there are only two prime matrices.

DID SOMEONE ORDER A PRIME...*the waiter squints*...m...matrix?...wot? Is this a joke? What...what does that even mean...who...who ordered a Prime Matrix?? glances under plate cover AHHHHHHHH!!! The waiter was horrified to find out that if they didn't watch this video till the end they will NEVER KNOW what was under the cover.(If you don't think that Chalk is behind this scheme, this cruel scheme pulled on an unsuspecting waiter well...I'll have you know that a pile of broken bricks could outwit you)
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[@txikitofandango]

I'm loving this 2x2 matrix arc

[@txikitofandango]

You get so much understanding by constraining to 2x2 structures of nonnegative integers

[@urekah37]

I was thinking the same it allows you to break things down into their base structure.

[@wolliwolfsen291]

The Q inverse is twice wrong at +/- 14:00, but the calculation continues with the correct Q inverse.

[@ingiford175]

Came here to say this.

[@krisbrandenberger544]

Yes. The lower right entry of Q inverse should be 1, not 0.

[@stavrospavlou5873]

There is a mistake in the inverse matrix of Q. The (2,2) element should be 1, not 0.

[@adamonus6473]

Its a subtle point, but I really enjoy that the proof at 12:00 shows that not only is every non-identity A a multiple of P or Q, but it is always a *left*-multiple of P or Q, so you only need to consider cases 1 and 3 at 20:30

[@krisbrandenberger544]

@19:32 The top case should be d>c not d

[@amaarquadri]

On the board at 10:21, you forgot the case where x=0 instead of c=0. I'm guessing those would also give you the trivial factorizations.

[@oliviermiakinen197]

Yes I had the same remark. Then I did the work and found that it leads to a product of matrices whose determinants are -1 instead of 1. So they are outside our domain. However you are right this case should not be forgotten.

[@quark67000]

0:03 For French people (and probably people from other countries), the notation ℤ⩾0 can be simplified with the notation ℕ, because in French, the ℕ set contains the number 0 (and we uses ℕ* for the set without the number 0).

[@Errenium]

my preferred notation convention. bless the French

[@user-jc2lz6jb2e]

"Primes" is the wrong term here. The property that "p = ab implies that a or b is 1" is called "irreducibility" (well, not exactly, because the actual definition says a or b is a unit), which is not the same as primality. Primality means if p divides ab, then p divides a or b. These two notions coincide in a unique factorization domain, like Z. These notions are actually for rings and not groups, but you could extend the set M to be of all 2x2 matrices over Z and still talk about these.

[@Anonymous-zp4hb]

"primes" gives the video a snappier title.

[@cxpKSip]

​In basic number theory courses, we use the irreducible definition as our definition of primes as fact in $\mathbb{Z}$, and the go on to prove the prime "definition" as Euclid's Lemma.

[@mtaur4113]

I thought in most algebra settings, irreducible was a weaker condition than prime. The definition given for prime is usually called irreducible. I forget when the two are equivalent, it could be the case here. But still.

[@adamc973]

​@blueredandyellow8389 We can say something quite strong about when matrix rings are integral domains. Rings of matrices with size > 1, over a non-trivial ring R, are never integral domains, since we can always take: M := 0 1 0 0 and then M^2 = 0.

[@thehockyebox4517]

Solution to the homework: generally the matrix ((x, x+1), (x-1, x)) with x on the diagonal and x+1, x-1 on the off-diagonal will have a decomposition PQ^{x-1}P. In our case x=7.

[@calvincrady]

10:05 The proof of this is incomplete; you also have to consider the case where x=0. In that case we get that the two matrices are ((1 1) (1 0)) and ((0 1) (1 0)), whose determinants are -1 and are therefore not elements of M.

[@talinuva]

In general, a matrix of the form [[n+1,n+2][n,n+1]] can be factored as PQⁿP, and the transpose is factored as QPⁿQ.

[@jgilferez]

We can compute Q^-1P(a b \\ c d) = (a+c b+d \\ -a -b). This shows that if X is in M, then Q^-1PX is not, or equivalently, for all X,Y in M, PX =/= QY. (Analogously, for all X,Y in M, XP =/= YQ. ) From this we can deduce that the factorization of a matrix in M as a product of P’s and Q’s is unique.

