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@aamir1756 Жыл бұрын
Please add more videos on vector space or on linear algebra.. I will be very thankful..
@TheRastaDan4 жыл бұрын
"but it's a very rewarding challenge to try to convince yourself this is true" - this perfectly describes how learning maths is like
@gaaraofddarkness4 жыл бұрын
I just love the way this channel is teaching abstract algebra. Whoever scripted this is a genius. Thanks a lot!!!
@nomukun1138 Жыл бұрын
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.
@ilguerrierodragone12911 ай бұрын
So you can also write a "circle of sevens"?
@nomukun113811 ай бұрын
@@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.
@ilguerrierodragone12911 ай бұрын
@@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?
@nomukun113811 ай бұрын
@@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.
@ilguerrierodragone12910 ай бұрын
@@nomukun1138 thank you very much, I didn't know that
@manan44365 жыл бұрын
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....
@navjotsingh22514 жыл бұрын
Same. Once I get enough money I’m going to give back to the open source community (this includes freely available KZbin video and free software I used during my education)
@kilian82504 жыл бұрын
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.
@atrophysicist8 ай бұрын
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?
@kilian82508 ай бұрын
@@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.
@welovfree7 жыл бұрын
Thank you very much for this math videos, and we want more :)
@ericdew20215 жыл бұрын
For rings Z/, the units are those numbers m that are relatively prime to n.
@viktyusk4 жыл бұрын
Z/30Z has the smallest percentage of units (8 units = 27 %).
@findyourownusername4 жыл бұрын
Seems Z/nZ , with n being a multiple of 6, has significantly lower percentage of units. Check Z/24Z, Z/36Z etc..
@Grassmpl2 жыл бұрын
Min phi(n)/n for n up to 50.
@Yougottacryforthis Жыл бұрын
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)!
@nomukun113811 ай бұрын
@@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.
@jimnewton45344 жыл бұрын
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.
@terryendicott29397 жыл бұрын
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.
@pritamb48517 жыл бұрын
I would also like to watch these
@theultimatereductionist75925 жыл бұрын
"all would be ideal" PUN! Ha!
@randomdude91355 жыл бұрын
*modules*
@s.k.potdarpotdar83775 жыл бұрын
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.
@Grassmpl2 жыл бұрын
@@s.k.potdarpotdar8377 very desperate aloo masala.
@kapilvishwakarma87272 жыл бұрын
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%
@nicolasmenet24713 жыл бұрын
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.
@christianshewmake35444 жыл бұрын
@Socratica Just visualized the multiplication table for the rings Z/nZ for n in {1,400}. Insanely beautiful. Units here: kzbin.info/www/bejne/f2nZiX6KoZijd68 Full table here: kzbin.info/www/bejne/qoasoqaPnp2bfpY
@Socratica4 жыл бұрын
Neato! Thanks for sharing! 💜🦉
@pritamb48517 жыл бұрын
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.
@maxpercer71194 жыл бұрын
why do we assume the ring has 1 in it?
@hybmnzz26583 жыл бұрын
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
@RafaelMartinez-ih9hd3 жыл бұрын
@@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"?
@Paula-ye5sq7 жыл бұрын
Poxa, adoraria uma versão em pt desse! Adorei, mas o inglês me limitou um pouquinho
@kapilvishwakarma87272 жыл бұрын
Very Abstract from all Videos on KZbin.
@غبيفيالرياضيات7 жыл бұрын
Good video and good explaination i love it ♡
@Treviscoe4 жыл бұрын
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.
@navjotsingh22514 жыл бұрын
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 😁
@MathsWithAsad4494 жыл бұрын
Thanks, you make great videos, great content
@Mrpallekuling Жыл бұрын
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%.
@cameronspalding97923 жыл бұрын
@6:33 to prove that: consider the matrix [1,0;n,1]
@PunmasterSTP3 жыл бұрын
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!
@solewalk3 жыл бұрын
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.)
@gajananvanjari3225 жыл бұрын
Thanks.... And please don't stop uploading videos.
@davidwilkie95515 жыл бұрын
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..
@Yougottacryforthis Жыл бұрын
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.
@ysteinlangaker47304 ай бұрын
Ok. BUT: All you Americans are talking far too fast! Too much stuff is crammed into too little time!
@shachichoudhary85442 жыл бұрын
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
@twistedlot6 жыл бұрын
I
@OrdenJust7 ай бұрын
Excellent video. BTW, it seems that another notation for the set of units in a ring is U(R).
@naanungamulla6528 Жыл бұрын
Polynomial ring related videos upload mam.
@mathematicalexpert2082 жыл бұрын
helo
@divyanshukarn9967 жыл бұрын
This was really helpful. Thanks :)
@jokubaszitkevicius8243 Жыл бұрын
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!
@MuffinsAPlenty Жыл бұрын
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.
@Yougottacryforthis Жыл бұрын
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)!
