You can simply use CRT in the ring Z_n to find the solutions of the equation x^2 = x. All of them have the form (b_1, b_2, ..., b_k), where all the b_k's are either 0 or 1, and n = p_1^{a_1}*p_2^{a_2}...p_k^{a_k} is the factorization of n. For example, 36 = 4*9 = 2^2*3^2. Then (0, 0) corresponds to 0, (1, 1) corresponds to 1, (0, 1) corresponds to 28, i.e. 28mod4=0 and 28mod9=1, and (1, 0) corresponds to 9, i.e. 9mod4=1 and 9mod9=0.
Although I'm not technically an algebraist, back in the day, I learned to love the subject using the first edition of John Fraleigh's text in my first course in abstract algebra at Wilkes University (1967). Both he and my professor expressed the joys of this subject. This video, like most, was great fun. Thanks
I also saw Fraleigh's text in 1968 but we used McCoy’s. The former was beautiful with coloring if you recall the first time anything like that was done in an advanced book. However Fraleigh's book was full of serious theoretical mistakes that were eventually corrected! Fraleigh’s father had a PhD in mathematics and was a professor but Fraleigh went to Princeton but couldn't get his doctorate. I don't think he ever did get it elsewhere.
@@roberttelarket4934 I do recall some problem earlier on with permutations and later when I was doing a 2-sem algebra course in graduate school, I tried to use Fraleigh's text as a reference for something and found the explanation at least to be confusing if not wrong, but that was the fall of 1969. I ended up in algebraic topology as a specialty. I do know that book was published with revisions up into the 200xs.
@@kappasphere Nope. Michael clearly says that "when you divide 16 by 7 you get a remainder of 9". Since when is the remainder greater than the divisor? This is definitely wrong (though 9 and 2 are of the same class, but that's not what he said).
@Arjan Hulsebos 2 is just a representative of a congruence class. while it is standard to use 2, there is nothing incorrect with using 9 or any other representative
@Arjan Hulsebos you can make the same argument for 16 then and call it incorrect. As @notfeelin said, 16 and 9 are just a representative of the congruence class 2.
The elements of Z/nZ are not integer remainders (like 0, 1, ..., 6). The elements are cosets of an ideal of Z, namely of nZ. That's why we use the notation Z/nZ: Z is the ring of integers, and nZ is the ideal {... -3n, -2n, -n, 0, n, 2n, 3n...}. Z/nZ is the quotient ring of Z by the ideal nZ. This works the same way as it works for groups and normal subgroups: the elements of Z/nZ are the cosets of the ideal, and addition/multiplication is defined by taking any representative integers from the cosets, adding or multipliying them, and then taking the result to be the coset containing that integer. It is important to note that this addition and multiplication is well defined (exactly because nZ is an ideal, similar to how this only works for normal subgroups). What this means is that it doesn't matter which representatives you use, you will always get the same coset as a result. For example, in Z/7Z, we may want to compute (5+7Z)*(4+7Z). To do this, we could use the representatives I've given here: 5*4 = 20, so the result is the coset 20+7Z, in other words the set {..-8, -1, 6, 13, 20, 27...}. So this is also equal to the coset 6+7Z, or -1+7Z, or any other representative +7Z. Alternatively, we could have used different representatives from the beginning: 5+7Z = -2+7Z, and 4+7Z = -3+7Z for example. So we could instead compute (-2)*(-3) = 6, and we again get the result 6+7Z, the same coset as above. Hopefully this makes it clear that it does not matter which representatives you use, because the elements are actually cosets of an ideal, not integers.
Did I miss it, or did he not actually do the problem in the thumbnail? Surely the ring in which 0/2 = 3 is Z6. Also, what is the difference in these integer rings and modulus operations? Why are they treated differently? Why is it not 0/2 is congruent to 3 (mod 6)?
What elementary school did you go to Michael? We never had additive inverses in elementary schools!!! School was completely different in the U.S. in the late 1950's! It was terrible even in junior high school(now called middle school).