[@ingiford175]

That was what I was wondering

[@Uranyus36]

this is a very very cool idea about implementing primeness into things beyond numbers. one immediate question coming out of my mind is that is the factorisation unique? that means for whatever prime matrices we use to factorise the non-prime matrices from whatever direction, it will end up with the same products of P's and Q's

[@marinchan6334]

You will get to know all this details in ring theory course,where we genralise defintions of primes and irreducible elements.

[@stanleydodds9]

This is the right question to ask, because it really brings up the point that what he gave here is not the definition of a prime, but rather (close to) the definition of an irreducible element. They become equivalent ONLY if factorisation is unique, or in other words, when the ring is a unique factorisation domain. In general, and irreducible element is one where its only factorisations are a unit times something (a trivial factorisation, since you can factor a unit out of anything). A prime is an element such that if it divides any product of two elements, then it must divide at least one of the two. In both cases, you also define them to exclude the trivial case of being a unit themselves. An example of something that is irreducible but not prime would come from the classic example of an integral domain that is not a unique factorisation domain. In almost any course, the example you'll be given is Z[sqrt(-5)], i.e. numbers that look like a+b*sqrt(-5). Lets look at 6. We can write 6 = 2*3, but we can also write 6 = (1+sqrt(-5))*(1-sqrt(-5)). It's easy to verify that all of 2, 3, 1+sqrt(-5), and 1-sqrt(-5) are irreducible: the trick is to define a norm which preserves multiplication and maps to the integers. In this ring, the norm to use is N(a + b*sqrt(-5)) = a^2 + 5b^2. You'll find that the norms of the 4 factors above are 4, 9, 6, and 6 respectively, and that there are no elements with norm 2 or 3. So any factorisation must include an element with norm 1, and the only such elements are the units 1 and -1. But in this proof that they are irreducible, we now also have an easy way to show that none of them are prime. 2 divides into 6 = (1+sqrt(-5))*(1-sqrt(-5)), but 2 divides neither of these factors (which you can check by it's norm, 4, not dividing either norm of 6). Similarly, 3 does not divide either factor despite dividing the product, and using the other factorisation, the other pair of irreducibles are also not primes. So just from having a ring which is not a unique factorisation domain, we get for free several elements which are irreducible, but not prime.

[@Uranyus36]

Very interesting! I thought the things shared in this video are merely some fun implementation of an existing idea to somewhere else, but it turns out to be a proper serious math subject. It makes me want to look into ring algebra more. Thank you both for the insights and thorough explanations.

[@marinchan6334]

@@Uranyus36 But just to remind you the space SL(2,Z_(>=0)) which is equal to M in the above video is a group with respect to matrix multiplication( not a ring as it is not closed under normal matrix addition)

[@filbranden]

Great video! I think one interesting question left unexplored here is whether the factorization into primes is unique or not. Either prove it's unique, or show counter examples. I guess there's something to be said about commutativity between P and Q such as (P^n Q^m) = (Q^x P^y) which would indicate there are multiple factorizations, or showing there are no such (m, n, x, y) for which that is true would indicate that would not be possible. (So, without digging further, my instinct is that factorizations are probably unique here.) In any case, great video, very instructive, quite a nice exploration of a unique and pretty interesting ring!

[@jgilferez]

Is it a ring, though? ;-)

[@jgilferez]

Indeed, P^nQ^m = (mn+1 n \\ m 1) and Q^xP^y = (1 y \\ x xy+1). So, they are different, for n,m,x,y>0 (and trivially equal if some of these numbers is 0). But, what about other longer terms?

[@IIIoneHrayep]

This matrix can be used for representation of dual numbers: x + e*y, where e^2=0 and e is not a zero. You can look for more details in wiki. I used another representation of these numbers in order to define norm(distance) for dual numbers in deep learning (here you can read more doi:10.1109/AVSS56176.2022.9959227 ).

[@xnick_uy]

Such a cool result! This factorization could be used to enconde any binary string of abritrary length as a 2x2 matrix in M. Maybe it is already used for that, somewhere out there, but I have no idea. 14:50 Shouldn't we also consider the case such that all the entries are in fact natural numbers but the determinant is not equal to 1?

[@megaing1322]

It can't happen that the determinate is not equal to 1, since it's the product of two matrices whos determinate is equal to 1.

[@aadfg0]

We can already encode binary strings as numbers, which are 1x1 matrices.

[@xnick_uy]

@@megaing1322 I see!! Thank you very much for the clarification.

