers, and thus the only two numbers with a whole number inverse, thus they are the only possible choices for a.
im a bit confused on matrices because as someone who learned about linear algebra and it was my introduction to matrices i dont see how they are used in any different fields. You said "if you view matrices as linear transformations" so there are numerous ways to view them? What are some of them? And what are some resources i can see to learn about matrices in the general sense and not just linear transformations?
@@atrophysicistyou can also view them as just ”lists of numbers”. One thing about matrices that’s used a lot in other fields are the linear algebraic groups, which are basically algebraic groups that are sub-algebraic groups of GL(n,R) for some ring R, i.e. the group of invertible n x n matrices over R. Here the connection with lienar transformations becomes weaker, in particular when R is a ”bad” ring. These groups show up in many different places.
How deeply are you going to go into ring theory? -- GCD, PID, Noetherian Rings --- all would be ideal. Seriously it would be nice to mention ideas and link them back to modules. I really like this series.
hi soctatica madam how are u.today i watched "units in rings (abstrct algebra)it is very i teresting.can u publish these subjects in text books written by u?it is not possible tu grasp certain points irealy appreciate your calibre.bye madam.
If I'm capable, then I would definitely support,, I'm still a student. but in near future I'll sure ..I made a list of channels for that. like 3b1b, two minutes paper, khan academy, long list....
Same. Once I get enough money I’m going to give back to the open source community (this includes freely available TH-cam video and free software I used during my education)
It is interesting how many problems are easy to state and understand, but so difficult to solve. I didn't know the characterization of the units of a ring was such a hot topic in algebra.
Ans. 1.We can Compute Persentage Of All(Z/nZ) using Formula ={π(n)/n}×100 Where π(n) is euler function. 2. Z/30Z Has the smallest %of units that is 26.6666666..%= app 27%
Thank you so much for creating these videos. I've just started my masters in math and need to get my knowledge up to speed. Could you do a video on semi-direct product of groups. Also maybe some videos on lie groups/lie algebras. Thanks, love your videos.
I think this video was like a "unit university". All puns aside, Socratica is amazing and I plan to watch the rest of the abstract algebra videos today!
Regarding the challenge that in a ring of matrices with integer elements the units are the matrices with determinant +-1 I think I found a short proof in case anyone is interested :) Firstly, any square matrix with nonzero determinant has an inverse. So we only need to verify that all elements of the inverse matrix are indeed integers. The inverse of a matrix is given by its adjugate divided by its determinant. en.m.wikipedia.org/wiki/Adjugate_matrix The adjugate in turn is the transpose of its cofactor matrix. The cofactor matrix is the matrix which has a minor (subdeterminant of the matrix) at each element’s position. Since the determinant of a (sub)matrix with integer elements is always an integer we know that the minors are all integers. Thus the cofactor matrix and the adjugate matrix both have only integer elements. Dividing a matrix with integer elements by the determinant (of the original matrix) yields an inverse with integer elements if and only if the determinant is +-1 thus concluding the proof of the units being the matrices with determinant +-1.
Great video. Seems like that were people including myself wondering why 1 exists in the first place at 2:26. It might have been better to show beforehand that: Let x be an element of R^x (set of units) of a ring R. Since x is a unit, by definition of a unit, R also has x^-1 which is the inverse of x. Since R is closed under multiplication, x multiplied by x^-1 is in R. Since x multiplied by x^-1 is 1, thus 1 is in R. (shown existence of 1 in case R^x is not empty.)
@Socratica Just visualized the multiplication table for the rings Z/nZ for n in {1,400}. Insanely beautiful. Units here: th-cam.com/video/I8vXKTl3pAM/w-d-xo.html Full table here: th-cam.com/video/tUIqsYi8hH4/w-d-xo.html
My proof was as follows: The number of co-primes to n is Euler(n), therefore the percentage of units is Euler(n)/n. Euler(n) is given by n*Pi(1-1/p) aka the product of the distinct primes that divide n. So the percentage will just be Pi(1-1/p), so we reduced our problem to find which number between 1 and 50 has the minimum product. We can see from the formula that the more distinct prime factors the lesser the percentage (since trivially 1-1/p < 1) therefore we need to find the number with the most distinct prime factors. This is still not an easy task but not insurmountable. My continuation from here was to look at the largest p(n)! that satisfies p(n)!