So in R, x^2=x returns 2 solutions {0,1} aka zero divisors; in Z36 we get 4 {0,1,9,28}, but I noticed that in Zn (at least up to 10000) zero divisors are limited to 2^n (up to a max of 32) in number. Contrastingly x^3=x yields [1,2,3,5,6,9,15,18,27,45,54,81,135,162,243,405] as the possible sizes of the sets of zero divisors. Not sure what the consequence (or pattern) is, but...?
This is my first exposure to"zero divisors". What a cool concept! Thank you, Professor Penn. I imagine one use for zero divisors in (large) prime-number factorization techniques, and perhaps another in accounting for null spaces in higher geometries past the 4th or 5th spatial dimensions.
8:31 The epsilon of the dual numbers is another nilpotent (nilsquare) element. You can write it as a matrix, too. And I saw in Wikipedia's article "Smooth infinitesimal analysis" there are "nilsquare or nilpotent infinitesimals".
I'm sure you mentioned that ab-ac=0 or a(b-c)=0 equivalent to b=c provided there are no zero divisors. 3(b-c)=0 does NOT imply b=c modulo 6 since 3x2 = 0, 3 has no multiplicative inverse When we learned about integral domains, a standard exam question was to prove cancellation fails.
It's not a *big* blunder, considering that he obviously means the congruence class of 9, which is the same as the congruence class of 2. I'd call it a minor mistake, brought on by his committing himself to defining the underlying set of Z_7 to be {0, 1, 2, 3, 4, 5, 6} (chosen representatives of their congruence classes).
@@toddtrimble2555 It's not about Z7 or congruence classes. His big blunder is saying "If you divide 16 by 7 you get a remainder of 9." That's just wrong.
@@michaelpark It's wrong, yes, as I also said, but to call it a "big blunder" (which makes it sound like the whole video comes crashing down because of it) is an exaggeration. To repeat: not much more than a minor slip, such as we all make from time to time when we're focusing on bigger issues, and easily rectified along the lines I mentioned.
@@toddtrimble2555 In what sense does calling something a 'bug blunder' imply the video comes 'crashing down?' It doesn't. It was a big blunder because it's very obvious that he was wrong. The degree to which this affects the greater video is not relevant.
Another counterintuitive result from zero divisors are products being identical to each other with identical as with non-identical factors: Suppose a,b∈R with a≠0, b≠0 and ab=0. Further suppose c,d∈R with c≠0, d≠0, and c-d=b and since b≠0, c≠0, d≠0 and c-d=b therefore c≠d. So ab=0 ⇔ a(c-d)=0 ⇔ ac-ad=0 ⇔ *ac=ad with a≠0 and c≠d,* which is in one sense counterintuitive and in another one it is quite intuitive. (2·c=2·d if and only if c=d, but a quadratic matrix A might have a non-trivial kernel such that Ac=Ad=0 with c and d being elements of its kernel with c≠d.) Take for example that 2×2 matrices example in this video at 6:45: (1 2) (-2 -2) = (1 2) [(0 -5) - (2 -3)] = (1 2) (0 -5) - (1 2) (2 -3) = (0 0) (2 4) (1 1) = (2 4) [(6 -6) - (5 -7)] = (2 4) (6 -6) - (2 4) (5 -7) = (0 0) ⇔ (1 2) (0 -5) = (1 2) (2 -3) = (12 -17) ⇔ (2 4) (6 -6) = (2 4) (5 -7) = (24 -34)
Yes. That's the trivial ring which has only 1 element. And 0 = 1 in that ring of course ;) Which is why many proofs exclude the trivial ring (also called the Zero Ring it seems) because they often assume that 0 does not equal 1.
x^2 = x -> x^2 - x = 0 -> x^2 - 2x(1/2) + 1/4 = 1/4 -> (x-1/2)^2 = 1/4, thus if you can find any element where it squares to 1/4 you can simply add 1/2 to get an idempotent element. for the case of 1/2 and -1/2, they square to 1/4, so adding 1/2 to each gives 1 and 0, the common solutions. in the split-complex numbers, j^2=1 and so j/2 squares to 1/4, giving another idempotent (1/2 + j/2) which squares to itself; this solution also demonstrates another way of discovering idempotent terms: (1/2+j/2) = (1+j)/2, and (1+j) squares to 2(1+j), thus for all x^2 = ax, you can define y = x/a and find that y^2 = x^2/a^2 = ax/a^2 = x/a = y.