[@АндрейДенькевич]

so have an idea. Let introduce (-a)-ary d-digit number system with only difference ,inversed carry , not a^n as usual but a^(-n). Infinity_a(biggest number) of (-a)-ary infinity-digit number system equals a. Infinity_a_d of (-a)-ary d-digit number system equals (a-1)*Sum a^(-i); for i:=0 to d; or infinity(-a,d)=(a^(d+1)-1)/a^d. And so we begin at point. it generates 3 types of 1-oorder shapes: Examples(pascal triangles) is for d=0..4. 1) d-vertex simplex (2^n-shape, 2^n-ary d-digit number system). 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 2) d-donut (even-shape, even-ary d-digit n/s). d--donut_4 for example: 1 2 2 4 8 4 8 24 24 8 16 64 96 64 16 3) d-cube. Counter in (exterior xor interior): 3.1) positive curvature (in exterior) odd-shape, odd-ary d-digit n/s). d-cube_3 for example: 1 2 1 4 4 1 8 12 6 1 16 32 24 8 1 3.2) negative curvature(in interior)(-odd)-shape, (-odd)-ary d-digit n/s). dark d-cube_3 for example: 1 1 2 2 5 2 4 12 9 2 8 28 30 13 2 1 1 1 1 1 * 1 1 = 2 1 1 2 1 1 2 1 4 4 1 1 3 3 1 1 3 3 1 8 12 6 1 1 4 6 4 1 1 4 6 4 1 16 32 24 8 1 1 1 1 1 1 * 2 1 = 3 1 1 2 1 4 4 1 9 6 1 1 3 3 1 8 12 6 1 27 27 9 1 1 4 6 4 1 16 32 24 8 1 81 108 54 12 1 1 1 1 2 2 * 2 1 = 6 2 4 8 4 4 4 1 36 24 4 8 24 24 8 8 12 6 1 216 216 72 8 16 64 96 64 16 16 32 24 8 1 1296 1728 864 192 16 1 1 1 2 1 * 2 1 = 4 1 4 4 1 4 4 1 16 8 1 8 12 6 1 8 12 6 1 64 48 12 1 16 32 24 8 1 16 32 24 8 1 256 256 96 16 1 So we receive following aryphmetic: in Pascal triangle wich represent number system, shape: 1) c[d,i]=c[d-1,i]*c[2,1]+c[d-1,i-1]*c[2,2]. 1.1) c[d,1] means most curved face,vertexor exterior(if c[2,1]=1). 1.2) c[d,d] means least curved face, interior . 2) c[1,1]^(1-1)=1 is carry into first digit . (c[2,1]+c[2,2])^(2-1) is carry into second digit (c[d, 1]+..+c[d,d])^(d-1) is carry into d digit. 3) (n-1)-order n/s nested into (n)-order n/s infinity + carry into first digit of (n-1)-order n/s equals carry into first digit of (n)-order n/s. sphere (1;1;..); 1-ary n/s. Point without exterior and without interior. simplex^1=simplex_2; (1;1 1;..). 2-ary n/s. Point with (exterior xor interior). if d=1(1-vertex simplex) then Counter in (interior xor exterior) else Counter dislocated and counts exterior and interior. simplex^2=simplex_3(cube_3 or common cube); (1;1 1;..)*(1;1 1;..)=(1;2 1;..); 3-ary n/s. Counter in exterior because 2>1. 2 vertices 1 interior(edge),1-chain. simplex^3=simplex_4; (1;1 1;..)*(1;2 1;..)=(1;3 1;..); 4-ary n/s; 3 vertices 1 interior . simplex^4=simplex_5; (1;1 1;..)*(1;3 1;..)=(1;4 1;..); 5-ary n/s. 4 vertices 1 interior . donut^1=donut_4; (1;2 2;..) 2 vertices 2 interiors(edges), 2-ring. donut^2=donut_10; (1;2 2;..)*(1;2 2;..)=(1;2^2+2^1 2^2;..); 10- ary n/s. 6 vertices 4 interiors . donut^3=donut_22; (1;6 4;..)*(1;2 2;..)=(1;2^3+2^2+2^1 2^3;..); 22-ary n/s. 14 vertices 8 interiors . donut^4=donut_46; (1;14 8;..)*(1;2 2;..)=(1;2^4+2^3+2^2+2^1 2^4;..); 46-ary n/s. 30 vertices 16 interiors . donut^1*simplex^1=donut_6(1;4 2;..);6-ary n/s simplex^1*donut^1=cube_5(1;3 2;..);5-ary n/s. sphere*(1;a b;..) *sphere^T=(1;1;..)*(1;(a+b);(a+b)^2, (a+b)^3;..) =infinity of (a+b) -ary n/s divided by (a+b-1).