@@YougottacryforthisThat suggests that Z/210Z should also have a small number of units. (Edited for skipping nonprime coprimes) There are some coprime values that aren't actually primes, for example 11*11=121 is coprime with 210. There aren't any natural numbers coprime with 30 and less than 30 that aren't also primes, but that's not the case for larger primorials. Wolfram Alpha says there are 46 primes less than 210, but composite numbers with all prime factors greater than 7 will also be coprime. These are all two-factor composite numbers (because the smallest three-factor composite with factors greater than 7, 11^3=1331, is too large) and have the factors between 11 and 19 (because anything greater than 11*19=209 is too large) 11*11 = 121 11*13 = 143 11*17 = 187 11*19 = 209 13*13 = 169 13*17 = 221 (too large) (all other combinations too large) 46 primes less than 210 = 2*3*5*7 +1 one is not coprime -4 primes that are factors of 210, so not coprime +5 composite numbers that are coprime Therefore Z/210Z should have 48 units for a percentage of 22.9% Z/2310Z should also have a lower percentage of units than any smaller modulo ring.
The number of co-primes to n is Euler(n), therefore the percentage of units is Euler(n)/n. Euler(n) is given by n*Pi(1-1/p) aka the product of the distinct primes that divide n. So the percentage will just be Pi(1-1/p), so we reduced our problem to find which number between 1 and 50 has the smallest product. We can see from the formula that the more distinct prime factors the lesser the percentage (since 1-1/p < 1) therefore we need to find the number with the most distinct prime factors.
If the ring R does not have an identity element then it does not make sense to even question which elements have inverses. Thus it does not even make sense to speak about R^x
@@hybmnzz2658 OK. That makes sense. But it is NOT a requirement for a ring to exist 1, known as *unity*, or *multiplicative identity*,( at least according to some authors) as it is a requirement, nevertheless, to exist a 0, or zero *additive identity*... Therefore, should not change the slide at 2:23 for saying "for any ring with unity"?
At 3:12, I think we also want to say: 1. x,y in R implies there exist x^-1 = a, and y^-1 =b; 2. by definition of multiplicative inverse, we have a^-1 = x, b^-1 =y; 3. Thus, there exist inverses for a and b ( for x^-1 and y^-1), so a,b must also be in R^x (set of units in R); 4. Thus, since we can multiply elements of R^x, we can do: b.a = (y^-1).(x^-1), and HENCE, x.y has an inverse (I.e. SHOW why (y^-1)(x^-1) should also exist in R^x
Couldn't you have uploaded this a few days back. I had an Abstract Algebra exam and I messed up right on this topic :( costing me the pass of that exam. Well this ain't my lucky month anyway so far so.. I guess I'd use this clarifications in the future. Thanks for doing these amazing videos !
I saw a critique of this type of presentation, that it's very clear logical wording, but only students with linguistic aptitude retain the information, ie it's content without an accompanying learning by doing technique, context. I'm interested in the subject because it's relevant to e-Pi-i resonance imaging of axial-tangential orthogonality spacing in-form-ation by "algebraic" logarithmic numberness in potential prime and cofactors of phase-locked multi-phase state holography. But I don't have the linguistic skills, so here's the handover to real Mathematicians who do..
She goes very fast through the material (for me anyway; maybe I'm not all that bright or it's early in the morning). Nice dig at the end for non-subscribers.
It’s nothing to do with intelligence, don’t put yourself down. She is going fast, these videos are more like reviews. Keep your notes and textbooks at hand to go through at your own pace. Also, feel free to pause the video and rewind it until you understand, I believe in you 😁
I dunno how to prove it but it seems to me that an element from ring Z/nZ mod n have inverses if and only if they are coprime to n. How do I prove this fact. Kinda looks similar to Bezout lemma: ax=1(mod n) where x is inverse of a => ax+kn=1(mod n) for some integer k and by bezout lemma we have that a and n are coprime because sum of their different products with integers is 1 (which correspons to gcd(a,n)=1) but still I am not convinced that this is technically proved. Because if I know that gcd(a,n)=1 => ax+ny=1 for some integers x,y follows from bezout lemma but does it work other way around? does this lemma hold if and only if? If that so, then it proves my first statement, otherwise, I had an idea to rewrite: ax=1 (mod n) => ax-1=0 (mod n) => ax-1=n*k for some k => ax-nk=1 and if a and n has divisor d, then d also divides ax-nk and hence divides right hand side, i.e. 1. So d must be 1 and gcd(a,n)=1. But still something tells me that it cannot be that simple. Please, correct my mistakes in "proofs". I am very fond of this channel, especially, for abstract algebra/algebraic structures because it is relevant for me as I have a module on it this semester. Take care!
