Chaos

My worry is that technically, there are an infinite amount of Aons, right? You can add modifiers to any base Aon and it is a "different" Aon that has a different effect. Just as there effectively infinite Commands, there'd be infinite Aons. Two actually (that isn't the best image, but whatever), but that comment about four makes a compelling case. By the way, Reader, my favorite part of this theory is the comment about the Ten Heightenings. I don't think anyone picked up on the significance of the "ten" there before you, and that's seriously cool. I just wish the Selian magics would conform to this scheme; I wouldn't hesitate to espouse it if they did.

For Engraving, I fully expect that there are magics powered by a Splinter. Splinters can have an intent, so I have no reason not to suspect that this intent can't lead to a magic (with respective interaction with a Shardworld). Since we know there are more than just the core Shardworlds, I'm sure a magic like this exists. However, such magics would be a lot rarer, simply because there's a lot less energy in a Splinter than a Shard. It'd be less likely for a magic user to come into being. This would be so amazing. Fear or something like that would be a really cool Shard. However, Art isn't really an intent. Plus, Warbreaker already has color and art central to its magic, so I can't expect Brandon retread that again.

I don't know, man, maybe Ruin and Preservation agreed to just stuff it and do Feruchemy this way because they thought Feruchemy would be awesome? Seriously though, you have a really good point about the storing Preservation, because that isn't how Allomancy works. And Ruin would want power destroyed every time you tapped. It's like, hey, the process of storing isn't exclusively of Preservation, and tapping isn't exclusively of Ruin Like, almost as if their magic is balanced, and when you store, you simultaneously "use" Ruin and Preservation, and likewise for tapping. I guess really, technically speaking, if storing was totally of Preservation, you'd eat some metals and then to tap (because it's of Ruin), you'd need to materialize a spike in the Spiritual Realm, and spike that attribute to come back. Then you stare at yourself and you think, "Man, Shadesmar blood is really weird." Okay, so thanks for reminding me why both of these ideas are kind of ridiculous (Sorry, Windrunner.) EDIT: Upvote for you using surge and its synonyms I'd give you more if you became surge-monogamous, though, and stayed faithful to the word surge

I'm not totally sure at the moment. I think about it and ask KChan if she has some bright ideas. It's hard because the off-topicness, even if it was separate from the topic here, is still really random, so it's hard to encapsulate it in a new thread title.

Huh. I don't remember reading this before, but if I did, I'm a lot more receptive to that idea now. What Windrunner says below is true. Think about it this way, and separate the effects of the magic from how you get it. If you think about Hemalurgy, its effect is really not Ruinous at all--in fact, it gives you more power, and is additive in a way. However, how you get the power is certainly destructive and ruin-like. As it turns out, a magic being "of Ruin" or "of Preservation" or "of Honor" only means that the way you access the power is in line with the Shard's intent. This is (essentially) what the Principle of Intent says, though I need to clean up and revise it for clarity. But we see this working all the time. For Awakening to work, you have to Endow the power--this is separate from the actual effects of the Commands, which is more like "fetch keys" or whatever. Only the way the power is accessed needs be of Endowment. With Honor, the power is gained through some sort of oath or bond. With Devotion, everyone who the Shaod took was incredibly devoted to something. And with the Dakhor and Dominion, the power you get imbued with through the Dahkor ritual requires sacrifice and "dominating" that other person. It all makes sense. Now, you might ask, how the crap is the way you get Allomancy in line with Preservation? Well, think about it. With Hemalurgy, that fragment of the soul that is stolen is reduced/Ruined. With Feruchemy, your body's power is neither Preserved nor Ruined, but it is utilized. When you use Allomancy, however, neither your body or spirit is utilized at all--your body, and the fragment of Preservation inside you, is totally being Preserved. The direct consequence of this is that, because Preservation's magic will not let you destroy yourself in the process, you must draw on an external source: Preservation itself. It's a weird explanation, but in fact the canonical one. I agree. If this is how Feruchemy works, I agree with Windrunner's explanation more. You're "preserving" the power for later. Makes sense to me. That said, I'm still not totally convinced that it's necessary to think about Feruchemy in this way, but it's certainly a possibility.