[@kianushmaleki]

Interesting video. Thanks for the upload. When I heard you say x, y, z, and w i realized that you are a mathematician and I am a physicist. I expected x, y , z, and t. The lovely t 🙂

[@nHans]

w = ict. Does that help?

[@MrFlaviojosefus]

But, @ 15:20 (approximately), you must acknowledge that P inverse and Q inverse are already not inside M. M is defined as a, b, c, and d inside Z >= 0. P inverse and Q inverse contain negative entries violating this rule.

[@noahniederklein8081]

Shouldn't the first condition be d≥c and b≥a? Since d-c must be nonnegative in the matrix at the right? 19:39

[@petver29]

This is a very interesting example of non-commutative factorization domain.

[@jgilferez]

Except that it’s not a ring. (At least not with the usual addition of matrices.)

[@petver29]

@@jgilferez right. It is a monoid. What he has shown is that this monoid is free generated by two elements.

[@jgilferez]

Well, he didn’t show that the factorizations are unique (although, I could believe they are). So, to prove that it’s indeed the free monoid with two generators we still need to see that.

[@PrzemyslawSliwinski]

Slightly off-topic: some examples of the 2x2 matrices with a prime determinant can trivially be constructed*. Is this the case for larger ones too? --------------------- * However, if in a determinant formula, ad - bc = P both bc and P are primes, then is such a construction equivalent to the Goldbach conjecture?

[@mechmaker9346]

Definition of prime is actually a definition for irreducible element here. And it has such form just because all invertible matrices are identity matrice.

[@lukasschmitz9030]

A general trend I have noticed is that anglophone mathematicians rarely (if ever) use \mathbb{N} to refer to the natural numbers and tend/seem to write \mathbb{Z}\geq 0 to refer to the (non-negative depending on whether you define the natural numbers starting with 1 or 0) natural numbers. In the same way one more often Heras them speaking of "non-negative integers" or "positive integers" instead of just "natural numbers". In Germany, however, mathematicians do make use of these distinct "symbols" and you will rarely hear somebody say something like "non-negative integers" and much more often just "natural numbers". Does anybody know whether this is just a KZbin thing or whether this is an actual, "cultural" difference? If so does anybody know why that is?

[@jgilferez]

I don’t think this is just a KZbin thing. Considering 0 a natural number is not completely standard everywhere (as I wish it was). The expression “nonnegative integer” makes the reference clear, without need of further clarification on whether you’re including 0 as part of the natural numbers or not.

[@lukasschmitz9030]

@@jgilferez, you could just write \mathbb{N}_0 to clarify that you are including zero. Still, in Germany people usually use \mathbb{N} or \mathbb{N}_0.

[@jgilferez]

Yes, you could use these. But notice that N_0 is not self-explanatory. Admittedly, there are not many reasonable readings, but it’s not so obvious as Z_\geq 0.

[@lukasschmitz9030]

@@jgilferez, sure but that doesn't answer my question of why there is/seems to be such a cultural difference in usage.

[@jgilferez]

Correct, it doesn’t answer that question. 😅

[@samwalko]

23:00 I'm noticing a suspicious number of Fibonacci numbers. I believe F(n)*F(n+3)+1 = F(n+1)*F(n+2), giving an easy way to construct a few infinite families of matrices in this set. I won't prove it, but I did check it for n

[@PhilBoswell]

Where's the video about matrix multiplication promised at about 6:00❓

[@alexdemoura9972]

Video title: *Why do we multiply matrices the way we do??* KZbin video code = cc1ivDlZ71U After YT address use: *watch? v = cc1ivDlZ71U* Sorry, no links are allowed in YT comments.

[@MichaelPennMath]

Here you go: kzbin.info/www/bejne/mZSUmql6ob9qZ7c

[@iabervon]

It'd be interesting to talk about induction in M. Presumably, for a proposition F, if F(I) is true and F(A)=>F(PA) and F(A)=>F(QA), then F(B) for all B in M. It would be interesting to prove that this weird induction is valid, and prove something using it.