You are correct, and indeed, Bézout's Lemma is the way to go. But for the other direction of the proof, you need something a tiny bit stronger than Bézout's Lemma. For the other direction of your proof, the thing to realize is the following: if there exist integers x and y with ax+ny=c, then c is a multiple of gcd(a,n). Using this fact, we can conclude that if there exist integers x and y with ax+ny=1, then gcd(a,n) = 1. I'll let you try to do the rest of the proof from there. But feel free to comment back if you want an additional hint.
The number of co-primes to n is Euler(n), therefore the percentage of units is Euler(n)/n. Euler(n) is given by n*Pi(1-1/p) aka the product of the distinct primes that divide n. So the percentage will just be Pi(1-1/p), so we reduced our problem to find which number between 1 and 50 has the smallest product. We can see from the formula that the more distinct prime factors the lesser the percentage (since 1-1/p < 1) therefore we need to find the number with the most distinct prime factors. This is still not an easy task but not insurmountable. My continuation from here was to look at the largest p(n)! that satisfies p(n)!
It's not a convention - it's a fact. Say you have x.y. Then if you multiply on the left by y^-1.x^-1, you get y^-1.x^-1.x.y. See how we have x^-1.x in there? That can become the identity element, leaving us with y^-1.y, which also becomes the identity element. So y^-1.x^-1.x.y = identity. Similarly, multiplying x.y on the right by y^-1.x^-1 leaves us with x.y.y^-1.x^-1. Then the y.y^-1 is the identity, leaving x.x^-1, which is also the identity. On the other hand, if you were to multiply x.y on the right by x^-1.y^-1, you get x.y.x^-1.y^-1. Can we simplify this at all? It's not clear. If multiplication is _commutative,_ then x^-1.y^-1 = y^-1.x^-1. But if multiplication is not commutative, then these two expressions are not equal, and (x.y)^-1 = y^-1.x^-1.
Nice. Can u suggest me a book for solving problems in topics in Algebra. Already theory part in I N Herstein. Only problems oriented based on groups and its content, Rings and its content, Vector space.
Thank you very much for these great videos and your effort in general. I have one question: Depending on the chosen definition of a ring, the identity element 1 is not necessarily an element of the ring. But that would imply, that the set of units of R is only a group, if 1 is in R, and not in general, right? The proof shown in this video assumed that 1 is in R.
The concept of a unit doesn't make sense unless 1 is in R. After all, a unit is an element that has a multiplicative inverse, which is an element that multiplies something to 1.
@@MuffinsAPlenty This is a reasonable argument. But there are definitions of a ring, which only demand the set with multiplication to be a semigroup, which does not necessarily have a neutral element. That's why I think it is sensible to mention it.
That is quite the list: the aforementioned ring is the ring of all polynomials evaluated at root 2. A more intuitive presentation is a+b✓2 for integers a and b. This has a lot of units like (✓2 +1)^n for all positive integers n
40298 tesuya factorization is unique up to order means that if we consider the prime factorization of 30, 30=2×3×5, but can also be written as 30=3×5×2 or 30=5×2×3, these products are all the same and so we say the factorization is unique "up to" order, so we don't consider writing the product with it's terms scrambled up as counting as another factorization. 😃
UMU-i-D you should try watching from the start of the Abstract Algebra playlist You’ll have a better chance of understanding what she’s talking about :D
6:39 Units of ℤ/2ℤ : 0·0 = 0 ; 0·1 = 0 ; 1·0 = 0 ; 1·1 = 1 so 75% of units in the ring Units of ℤ/3ℤ : 0·0 = 0 ; 0·1 = 0 ; 0·2 = 0 ; 1·0 = 0 ; 1·1 = 1 ; 1·2 = 2 ; 2·0 = 0 ; 2·1 = 2 ; 2·2 = 1 ; so 5/9 = 55.55 % of units in the ring Units of ℤ/4ℤ : 0·0 = 0 ; 0·1 = 0 ; 0·2 = 0 ; 0·3 = 0 ; 1·0 = 0 ; 1·1 = 1 ; 1·2 = 2 ; 1·3 = 3 ; 2·0 = 0 ; 2·1 = 2 ; 2·2 = 0 ; 2·3 = 2 3·0 = 0 ; 3·1 = 3 ; 3·2 = 2 ; 3·3 = 1 so 8/16 = 50 % of units in the ring I don't know how to do to prove it, but I suspect ℤ/2ℤ to have the biggest percentage of units in the ring and ℤ/50ℤ the smallest
Просто очень и примеры очевидные. Уровень колледжа или 1 курса университета. Жду видео про группы обратимых элементов локальных колец или коммутативных конечных колец с единицей.
