there are only two prime matrices. 

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

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

Came here to say this.

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

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

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.

my preferred notation convention. bless the French

​@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.

"primes" gives the video a snappier title.

​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.

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

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.

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.

@@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)

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

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

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.

That was what I was wondering

w = ict. Does that help?

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.

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

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

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).

Is it a ring, though? ;-)

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?

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

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

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

Here you go: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-cc1ivDlZ71U.html

I don’t think this is just a RU-vid 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.

@@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.

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.

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

Correct, it doesn’t answer that question. 😅

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.

yes but then the factors also aren’t in M

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

@@meri7108 oh, right

Oh yes

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

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.

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

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.

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

yes but then the factors also aren’t in M

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

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

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

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

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

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

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).

@@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.

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

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.

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