[@mrminer071166]

Is there then no obligation for the primes to multiply together, somehow, to GENERATE all the composites?

[@user-ej7sr3ow8b]

10:20 but if we let x is zero for P, then we will get a or w = 1 and another is 0, then b=c=y=z=1 and d=x=0, which the two matrices will still have a product P, but neither is P or I

[@meri7108]

yes but then the factors also aren’t in M

[@TheEternalVortex42]

Those matrices have determinant -1 so aren't in M

[@user-ej7sr3ow8b]

@@meri7108 oh, right

[@user-ej7sr3ow8b]

Oh yes

[@VideoFusco]

I believe that having only the matrices P and Q as prime is a consequence of having restricted the set only to matrices with determinant 1. What happens if you expand the set M to matrices with determinant in N or in Z?

[@SimonClarkstone]

22:37 That matrix contains only Fibonacci numbers, which reminded me where I'd seen P and Q before: You can use one of them to calculate Fibonacci numbers via matrix exponentiation.

[@shanathered5910]

is this also true for other unique factorisation domains like the Eistenstein integers or the Gaussian integers?

[@pierreabbat6157]

Those are complex numbers, which cannot be ordered, so the condition that all elements be nonnegative cannot be stated.

[@tessapugh868]

i'm worried about your descriptions bc yk its fine

[@seanbastian4614]

10:00 You forgot the case when we suppose x=0. In fact, you get two matrices in M which are not the identity. You end up having 2 cases: (a,b,c,d,w,x,y,z)=(1,1,1,0,0,0,1,1) or (a,b,c,d,w,x,y,z)=(0,1,1,0,1,0,1,1). When you write these matrices, none of them is the identity matrix.

[@meri7108]

yes but then the factors also aren’t in M

[@freewilly1337]

Yeah he forgot that case but I strongly believe both the P and Q are prime. With the last proof even the only primes. I do have an idea which requires no direct computation but is a bit more technical: You can define a partial order on M; A

[@TheEternalVortex42]

Those matrices have determinant -1 so aren't in M

[@TheEternalVortex42]

You're right he should have addressed this case in the video though.

[@seanbastian4614]

You all are right but he still should have checked it or left it as homework

[@vaevfunc]

I am a bit confused by this definition of being a prime matrix. If we take a one-dimensional matrix, i.e. a number, and say that it’s determinant is 1, then we end up with a single element. I think it could be more generalizable to define a prime matrix as one having positive integer determinant with no divisors in the same set of matrices (even though probably it doesn’t have an interesting structure, lol)

[@innokentiyromanchenko1450]

can i successfully achieve same results for bigger matrices? i mean finite amount of prime and hence factorization

[@halfe8128]

Was this recorded in the last few days? Room looking a smoky orange...

[@MichaelPennMath]

It's "warmth" from post-production additions. -Stephanie, Editor

[@ZekeRaiden]

Interesting. My understanding of the fundamental categories of number in number theory was: "Unit": Numbers which are equal to their own multiplicative inverse, or equivalently, numbers which have exactly one factor (the number itself.) 1 is the only unit in the naturals, integers, rationals, or reals. If there is only a single unit ("the" unit), then it is also the multiplicative identity element for a set. "Prime": Numbers which have exactly two integer factors, the multiplicative identity and the number itself; or equivalently your definition (if p=ab, then p=a and b=1, or p=b and a=1.) "Composite": Numbers which can be written as a product of at least two other elements,, neither of which are the multiplicative identity (or an equivalent definition built off of your definition of prime.) "Zero divisor": Numbers which have no multiplicative inverse. If only one zero divisor exists, then it is also the additive identity element for a set. This exhaustively categorizes all sets that behave analogously to the integers.

[@jgilferez]

These are not the standard definitions, nor they are equivalent to them. In a (commutative) ring (with unity), we can distinguish the following kinds of elements: Unit: any element with a multiplicative inverse, such as -1 in Z, and i in Z[i]. Irreducible: any nonzero element a that is not a unit and so that for every decomposition a = b•c, either b or c is a unit. (Composite: any nonzero element that is neither a unit nor irreducible.) Prime: any nonzero element a that is not a unit so that if a is a factor of b•c, then a is a factor of b or a is a factor of c. Zero divisor: any element a so that there is another element b such that a•b = 0.