No, it isn't inaccurate. By definition, units are not considered to be prime elements of a ring. Now, the reasons why 1 isn't considered a prime number (in the natural numbers) are pretty much the same reason why units are not considered prime elements (in an arbitrary commutative ring), but that doesn't make the statement inaccurate.
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Please add more videos on vector space or on linear algebra.. I will be very thankful..
"but it's a very rewarding challenge to try to convince yourself this is true" - this perfectly describes how learning maths is like
Z/12Z is a very cool group in relation to music, since the octave is commonly divided into 12 equal steps. The only two units beside the obvious 1 and -1 are 5 and 7, which correspond to the perfect fifth and its inverse, the perfect fourth. Because the perfect fifth is a unit, you can use it to generate the entire scale. That's the "circle of fifths". You can't use any interval besides the fifth and the step (and their inverses) to organize every possible key signature.
So you can also write a "circle of sevens"?
@@ilguerrierodragone129 Basically, yeah! Instead of a "Circle of Fifths" you could call it a "Circle of Seven Half-step Jumps". The interval of seven half-steps is called a Fifth by musicians because it's the fifth note in the major scale. (and it doesn't help that each of the 12 equal steps in an octave are called "half-steps") It doesn't really make sense from a math perspective.
@@nomukun1138I'm sorry i didn't understand very well. I've read your comment again and I think it's not correct that -1 is in Z/12Z, also there is 11 which is a unit too in Z/12Z. Said that i want to understand better your concept, I don't know very well what is the cricle of fifths in music, could you explain it to me? I think that 7 is not the inverse of 5 in Z/12Z. What is the perfect fourth? And how i organize the signatures based on the circle of fifths?
@@ilguerrierodragone129 Music terminology is extremely confusing.
-1 in the context of Z/12Z is the same as 11. I thought calling it -1 emphasized that it steps backward through the ring. It's the same as negative one.
7 is the additive inverse of 5 in Z/12Z. The word "inverse" usually means the "multiplicative inverse" but I meant "additive inverse". In music theory, the Perfect Fourth Interval (5 half-steps) is called the inverse of the Perfect Fifth Interval (7 half-steps), that's the reason I used the word "inverse" without considering that it would be confusing.
Let me try to explain the circle of fifths...
Every melody or piece of music (in this traditional European system of music) has a Key Signature describing which notes are used most frequently. A melody with the Key Signature of C would have the notes C D E F G A B, and a melody in the Key Signature of G would have the notes G A B C D E F#. The Key Signature of C# includes the notes C# D# E# F# G# A# B#, which has no overlap with the Key of C, even though the root notes are only one space apart. Playing a melody in C# at the same time as a melody in C sounds horrible because none of the notes match. However playing a melody in G at the same time as a melody in C will often sound very good, because many of the notes in the Key Signature are the same.
(and because 7 generates Z/12Z, every possible key signature is some precise number of Fifths away from every other key signature, and this number of Fifths is a good measure of similarity of the key signatures)
The Key of C is "closer" in sound to the Key of G even though the root notes are a Fifth apart. The Key of C is "distant" in sound from the Key of C# even though the root notes are only one half-step apart. We say that C and G are "neighbors on the Circle of Fifths". Changing a piece of music by seven half-steps, corresponding to a Fifth, sounds pleasant and changing it by one half-step sounds rough.
@@nomukun1138 thank you very much, I didn't know that
I just love the way this channel is teaching abstract algebra. Whoever scripted this is a genius. Thanks a lot!!!
6:20 If you view the matrices as linear transformation, the determinant is the factor by which it scales the plane. If the matrix scales the plane by any other factor than 1 (or -1), say a, the inverse would have to scale back space by 1/a (a-inverse). Since 1 and -1 are the only two unit elements in the ring of whole numbers, and thus the only two numbers with a whole number inverse, thus they are the only possible choices for a.
im a bit confused on matrices because as someone who learned about linear algebra and it was my introduction to matrices i dont see how they are used in any different fields. You said "if you view matrices as linear transformations" so there are numerous ways to view them? What are some of them? And what are some resources i can see to learn about matrices in the general sense and not just linear transformations?
@@atrophysicistyou can also view them as just ”lists of numbers”. One thing about matrices that’s used a lot in other fields are the linear algebraic groups, which are basically algebraic groups that are sub-algebraic groups of GL(n,R) for some ring R, i.e. the group of invertible n x n matrices over R. Here the connection with lienar transformations becomes weaker, in particular when R is a ”bad” ring. These groups show up in many different places.