Hi, We can't answer to your "shorts" because there is no comment box in the shorts. Any way I answer to your last one here : I prefer the method without fraction decomposition.
In Spanish we often say 2 + 2 = 4 (for obvious things to be deducted). And I remember a maths teacher who said that we shouldn't say this, because in rings it isn't true. I have some very basic programming knowledge, and this video made me remember about the overflow of a variable, let's say you have a 8 bit unsigned integer, then if you add 1 to 255,you get 0.
Well you do get 0, but the carry flag should be set too. The result of the addition does loop back to 0 because of register - accumulator width sizes having the property of 2^(n-1) bit patterns before resetting. Then the carry flag as well as other bit flags can be used to determine the next instruction. This is just the innate nature of a finite state machine. This is what happens when you work with discrete values within computations as opposed to working with complete analog systems. And for this reason is why we have IEEE...
The important to realise for Z_36 is that 36 has a prime factorization of [2^2]*[3^2]. Any numbers that are multiples of 6 will automatically get the quality of being nilpotent in Z_36 (i.e. their powers will be identical to 0). There are numerous examples of other numbers in Z_36 that loop when they reach certain powers of themselves - most notably powers of factors of the prime divisors. In additions, others have powers that are identical to 1. The main takeaway is that numbers in Z_36 are equivalent to congruences modulo 36, and so most of these numbers obey modular principles, such as Fermat's Little Theorem.
Furthermore, the only numbers in Z_36 that satisfy have a particular form. The numbers 0 and 1 are trivial, but all other numbers take the form 4|x and 9|(x - 1), OR 9|x and 4|(x - 1). If we look at x2 = x, we can factorize this as (x)(x - 1) = 0. 0 and 1 are the normal solutions for this equation, but as there are now zero divisors to consider, these are not the only solutions. In Z_36, parity is preserved (0 is even, 36 = 0 in Z_36, 36 is even, therefore all even numbers in Z are even in Z_36), we know that either x or (x - 1) is even, and the other must be odd. To be a pair of zero divisors in Z_36, the product must be a multiple of 36 in Z. With one factor being even, we know this must be a multiple of 4. The other factor must be a multiple of 9, as it is not possible to find two consecutive numbers that are multiples of 3 in this ring. Two examples of numbers where x^n = x in Z_36 are 9 and 28.
In the 2x2 Matrix example A and B are zero divisors because det(B) = 0 and therefore B has no inverse? Is it correct that in the set of 2x2 invertible matrices there are no zero divisors?
Since det(AB)=det(A)*det(B), if both A and B are invertible, none of det(A) and det(B) can be 0, so can't be det(AB), and hence AB is invertible and cannot be 0.
@@ianfowler9340 I think they both have to be singular. Well, one could be the zero matrix, and the other something else, but we're not interested in that!
Can a ring be said to predict zero products of non-zero factors when it contains subsets in which one of the elements, while not being the 0 of the ring, has the usual properties of 0 when considered only within that subset?
Nah, don't think that works. Let a be in the subset, and 0' be our zero in the subset, then: 0' = 0'+0' = -a*0' + a*0' = (-a+a)0' = 0*0' = 0 This does assume -a is also in the set, but I think you're certainly losing some of your zero properties if you lose it. This is pretty similar to the classic 'can your set have two distinct identities', to which the answer is always no.
The quaternions are a non-commutative division ring. Although there are no zero divisors, there are infinitely many square roots of -1, lying on two circles. Would you be interested in talking about quandles?