You... kind of have an odd sense of humor Hopefully you can understand why being facetious sounded a lot more offensive than hilarious. You can't simply go from one kind of humor to a totally different one (joking and intentionally being inappropriate) and expect humor to be totally transitive. Honestly, looking at the post that was downvoted, it's really hard to find it funny. It feels a lot more attack than keeping things in the spirit of the thread. A simple and polite, "hey guys, can we keep it on topic or maybe split the topic? This really isn't a General Theories topic any longer" doesn't make you seem like a jerk in any capacity. In fact, that makes you seem even better, because you care about the overall organization of the forum. Instead, you chose to be intentionally inappropriate in a weird joke. That's a little less cool. Since this is the second time I've seen you be intentionally offensive for the sake of a laugh, I'd really like you to think very hard if that type of humor will actually have the effect you desire. That humor does have a place, but honestly, I can't think of a single context in terms of this community where that would be preferred over being polite. So... just watch out for that, okay? If someone could find it offensive, that may not be the best idea, you know?

I saw this idea in a previous thread, and I really like it. But how do you explain Elantris? Devotion and Dominion are certainly complementary, but there are lot more than sixteen Aons. Can you explain this fifth BioChromatic entity for me? We haven't seen this in the book, I'm pretty sure...

Relax, man. There's nothing wrong with a bit of looseness. You (as well as everyone else) always has the ability to add something on-topic to the discussion. I find that politely asking to get back on topic is a lot more effective. I personally don't have a problem with some off-topic fun, but if you want to steer the discussion to something more productive, you gotta admit there are a lot more polite and effective ways to do so, you know?

http://www.moviewavs.com/php/sounds/?id=bst&media=WAVS&type=Movies&movie=Star_Wars_Episode_IV_A_New_Hope&quote=weredoomed.txt&file=weredoomed.wav Also. http://www.youtube.com/watch?v=USHHQRodF88

I agree completely. We're doomed. I think that's likely as well. Recent quotes directly contradict the idea that Odium has multiple Shards. He doesn't want multiple Shards, because their intents would affect him. He has a Shard that is well matched for his personality, and doesn't want to change. Instead of collecting Shards, he wants to be the only Shard left.

That's so awesome! Congrats! Yes it is, actually. Go into edit your profile and go to the Settings tab. By the time zone, there's a DST check box. If the box is checked, try uncheck and save, then go back in and check it again. I know for me, DST is working, so I'm sure that's it I would like to announce that I just wrote eight pages (single spaced) about complex analysis for the Chaos's Math Help thread. If only my novel was so easy to write so much.