For rings Z/, the units are those numbers m that are relatively prime to n.
How deeply are you going to go into ring theory? -- GCD, PID, Noetherian Rings --- all would be ideal.
Seriously it would be nice to mention ideas and link them back to modules.
I really like this series.
I would also like to watch these
"all would be ideal" PUN! Ha!
*modules*
hi soctatica madam how are u.today i watched "units in rings (abstrct algebra)it is very i teresting.can u publish these subjects in text books written by u?it is not possible tu grasp certain points
irealy appreciate your calibre.bye madam.
@@s.k.potdarpotdar8377 very desperate aloo masala.
If I'm capable, then I would definitely support,, I'm still a student. but in near future I'll sure ..I made a list of channels for that. like 3b1b, two minutes paper, khan academy, long list....
Same. Once I get enough money I’m going to give back to the open source community (this includes freely available TH-cam video and free software I used during my education)
It is interesting how many problems are easy to state and understand, but so difficult to solve. I didn't know the characterization of the units of a ring was such a hot topic in algebra.
Ans.
1.We can Compute Persentage Of All(Z/nZ) using
Formula ={π(n)/n}×100
Where π(n) is euler function.
2. Z/30Z Has the smallest %of units that is 26.6666666..%= app 27%
Thank you so much for creating these videos. I've just started my masters in math and need to get my knowledge up to speed. Could you do a video on semi-direct product of groups. Also maybe some videos on lie groups/lie algebras. Thanks, love your videos.
I think this video was like a "unit university". All puns aside, Socratica is amazing and I plan to watch the rest of the abstract algebra videos today!
Thank you very much for this math videos, and we want more :)
Regarding the challenge that in a ring of matrices with integer elements the units are the matrices with determinant +-1 I think I found a short proof in case anyone is interested :)
Firstly, any square matrix with nonzero determinant has an inverse. So we only need to verify that all elements of the inverse matrix are indeed integers.
The inverse of a matrix is given by its adjugate divided by its determinant.
en.m.wikipedia.org/wiki/Adjugate_matrix
The adjugate in turn is the transpose of its cofactor matrix. The cofactor matrix is the matrix which has a minor (subdeterminant of the matrix) at each element’s position.
Since the determinant of a (sub)matrix with integer elements is always an integer we know that the minors are all integers. Thus the cofactor matrix and the adjugate matrix both have only integer elements. Dividing a matrix with integer elements by the determinant (of the original matrix) yields an inverse with integer elements if and only if the determinant is +-1 thus concluding the proof of the units being the matrices with determinant +-1.
Great video. Seems like that were people including myself wondering why 1 exists in the first place at 2:26. It might have been better to show beforehand that:
Let x be an element of R^x (set of units) of a ring R. Since x is a unit, by definition of a unit, R also has x^-1 which is the inverse of x. Since R is closed under multiplication, x multiplied by x^-1 is in R. Since x multiplied by x^-1 is 1, thus 1 is in R. (shown existence of 1 in case R^x is not empty.)
@Socratica
Just visualized the multiplication table for the rings Z/nZ for n in {1,400}. Insanely beautiful.
Units here: th-cam.com/video/I8vXKTl3pAM/w-d-xo.html
Full table here: th-cam.com/video/tUIqsYi8hH4/w-d-xo.html
Neato! Thanks for sharing! 💜🦉
Z/30Z has the smallest percentage of units (8 units = 27 %).
Seems Z/nZ , with n being a multiple of 6, has significantly lower percentage of units. Check Z/24Z, Z/36Z etc..
Min phi(n)/n for n up to 50.
My proof was as follows:
The number of co-primes to n is Euler(n), therefore the percentage of units is Euler(n)/n.
Euler(n) is given by n*Pi(1-1/p) aka the product of the distinct primes that divide n. So the percentage will just be Pi(1-1/p), so we reduced our problem to find which number between 1 and 50 has the minimum product.
We can see from the formula that the more distinct prime factors the lesser the percentage (since trivially 1-1/p < 1) therefore we need to find the number with the most distinct prime factors.
This is still not an easy task but not insurmountable. My continuation from here was to look at the largest p(n)! that satisfies p(n)!
@@YougottacryforthisThat suggests that Z/210Z should also have a small number of units.