"two circles"? Every normalized quaternion without a scalar term is a square root of -1, which is equivalent to the set of points on a 2-sphere (the normal 3D sphere). It's also the set of normalized line reflections aka normalized 180° rotations. I suppose you could plot it as a pair of points, one on a circle, and one on a semicircle, but that's just longitude and latitude.
It's not that kind of ring ;-) "The term "Zahlring" (number ring) was coined by David Hilbert in 1892 and published in 1897.[13] In 19th century German, the word "Ring" could mean "association", which is still used today in English in a limited sense (for example, spy ring),[14] so if that were the etymology then it would be similar to the way "group" entered mathematics by being a non-technical word for "collection of related things". According to Harvey Cohn, Hilbert used the term for a ring that had the property of "circling directly back" to an element of itself (in the sense of an equivalence)." en.wikipedia.org/wiki/Ring_(mathematics)#History
So is it fair to say that zero divisors always come in pairs? Something like factors of an integer. Also, 4 is not a zero divisor in zed7 because the only way to get 4x = 0 where x is in zed 7 is to have x = 0 and the "and" in the definition requires x to be non-zero?
@@ianfowler9340 Very good insight! Yes, that's right. If p is prime, then Z_p has no zero divisors. And the reasoning you were going through in your post can be used to show why.
The set of complex numbers _does_ form a ring under addition and multiplication. You may have heard people refer to it as a "field". A field is a special type of ring, so it's still a ring.
At 13:37, another solution is 28. The solutions on [0..35] are 0, 1, 9, 28.
Shouldn't 4x4=2 in Z7? How do you get to 9 in Z7?
That's what I was thinking, lol
I guess if you define 9 to be in Z7 then 9 = 2. Otherwise yeah I'm stumped too lol
that's the same thing mod 7
yes
9 isn't even an element of Z7.
"... that's because if you divide 16 by 7 you get a remainder of 9"... didn't you mean a remainder of 2, instead?
They're the same
9 is not in Z7... should be 2 no?
Yes, you're right.
Right, he made a (silly) mistake
A ring with no zero divisors is a domain meaning we can solve using the zero product rule.
You can simply use CRT in the ring Z_n to find the solutions of the equation x^2 = x. All of them have the form (b_1, b_2, ..., b_k), where all the b_k's are either 0 or 1, and n = p_1^{a_1}*p_2^{a_2}...p_k^{a_k} is the factorization of n. For example, 36 = 4*9 = 2^2*3^2. Then (0, 0) corresponds to 0, (1, 1) corresponds to 1, (0, 1) corresponds to 28, i.e. 28mod4=0 and 28mod9=1, and (1, 0) corresponds to 9, i.e. 9mod4=1 and 9mod9=0.
Although I'm not technically an algebraist, back in the day, I learned to love the subject using the first edition of John Fraleigh's text in my first course in abstract algebra at Wilkes University (1967). Both he and my professor expressed the joys of this subject. This video, like most, was great fun. Thanks
I also saw Fraleigh's text in 1968 but we used McCoy’s. The former was beautiful with coloring if you recall the first time anything like that was done in an advanced book. However Fraleigh's book was full of serious theoretical mistakes that were eventually corrected!
Fraleigh’s father had a PhD in mathematics and was a professor but Fraleigh went to Princeton but couldn't get his doctorate. I don't think he ever did get it elsewhere.
@@roberttelarket4934 I do recall some problem earlier on with permutations and later when I was doing a 2-sem algebra course in graduate school, I tried to use Fraleigh's text as a reference for something and found the explanation at least to be confusing if not wrong, but that was the fall of 1969. I ended up in algebraic topology as a specialty. I do know that book was published with revisions up into the 200xs.
3:56 4.4 is not 9; but 2 in Z7
9=2 in Z7, so it doesn't make a difference
@@kappaspherethere is no such thing as 9 in Z7
@@kappasphere Nope. Michael clearly says that "when you divide 16 by 7 you get a remainder of 9". Since when is the remainder greater than the divisor? This is definitely wrong (though 9 and 2 are of the same class, but that's not what he said).
