Zero divisors will change your view of arithmetic. 

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.

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.

9=2 in Z7, so it doesn't make a difference

@@kappascopezz5122there is no such thing as 9 in Z7

@@kappascopezz5122 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).

Yes, you're right.

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.

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.

Of course! There is no 9 in Z_7.

So fast 😮

They're the same

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.

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.

"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.

link (that way just in case youtube doesn't like actual links in comments) /watch?v=cc1ivDlZ71U

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...

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

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.

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.

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!

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.

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.

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.

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.

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@@Escviitash Thanks a lot!

It'a not zero in the sense that it's not an additive identity in the ring.

They are the same

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.

It's the same