(Edited for skipping nonprime coprimes)
There are some coprime values that aren't actually primes, for example 11*11=121 is coprime with 210. There aren't any natural numbers coprime with 30 and less than 30 that aren't also primes, but that's not the case for larger primorials.
Wolfram Alpha says there are 46 primes less than 210, but composite numbers with all prime factors greater than 7 will also be coprime. These are all two-factor composite numbers (because the smallest three-factor composite with factors greater than 7, 11^3=1331, is too large) and have the factors between 11 and 19 (because anything greater than 11*19=209 is too large)
11*11 = 121
11*13 = 143
11*17 = 187
11*19 = 209
13*13 = 169
13*17 = 221 (too large)
(all other combinations too large)
46 primes less than 210 = 2*3*5*7
+1 one is not coprime
-4 primes that are factors of 210, so not coprime
+5 composite numbers that are coprime
Therefore Z/210Z should have 48 units for a percentage of 22.9%
Z/2310Z should also have a lower percentage of units than any smaller modulo ring.
The number of co-primes to n is Euler(n), therefore the percentage of units is Euler(n)/n.
Euler(n) is given by n*Pi(1-1/p) aka the product of the distinct primes that divide n. So the percentage will just be Pi(1-1/p), so we reduced our problem to find which number between 1 and 50 has the smallest product.
We can see from the formula that the more distinct prime factors the lesser the percentage (since 1-1/p < 1) therefore we need to find the number with the most distinct prime factors.
Puzzle:
Each prime p has p-1 units.
Below composite number: # of units
4: 2 - 6: 2 - 8: 4 - 9: 6 - 10: 4 - 12: 4 - 14: 6 - 15: 8 - 16: 8 - 18: 6 - 20: 8 - 21: 12 - 22: 10 - 24: 8 - 25: 20 - 26: 12 - 27: 18 - 28: 12 - 30: 8 - 32: 16 - 33: 20 - 34: 16 - 35: 24 - 36: 12 - 38: 18 - 39: 24 - 40: 16 - 42: 12 - 44: 20 - 45: 24 - 46: 22 - 48: 16 - 49: 42 - 50: 20.
The smallest percentage of units: 30 (8 units, 0.2666%)
For numbers up to 100: also 60 and 90 have 0.2666%.
Thanks.... And please don't stop uploading videos.
The best abstract algebra explanation is here guys!
Z/ nZ has units which are as such: all natural no less than n and coprime with n. As 1,5,7,11 are the only no.s till 12 and coprime to this
Good video and good explaination i love it ♡
why do we assume the ring has 1 in it?
If the ring R does not have an identity element then it does not make sense to even question which elements have inverses. Thus it does not even make sense to speak about R^x
@@hybmnzz2658 OK. That makes sense. But it is NOT a requirement for a ring to exist 1, known as *unity*, or *multiplicative identity*,( at least according to some authors) as it is a requirement, nevertheless, to exist a 0, or zero *additive identity*...
Therefore, should not change the slide at 2:23 for saying
"for any ring with unity"?
Poxa, adoraria uma versão em pt desse! Adorei, mas o inglês me limitou um pouquinho
Excellent video. BTW, it seems that another notation for the set of units in a ring is U(R).
For the puzzle, the units are the elements of the ring which are relatively prime to n, ranging from 2 to 50, respectively.
At 3:12, I think we also want to say: 1. x,y in R implies there exist x^-1 = a, and y^-1 =b; 2. by definition of multiplicative inverse, we have a^-1 = x, b^-1 =y; 3. Thus, there exist inverses for a and b ( for x^-1 and y^-1), so a,b must also be in R^x (set of units in R); 4. Thus, since we can multiply elements of R^x, we can do: b.a = (y^-1).(x^-1), and HENCE, x.y has an inverse
(I.e. SHOW why (y^-1)(x^-1) should also exist in R^x
Thanks, you make great videos, great content
@6:33 to prove that: consider the matrix [1,0;n,1]
Couldn't you have uploaded this a few days back. I had an Abstract Algebra exam and I messed up right on this topic :( costing me the pass of that exam. Well this ain't my lucky month anyway so far so.. I guess I'd use this clarifications in the future. Thanks for doing these amazing videos !
I saw a critique of this type of presentation, that it's very clear logical wording, but only students with linguistic aptitude retain the information, ie it's content without an accompanying learning by doing technique, context.
I'm interested in the subject because it's relevant to e-Pi-i resonance imaging of axial-tangential orthogonality spacing in-form-ation by "algebraic" logarithmic numberness in potential prime and cofactors of phase-locked multi-phase state holography.