3:58 I think 4 time 4 should be 2. If you divide something by 7, remainder is always between 0 and 7
we can write it as 9 too as 9 and 16 both leave the remainder of 2 when divided by 7. So 9 and 16 are congruent to each other under mod 7.
@Arjan Hulsebos 2 is just a representative of a congruence class. while it is standard to use 2, there is nothing incorrect with using 9 or any other representative
@Arjan Hulsebos you can make the same argument for 16 then and call it incorrect. As @notfeelin said, 16 and 9 are just a representative of the congruence class 2.
The elements of Z/nZ are not integer remainders (like 0, 1, ..., 6). The elements are cosets of an ideal of Z, namely of nZ. That's why we use the notation Z/nZ: Z is the ring of integers, and nZ is the ideal {... -3n, -2n, -n, 0, n, 2n, 3n...}. Z/nZ is the quotient ring of Z by the ideal nZ. This works the same way as it works for groups and normal subgroups: the elements of Z/nZ are the cosets of the ideal, and addition/multiplication is defined by taking any representative integers from the cosets, adding or multipliying them, and then taking the result to be the coset containing that integer.
It is important to note that this addition and multiplication is well defined (exactly because nZ is an ideal, similar to how this only works for normal subgroups). What this means is that it doesn't matter which representatives you use, you will always get the same coset as a result.
For example, in Z/7Z, we may want to compute (5+7Z)*(4+7Z). To do this, we could use the representatives I've given here: 5*4 = 20, so the result is the coset 20+7Z, in other words the set {..-8, -1, 6, 13, 20, 27...}. So this is also equal to the coset 6+7Z, or -1+7Z, or any other representative +7Z.
Alternatively, we could have used different representatives from the beginning: 5+7Z = -2+7Z, and 4+7Z = -3+7Z for example. So we could instead compute (-2)*(-3) = 6, and we again get the result 6+7Z, the same coset as above. Hopefully this makes it clear that it does not matter which representatives you use, because the elements are actually cosets of an ideal, not integers.
@@addyraptor3675 He's right. Look at definition on the blackboard. 9 is not in Z7.
3:58 Shouldn't it be 4*4=2?
Of course! There is no 9 in Z_7.
In the range Zed-7 wouldn't 4 * 4 = 2 ? 9 being outside Zed-7 ?
Did I miss it, or did he not actually do the problem in the thumbnail?
Surely the ring in which 0/2 = 3 is Z6.
Also, what is the difference in these integer rings and modulus operations? Why are they treated differently? Why is it not 0/2 is congruent to 3 (mod 6)?
Will have to check out your abstract algebra course. Vcool!
Python 1-liner to compute the solutions of x^2 = x mod 36:
[i for i in range(36) if i**2 % 36 == i % 36]
=> [0, 1, 9, 28]
What elementary school did you go to Michael? We never had additive inverses in elementary schools!!! School was completely different in the U.S. in the late 1950's! It was terrible even in junior high school(now called middle school).
@ 3:59 wouldn't 4x4 be 2 in Z7 ?
I like this video a lot. It feels like a guided tour through these examples, similar to your videos on dual numbers.
So in R, x^2=x returns 2 solutions {0,1} aka zero divisors; in Z36 we get 4 {0,1,9,28}, but I noticed that in Zn (at least up to 10000) zero divisors are limited to 2^n (up to a max of 32) in number. Contrastingly x^3=x yields [1,2,3,5,6,9,15,18,27,45,54,81,135,162,243,405] as the possible sizes of the sets of zero divisors. Not sure what the consequence (or pattern) is, but...?
3:51 WOAH, WOAH, WOAH! 16 (mod 7)=2, so 4*4 DOES NOT EQUAL 9, BUT 2!!!
This is my first exposure to"zero divisors". What a cool concept! Thank you, Professor Penn.