[@ingiford175]

Quick nit: you have a-b>0, but do not explain why they can not be = 0, which is simple, if they are both zero, the determinate of the matrix is 0 and outside of M. Same with c-d

[@Risu0chan]

and that structure is called a (finitely generated, with rank 2) free group. [edit:group, not monoid]

[@arjunsigdel8070]

Is there any analogous to these matrices where entries are complex numbers?

[@brian554xx]

Zero asks "am I nothing to you?" Mathematician affirms.

[@АндрейДенькевич]

"1" asks "am i carry to you?" First digit affirms. Zero asks "am i carry to you?" "1" affirms.

[@rfvtgbzhn]

2:59 from where do we know "immediatly" that b, -b, c and -c must be greater or equal to 0?

[@MuffinsAPlenty]

The definition of the set M given at the beginning of the video.

[@rfvtgbzhn]

0:51 every number has a multiplicative inverse. I guess you mean an integer multiplicative inverse, which only -1 and 1 have.

[@zackarysemancik5491]

I'm a bit confused. How do we know that -b and -c are >= 0? Is it simply because A^-1 is an element of the set M?

[@spankasheep]

We require our inverse to be an element of the set, which (like you said) needs the entries of the inverse to be non-negative. Given the unique form of the inverse we get -b and -c bigger or equal 0

[@somecreeep]

7:08 I was confused why you only looked at the case c=0 and never x=0. Then I noticed that if x=0 then the determinant of that matrix is 1=-yz and since y,z>=0 it must be that a matrix with 0 in the diagonals can't be in M. Would have been nice of you to mention that for completeness, but I see now that you didn't simply fail to consider the other case due to forgetfulness.

[@ingiford175]

You get the same sort of logic with the other choice, just slight variation

[@somecreeep]

@@ingiford175 No you don't, you get an invalid solution that isn't in the original set M

[@johns.8246]

What are the "primes" of SL2(Z) in general? I guess a prime matrix A would be defined by A being divisible by only A, -A, I, and -I .

[@jgilferez]

In a group all elements have inverses, so everything is divisible by anything else: indeed, a = (a•b^-1)•b, so every b is a factor of a.

[@Cor97]

At 13 minutes. P inverse and Q inverse are not in M. So can you use this argument here?

[@csilval18]

Why isn't the matriz with only ones in the off diagonal a unit? It has an inverse, itself. Also, what happened to the determinant (1/(ad-bc)) when computing the inverse?

[@APaleDot]

We're only considering matrices which fall into the set M, which he defined as 2x2 matrices with determinant = 1 and positive integer entries. The matrix with ones on the off-diagonal has determinant -1, so it's not in this set, and when computing the inverse 1 / (ad-bc) just equals 1.

[@eugen9454]

Actually being prime is stronger than being irreducible which was shown.

[@emanuellandeholm5657]

This is the special linear group of degree 2 over the field of Z>=0, right? Just trying to sort out my terminology Edit: this is wrong, as demonstrated below

[@jgilferez]

Not really, the only element with an inverse is the identity matrix, as he shows in the video. So, all we can say is that it’s a (noncommutative) monoid generated by two elements (P and Q). Also, Z>=0 is not a field either, just a commutative semiring (in the sense of two commutative monoids on the same set, with one operation distributing with respect to the other).

[@emanuellandeholm5657]

@@jgilferez Thanks. I guess I kind of went with det != 0 implies inverses and ran with that, but over a subset of Z it's not that easy.

[@emanuellandeholm5657]

If the group was over (some nice subset of) the field of Q, would that make things special linear? :)

[@jgilferez]

Usually, “special linear group” refers to the set of square matrices of a fix size (say n) over a field (say F) with determinant 1, with the usual matrix multiplication, and it’s denoted SL(n,F) or something similar. The rationals don’t contain any proper subfield. So, unless one wants to extend the terminology for some reason, special linear groups with matrices of rational numbers will include matrices whose entries are all sorts of rationals.

[@emanuellandeholm5657]

@@jgilferez Thanks again! I'm learning so much tonight!

[@Maths_3.1415]

25:31

[@codatheseus5060]

I guess gaussian primes are a different thing But they have matrix representations

[@writerightmathnation9481]

Sorry, but the phrase “A is not a multiple of P and Q” is not the logical negation of “A is a multiple of P or Q”.

[@charleyhoward4594]

the only reason i didnt like this video was i didnt understand it

[@marinchan6334]

In 19:52 s d>c in 1st observation.