But I don't have the linguistic skills, so here's the handover to real Mathematicians who do..
Very interesting...thank you... from Bangladesh
Hello to our Socratica Friends in Bangladesh! We're so glad you're watching. 💜🦉
Nice ma'am I have no words
Thanks!
Thank you for your kind support! We so appreciate it!! 💜🦉
This was really helpful. Thanks :)
She goes very fast through the material (for me anyway; maybe I'm not all that bright or it's early in the morning). Nice dig at the end for non-subscribers.
It’s nothing to do with intelligence, don’t put yourself down. She is going fast, these videos are more like reviews. Keep your notes and textbooks at hand to go through at your own pace. Also, feel free to pause the video and rewind it until you understand, I believe in you 😁
Thanks for another great video
Nicely explain ...thank a lot Mam ...very helpful for my upcoming interview
Thank you for your videos, could you please share about first and second theorem of isomorphism..
I really need it, now for upcoming semester exam
Con esta profesora me enamoro de la matematica.
I
can you please let me what is the grometrical interpretation of rings
I dunno how to prove it but it seems to me that an element from ring Z/nZ mod n have inverses if and only if they are coprime to n. How do I prove this fact. Kinda looks similar to Bezout lemma:
ax=1(mod n) where x is inverse of a => ax+kn=1(mod n) for some integer k and by bezout lemma we have that a and n are coprime because sum of their different products with integers is 1 (which correspons to gcd(a,n)=1) but still I am not convinced that this is technically proved. Because if I know that gcd(a,n)=1 => ax+ny=1 for some integers x,y follows from bezout lemma but does it work other way around? does this lemma hold if and only if? If that so, then it proves my first statement, otherwise, I had an idea to rewrite:
ax=1 (mod n) => ax-1=0 (mod n) => ax-1=n*k for some k => ax-nk=1 and if a and n has divisor d, then d also divides ax-nk and hence divides right hand side, i.e. 1. So d must be 1 and gcd(a,n)=1. But still something tells me that it cannot be that simple.
Please, correct my mistakes in "proofs". I am very fond of this channel, especially, for abstract algebra/algebraic structures because it is relevant for me as I have a module on it this semester. Take care!
You are correct, and indeed, Bézout's Lemma is the way to go. But for the other direction of the proof, you need something a tiny bit stronger than Bézout's Lemma.
For the other direction of your proof, the thing to realize is the following: if there exist integers x and y with ax+ny=c, then c is a multiple of gcd(a,n). Using this fact, we can conclude that if there exist integers x and y with ax+ny=1, then gcd(a,n) = 1.
I'll let you try to do the rest of the proof from there. But feel free to comment back if you want an additional hint.
Well explain
The number of co-primes to n is Euler(n), therefore the percentage of units is Euler(n)/n.
Euler(n) is given by n*Pi(1-1/p) aka the product of the distinct primes that divide n. So the percentage will just be Pi(1-1/p), so we reduced our problem to find which number between 1 and 50 has the smallest product.
We can see from the formula that the more distinct prime factors the lesser the percentage (since 1-1/p < 1) therefore we need to find the number with the most distinct prime factors.
This is still not an easy task but not insurmountable. My continuation from here was to look at the largest p(n)! that satisfies p(n)!
brillianrt video.very helpful!
Which book best of Ring theory ... please tell me mam....
At 3:10, any reason for the convention, the inverse of x.y is y^-1.x^-1 and not written as x^-1.y^-1 ?
It's not a convention - it's a fact. Say you have x.y. Then if you multiply on the left by y^-1.x^-1, you get
y^-1.x^-1.x.y.
See how we have x^-1.x in there? That can become the identity element, leaving us with y^-1.y, which also becomes the identity element. So y^-1.x^-1.x.y = identity.
Similarly, multiplying x.y on the right by y^-1.x^-1 leaves us with
x.y.y^-1.x^-1.
Then the y.y^-1 is the identity, leaving x.x^-1, which is also the identity.
On the other hand, if you were to multiply x.y on the right by x^-1.y^-1, you get
x.y.x^-1.y^-1.
Can we simplify this at all? It's not clear.
If multiplication is _commutative,_ then x^-1.y^-1 = y^-1.x^-1. But if multiplication is not commutative, then these two expressions are not equal, and (x.y)^-1 = y^-1.x^-1.
Very Abstract from all Videos on TH-cam.
Thanks a lot!
Can you make video about UFDs, please?
Nice. Can u suggest me a book for solving problems in topics in Algebra. Already theory part in I N Herstein. Only problems oriented based on groups and its content, Rings and its content, Vector space.