I imagine one use for zero divisors in (large) prime-number factorization techniques, and perhaps another in accounting for null spaces in higher geometries past the 4th or 5th spatial dimensions.
8:31 The epsilon of the dual numbers is another nilpotent (nilsquare) element. You can write it as a matrix, too. And I saw in Wikipedia's article "Smooth infinitesimal analysis" there are "nilsquare or nilpotent infinitesimals".
14:24 Ah, good old zero divisors
So fast 😮
The card with the video mentioned at 7:17 does not appear
link (that way just in case youtube doesn't like actual links in comments) /watch?v=cc1ivDlZ71U
How do you make a channel ad-free? Not by like how you get money, but youtube doesn't let you have no ads.
Ugh, that thumbnail is awful. You should swap it out for something else.
You forgot to mention that rings need to have a 1....
At 3:50 4 x 4 is actually equal to 2 in Z_7. 9 is not in Z7.
It is
9 is in Z_7, it's just equivalent to 2 Z_7. It's like writing 15/3 in Q or R instead of 5, it's not wrong, just redundant
13:35 28^2 mod 36 is also 28.
I'm sure you mentioned that ab-ac=0 or a(b-c)=0 equivalent to b=c provided there are no zero divisors.
3(b-c)=0 does NOT imply b=c modulo 6 since 3x2 = 0, 3 has no multiplicative inverse
When we learned about integral domains, a standard exam question was to prove cancellation fails.
3:54 9 isn't even an element of Z7. This was a pretty big blunder. Instead, 4 • 4 = 2 in Z7.
It's not a *big* blunder, considering that he obviously means the congruence class of 9, which is the same as the congruence class of 2. I'd call it a minor mistake, brought on by his committing himself to defining the underlying set of Z_7 to be {0, 1, 2, 3, 4, 5, 6} (chosen representatives of their congruence classes).
@@toddtrimble2555 It's not about Z7 or congruence classes. His big blunder is saying "If you divide 16 by 7 you get a remainder of 9." That's just wrong.
@@michaelpark It's wrong, yes, as I also said, but to call it a "big blunder" (which makes it sound like the whole video comes crashing down because of it) is an exaggeration. To repeat: not much more than a minor slip, such as we all make from time to time when we're focusing on bigger issues, and easily rectified along the lines I mentioned.
@@toddtrimble2555 In what sense does calling something a 'bug blunder' imply the video comes 'crashing down?' It doesn't. It was a big blunder because it's very obvious that he was wrong. The degree to which this affects the greater video is not relevant.
Where is the link at the end of the video
تذكير رائع.عودة إلى سنوات الجامعة 1994-1998 قبل اليوتيوب
Another counterintuitive result from zero divisors are products being identical to each other with identical as with non-identical factors:
Suppose a,b∈R with a≠0, b≠0 and ab=0.
Further suppose c,d∈R with c≠0, d≠0, and c-d=b and since b≠0, c≠0, d≠0 and c-d=b therefore c≠d.
So ab=0 ⇔ a(c-d)=0 ⇔ ac-ad=0 ⇔ *ac=ad with a≠0 and c≠d,* which is in one sense counterintuitive and in another one it is quite intuitive. (2·c=2·d if and only if c=d, but a quadratic matrix A might have a non-trivial kernel such that Ac=Ad=0 with c and d being elements of its kernel with c≠d.)
Take for example that 2×2 matrices example in this video at 6:45:
(1 2) (-2 -2) = (1 2) [(0 -5) - (2 -3)] = (1 2) (0 -5) - (1 2) (2 -3) = (0 0)
(2 4) (1 1) = (2 4) [(6 -6) - (5 -7)] = (2 4) (6 -6) - (2 4) (5 -7) = (0 0)
⇔ (1 2) (0 -5) = (1 2) (2 -3) = (12 -17)
⇔ (2 4) (6 -6) = (2 4) (5 -7) = (24 -34)
wouldnt every operation in the Z1 world be 0? cuz 1 is a divisor of every number
Yes. That's the trivial ring which has only 1 element. And 0 = 1 in that ring of course ;) Which is why many proofs exclude the trivial ring (also called the Zero Ring it seems) because they often assume that 0 does not equal 1.