@sumiyafatima32192 жыл бұрын
can you please let me what is the grometrical interpretation of rings
@samiazaman52405 жыл бұрын
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
@manishayadav11194 жыл бұрын
Nice ma'am I have no words
@alexlhanghal36104 жыл бұрын
Thank you for your videos, could you please share about first and second theorem of isomorphism..
@alexlhanghal36104 жыл бұрын
I really need it, now for upcoming semester exam
@ajaymaths54513 жыл бұрын
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.
@S.S.T82734 жыл бұрын
Which book best of Ring theory ... please tell me mam....
@helloitsme75534 жыл бұрын
Fun fact: in modular arithmetic, units of Z/nZ never have any factor in common with n other than 1 or -1
@shashwatavasthi14234 жыл бұрын
Yeah and that is so because numbers that have a common factor with n are zero divisors
@siyamislam95962 жыл бұрын
Very interesting...thank you... from Bangladesh
@Socratica2 жыл бұрын
Hello to our Socratica Friends in Bangladesh! We're so glad you're watching. 💜🦉
@paullivi84842 жыл бұрын
For the puzzle, the units are the elements of the ring which are relatively prime to n, ranging from 2 to 50, respectively.
@lukemartin58502 жыл бұрын
Thanks!
@Socratica2 жыл бұрын
Thank you for your kind support! We so appreciate it!! 💜🦉
@kami-brawlstars96353 жыл бұрын
who searched "unital ring" and this came up
@medicallifewithjohn5 жыл бұрын
Mashallah
@Rsingh12 жыл бұрын
The best abstract algebra explanation is here guys!
@amornthepboonsamrannjit Жыл бұрын
An expensive one please.
@Justdoit-zm8nz11 ай бұрын
❤
@davidshechtman47465 жыл бұрын
Why can't the concept of associates apply to square roots? Instead of using complex numbers.
@STIVESification7 жыл бұрын
Thanks for another great video
@ranuaziz2941 Жыл бұрын
Well explain
@hoanganhnguyen9714 жыл бұрын
Thanks a lot! Can you make video about UFDs, please?
@shotacercva47582 жыл бұрын
brillianrt video.very helpful!
@leelathapa85183 жыл бұрын
Good mam
@aavellancursos Жыл бұрын
Con esta profesora me enamoro de la matematica.
@bancodrut7 жыл бұрын
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 !
@shijuzashimura28653 жыл бұрын
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 ?
@MuffinsAPlenty3 жыл бұрын
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.
@Darkmatter3217 жыл бұрын
I gave thumb up, but have no idea what you're talking about. Good video!
@missingno96 жыл бұрын
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
@Alqureshipansar2 жыл бұрын
Mam 'is every division ring is field 'and if yes then how to show??
@MuffinsAPlenty2 жыл бұрын
No. The real Hamilton quaternions are a division ring which are not a field.
@hypergration_htan2 жыл бұрын
Every 'Commutative' Division Ring is a Field
@pinkumia18314 жыл бұрын
Nc
@saurabhsingh-ow7ue4 жыл бұрын
thank you madam........
@dip2905 жыл бұрын
Nicely explain ...thank a lot Mam ...very helpful for my upcoming interview
@kausikdas2364 жыл бұрын
Mam I can't see for giving subtitles, plz remove it
@Socratica4 жыл бұрын
Hello! You can turn off subtitles on your end (click the cc or gear icon).
@rashmipandey87524 жыл бұрын
What mean the symbol z[√2] it is given that it is a ring and we have to find all unit of this ring .?
@shashwatavasthi14234 жыл бұрын
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
@أحمدالشمري-ش5ش2ل6 жыл бұрын
What is the difference between units and fields ?
@vincentv.39924 жыл бұрын
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.
@MuffinsAPlenty4 жыл бұрын
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.
@vincentv.39924 жыл бұрын
@@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.
@anam.caballerowilson94214 жыл бұрын
🐧
@faisalamakah72534 жыл бұрын
easy to understand
@jalmar402987 жыл бұрын
5:00 what dose "up to" mean?
@sharavanakumar27377 жыл бұрын
Hulk used to say to Captain America in Age of Ultron :D :D
@wes69727 жыл бұрын
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. 😃
@jalmar402987 жыл бұрын
I understand your explanation. thank you
@math137 жыл бұрын
thanks fanatic vedio
@VelMurugan-me3xh6 жыл бұрын
49
@pallavlearn53487 жыл бұрын
I love your videos :D
@kunslipper7 жыл бұрын
Thank you so much.
@guilhemescudero91145 жыл бұрын
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
@evzhuravlev2 жыл бұрын
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.
@adamrubinson68757 жыл бұрын
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).
@MuffinsAPlenty7 жыл бұрын
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.
@adamrubinson68757 жыл бұрын
Ah, I see. It's been a while since I studies abstract algebra. "Prime element" has a precise definition with respect to rings.