Thank you very much for these great videos and your effort in general.
I have one question:
Depending on the chosen definition of a ring, the identity element 1 is not necessarily an element of the ring. But that would imply, that the set of units of R is only a group, if 1 is in R, and not in general, right? The proof shown in this video assumed that 1 is in R.
The concept of a unit doesn't make sense unless 1 is in R. After all, a unit is an element that has a multiplicative inverse, which is an element that multiplies something to 1.
@@MuffinsAPlenty This is a reasonable argument. But there are definitions of a ring, which only demand the set with multiplication to be a semigroup, which does not necessarily have a neutral element. That's why I think it is sensible to mention it.
Thank you so much.
What mean the symbol z[√2] it is given that it is a ring and we have to find all unit of this ring .?
That is quite the list: the aforementioned ring is the ring of all polynomials evaluated at root 2. A more intuitive presentation is a+b✓2 for integers a and b. This has a lot of units like (✓2 +1)^n for all positive integers n
Fun fact: in modular arithmetic, units of Z/nZ never have any factor in common with n other than 1 or -1
Yeah and that is so because numbers that have a common factor with n are zero divisors
Polynomial ring related videos upload mam.
5:00
what dose "up to" mean?
Hulk used to say to Captain America in Age of Ultron :D :D
40298 tesuya factorization is unique up to order means that if we consider the prime factorization of 30, 30=2×3×5, but can also be written as 30=3×5×2 or 30=5×2×3, these products are all the same and so we say the factorization is unique "up to" order, so we don't consider writing the product with it's terms scrambled up as counting as another factorization. 😃
I understand your explanation. thank you
thank you madam........
Mam 'is every division ring is field 'and if yes then how to show??
No. The real Hamilton quaternions are a division ring which are not a field.
Every 'Commutative' Division Ring is a Field
easy to understand
What is the difference between units and fields ?
Good mam
I love your videos :D
Why can't the concept of associates apply to square roots? Instead of using complex numbers.
thanks fanatic vedio
I gave thumb up, but have no idea what you're talking about. Good video!
UMU-i-D you should try watching from the start of the Abstract Algebra playlist
You’ll have a better chance of understanding what she’s talking about :D
❤
6:39
Units of ℤ/2ℤ : 0·0 = 0 ; 0·1 = 0 ;
1·0 = 0 ; 1·1 = 1 so 75% of units in the ring
Units of ℤ/3ℤ : 0·0 = 0 ; 0·1 = 0 ; 0·2 = 0 ;
1·0 = 0 ; 1·1 = 1 ; 1·2 = 2 ;
2·0 = 0 ; 2·1 = 2 ; 2·2 = 1 ; so 5/9 = 55.55 % of units in the ring
Units of ℤ/4ℤ : 0·0 = 0 ; 0·1 = 0 ; 0·2 = 0 ; 0·3 = 0 ;
1·0 = 0 ; 1·1 = 1 ; 1·2 = 2 ; 1·3 = 3 ;
2·0 = 0 ; 2·1 = 2 ; 2·2 = 0 ; 2·3 = 2
3·0 = 0 ; 3·1 = 3 ; 3·2 = 2 ; 3·3 = 1 so 8/16 = 50 % of units in the ring
I don't know how to do to prove it, but I suspect ℤ/2ℤ to have the biggest percentage of units in the ring and ℤ/50ℤ the smallest
49
Mam I can't see for giving subtitles, plz remove it
Hello! You can turn off subtitles on your end (click the cc or gear icon).
Просто очень и примеры очевидные. Уровень колледжа или 1 курса университета. Жду видео про группы обратимых элементов локальных колец или коммутативных конечных колец с единицей.
An expensive one please.
Mashallah
helo
Nc
🐧
who searched "unital ring" and this came up
Ok. BUT: All you Americans are talking far too fast! Too much stuff is crammed into too little time!
Saying "1 is not considered prime because it is a unit" is inaccurate. There are better reasons why 1 is not considered prime (look on wikipedia).
No, it isn't inaccurate. By definition, units are not considered to be prime elements of a ring.
Now, the reasons why 1 isn't considered a prime number (in the natural numbers) are pretty much the same reason why units are not considered prime elements (in an arbitrary commutative ring), but that doesn't make the statement inaccurate.
Ah, I see. It's been a while since I studies abstract algebra.
"Prime element" has a precise definition with respect to rings.
Not Hindi language?
Perhaps he meant Hindi subtitles haha