x^2 = x -> x^2 - x = 0 -> x^2 - 2x(1/2) + 1/4 = 1/4 -> (x-1/2)^2 = 1/4, thus if you can find any element where it squares to 1/4 you can simply add 1/2 to get an idempotent element. for the case of 1/2 and -1/2, they square to 1/4, so adding 1/2 to each gives 1 and 0, the common solutions. in the split-complex numbers, j^2=1 and so j/2 squares to 1/4, giving another idempotent (1/2 + j/2) which squares to itself; this solution also demonstrates another way of discovering idempotent terms: (1/2+j/2) = (1+j)/2, and (1+j) squares to 2(1+j), thus for all x^2 = ax, you can define y = x/a and find that y^2 = x^2/a^2 = ax/a^2 = x/a = y.
(-2,1)/(-2,1) looks singular to me, isnt it "zero/like" ?
It'a not zero in the sense that it's not an additive identity in the ring.
At about 11:35 the blue g(x) should be minus x so -x for x
@7:57...Did he say he could look inside the MATRIX! He's the chosen one!
Hi,
We can't answer to your "shorts" because there is no comment box in the shorts. Any way I answer to your last one here : I prefer the method without fraction decomposition.
To the right of each short there is a button with a speech bubble, click it and a comment section similar to this will appear.
@@Escviitash Thanks a lot!
❤❤❤
In Spanish we often say 2 + 2 = 4 (for obvious things to be deducted).
And I remember a maths teacher who said that we shouldn't say this, because in rings it isn't true.
I have some very basic programming knowledge, and this video made me remember about the overflow of a variable, let's say you have a 8 bit unsigned integer, then if you add 1 to 255,you get 0.
Well you do get 0, but the carry flag should be set too. The result of the addition does loop back to 0 because of register - accumulator width sizes having the property of 2^(n-1) bit patterns before resetting. Then the carry flag as well as other bit flags can be used to determine the next instruction. This is just the innate nature of a finite state machine. This is what happens when you work with discrete values within computations as opposed to working with complete analog systems. And for this reason is why we have IEEE...
The important to realise for Z_36 is that 36 has a prime factorization of [2^2]*[3^2]. Any numbers that are multiples of 6 will automatically get the quality of being nilpotent in Z_36 (i.e. their powers will be identical to 0).
There are numerous examples of other numbers in Z_36 that loop when they reach certain powers of themselves - most notably powers of factors of the prime divisors. In additions, others have powers that are identical to 1. The main takeaway is that numbers in Z_36 are equivalent to congruences modulo 36, and so most of these numbers obey modular principles, such as Fermat's Little Theorem.
Furthermore, the only numbers in Z_36 that satisfy have a particular form. The numbers 0 and 1 are trivial, but all other numbers take the form 4|x and 9|(x - 1), OR 9|x and 4|(x - 1).
If we look at x2 = x, we can factorize this as (x)(x - 1) = 0. 0 and 1 are the normal solutions for this equation, but as there are now zero divisors to consider, these are not the only solutions. In Z_36, parity is preserved (0 is even, 36 = 0 in Z_36, 36 is even, therefore all even numbers in Z are even in Z_36), we know that either x or (x - 1) is even, and the other must be odd. To be a pair of zero divisors in Z_36, the product must be a multiple of 36 in Z. With one factor being even, we know this must be a multiple of 4. The other factor must be a multiple of 9, as it is not possible to find two consecutive numbers that are multiples of 3 in this ring. Two examples of numbers where x^n = x in Z_36 are 9 and 28.
In the 2x2 Matrix example A and B are zero divisors because det(B) = 0 and therefore B has no inverse? Is it correct that in the set of 2x2 invertible matrices there are no zero divisors?
But then you'd have to find the zero element
That's right! It's impossible for an element to both have an inverse and be a zero divisor.
Since det(AB)=det(A)*det(B), if both A and B are invertible, none of det(A) and det(B) can be 0, so can't be det(AB), and hence AB is invertible and cannot be 0.
@@samueldevulder So at least one of A,B must be singular.
@@ianfowler9340 I think they both have to be singular. Well, one could be the zero matrix, and the other something else, but we're not interested in that!
What is the rule for the multiplication shown at 6:09?
As he says it's component-wise. So if you have two points (a, b) and (c, d) then their product (a, b)(c, d) = (ac, bd)
@@APaleDot Thank you for it!
English is not my native laguage, I overlooked the expression.
Can a ring be said to predict zero products of non-zero factors when it contains subsets in which one of the elements, while not being the 0 of the ring, has the usual properties of 0 when considered only within that subset?
Nah, don't think that works. Let a be in the subset, and 0' be our zero in the subset, then:
0' = 0'+0' = -a*0' + a*0' = (-a+a)0' = 0*0' = 0
This does assume -a is also in the set, but I think you're certainly losing some of your zero properties if you lose it. This is pretty similar to the classic 'can your set have two distinct identities', to which the answer is always no.
The quaternions are a non-commutative division ring. Although there are no zero divisors, there are infinitely many square roots of -1, lying on two circles.
Would you be interested in talking about quandles?
"two circles"? Every normalized quaternion without a scalar term is a square root of -1, which is equivalent to the set of points on a 2-sphere (the normal 3D sphere). It's also the set of normalized line reflections aka normalized 180° rotations. I suppose you could plot it as a pair of points, one on a circle, and one on a semicircle, but that's just longitude and latitude.
4x4=2mod7
They are the same
9 z7 should be 2
It's the same
The word “ring” suggests a physical object shaped like a torus. What analogy suggests this label for these abstract sets?
It's not that kind of ring ;-)
"The term "Zahlring" (number ring) was coined by David Hilbert in 1892 and published in 1897.[13] In 19th century German, the word "Ring" could mean "association", which is still used today in English in a limited sense (for example, spy ring),[14] so if that were the etymology then it would be similar to the way "group" entered mathematics by being a non-technical word for "collection of related things". According to Harvey Cohn, Hilbert used the term for a ring that had the property of "circling directly back" to an element of itself (in the sense of an equivalence)."
en.wikipedia.org/wiki/Ring_(mathematics)#History
@@Nikolas_Davis interesting. Thanks very much.
9:36 rare RELU appearence
zero divisors are so fun
So does the ring of differentiable functions have any zero divisors? I suspect the answer is no, but I'm not sure how to prove that.
Try with f(x) = x² when x>=0, 0 otherwise and g(x)=x² when x
I think the ring of analytic functions C->C avoids 0 divisors the way you're thinking, but not with real numbers.
"Rings that don't occur in everyday arithmetical structures" - Mathematician with a clock allergy
Clocks are generally only the additive group, which is why you don't think of 4 am and 6 am as midnight divisors.
@@iabervon Give me an cyclic group and I'll give you surjective ring homomorphism from the integers to it.
So is it fair to say that zero divisors always come in pairs? Something like factors of an integer.
Also, 4 is not a zero divisor in zed7 because the only way to get 4x = 0 where x is in zed 7 is to have x = 0 and the "and" in the definition requires x to be non-zero?
i think so yeah
No for example in Z_4 we have that 2 is a zero divisor because 2x2 = 0. But there is no other zero divisor.
@@TheEternalVortex42 So if p is prime then Z_p has no zero divisors?
@@ianfowler9340 Very good insight! Yes, that's right. If p is prime, then Z_p has no zero divisors. And the reasoning you were going through in your post can be used to show why.
Why are the complex numbers not a ring?
They are a ring. In fact, like the real and rational numbers, they are a special type of ring called a field.
The set of complex numbers _does_ form a ring under addition and multiplication. You may have heard people refer to it as a "field". A field is a special type of ring, so it's still a ring.