Thanks! That means a lot Since I am going to be a graduate student in the fall, and I'll need to teach precalculus... Your question is a very good one. To explain it, I need to talk to you a little bit about abstract algebra (which Joe here loves, so he may correct me). "Abstract algebra" sounds pretty scary, but really, at its core, we look at the properties of familiar number system--like, say, the real numbers--and we ask, "Hey, I wonder if these properties hold for different, weirder sets." An example of properties that we like would be the distributive property. a(b+c) = ab + bc There are a lot of properties that the integers, real numbers, etc. that have properties like this that we just take for granted. In the language of abstract algebra, those a,b, and c's don't have to be numbers. They could be a set. They could be a lot of weird things. We essentially "abstract" out what the elements of the space (integers, reals, complex numbers, sets, other things) and we say, "Well, if we define addition and multiplication appropriately, hey! Lookit that! The distributive property holds." In an abstract sense, other, weird monkey-butt objects might still have properties that we really like. So we give fancy names to things that have properties we like, like groups, rings, and fields, and then abstract algebra is about finding theorems about all groups. Now, here's the important thing when we do this: I can't just give you a list of numbers. I also have to defined an operation to act on these numbers. Like, say, addition. Because, obviously, I can define addition a lot of stupid ways. In the language of abstract algebra, let's see how I can define a new operation, which I'll call + rather than +, because the former is our new addition. Let's say our set of numbers we are applying this operation to is the integers. a + b = a + b. Okay, this is dumb--I've just defined my "new" abstract operation to be the familiar addition. So for any two a and b in our set, well. You just freaking add them. But I don't have to defined them this way. You might be thinking, "Wow, we're going to define a new addition! That's going to be tough," but you have to realize mathematicians their stupidly pathological examples Take this addition, also over integers. a + b = a + b + 1 It's a new operation, really. I'm going to add the two numbers together, and add 1 to them. I could do wackier things. a + b = ab Whoa, this is addition? Well, sure, + is just a name for an operation. It does "something" on two numbers. That something could be pretty weird. (Sorry about my previous post, I didn't know I could do superscripts) How we define our operations makes a big difference if we get things like, the distributive property to hold. In fact, for the distributive property, I also need to define a second operation, "multiplication". Let's make a multiplication right now! a x b = 2ab So, the regular multiplication, multiplied by 2. Do you feel powerful, defining a new freaking multiplication? Well, you can certainly write things down. As we can see, you can define a lot of operations. But in abstract algebra, we don't want just any operation; we want the operations that actually give useful properties. Let's check the distributive property for this multiplication and the last addition that I mentioned. I bet you it's not going to work at all. a x (b + c) Note that I'm being very careful differentiating between my abstract operation and our "normal" operations. In this form, the distributive property would read like this: a x (b + c) = a x b + a x c. It's exactly the same as a(b+c)=ab+ac, but I'm using it with different addition and multiplication. In actual abstract algebra books, though, you have to realize that they will just the + sign down and you have to remember "Hey, this is the abstract operation". Mathematicians are lazy. a x (b + c) = a x (bc) There, I did my weird addition. Now, actually, I defined these operations on the integers. My operations don't apply on anything that isn't an integer. So I need to check that bc is also an integer. In fact, typically the distributive property is the last property we prove for "rings"--one of the earlier properties is closure, which means that for a and b in our set, a + b is also in that set. It's easy to see that for any integers, b^c needs not be an integer. It's certainly true in the positive case, 2^5 is sure an integer, and you won't get any decimals out. But what if I choose c to be a negative number? 2^(-1). Both are integers, and we get out 1/2, if you recall your exponent rules about negative exponents. 1/2 is certainly not an integer. So really, without closure we can't really do too much. So I'm going to change "space" that we are working on to be the positive integers (0,1,2,3...). This way, our "distributive property check" is actually defined (I kind of like things to be defined, don't you?) But, we'll show the distributive property still fails. a x (bc) = 2abc There, I applied our multiplication. Now, for the distributive property to hold, we need a x (b + c) = a x b + a x c. So we what we want is (2ab)2ac I think it's pretty clear that for that these weird-chull operations, there's no way I can doctor it so 2abc actually equals (2ab)2ac. So the distributive property fails. This means that, technically speaking, these operations don't form a "ring", but that's not really important that you know what a ring is here Now, why did I go through this excruciatingly long exercise, instead of writing my novel like I'm supposed to today? To simply show you that there's a lot of ways to define addition and multiplication. We want to pick the "good" additions and multiplications so that we get the properties we want. Let's finally talk about some freaking complex numbers. Complex numbers, a + ib where a and b are real numbers, are really just a set of two real numbers. They are ordered pairs. If you've had any matrices, we can write a + ib as a row (or column) vector as so: (a, . There, that's a vector. Ordered pairs are always vectors. Mathematicians like calling this R2 (instead of just "the real numbers", which they just call R), because hey, we have two dimensions of real numbers. I don't have just one real number, I have two. So, complex addition and multiplication are a bit different from our regular addition and multiplication. Our normal real number addition just takes two numbers and adds them. In the complex world, we are adding two pairs of real numbers. Four numbers to keep track of here. The short story is that there is only one way to define addition and multiplication to pairs of numbers in a way that gives us a "field". I'll tell you the punchline: the way we define multiplication on these pairs of numbers is actually going to correspond to using i. Fields are really cool--the real numbers with our usual operations are a field. Basically, fields have addition, subtraction, multiplication, and division. What is a field? I don't really want to write out the definition, because really, all these abstract algebraic structures are defined by a list of properties that we want. I'm lazy, so I won't list them, but you can look on Wikipedia about it. Just skip that first intro paragraph and go down to the Definition section. I may have to write down certain properties for my proof next, but I don't want to write them all out. Now, how would you define a + on pairs of numbers? Don't think too hard. The obvious way to do this would be to add each component of the vector: (a, + (c, d) = (a + c, b + d) This is exactly what you'd do in linear algebra/matrix theory. Nothing "complex" yet. But what normal vectors don't do is have a good multiplication. How the crap am I supposed to multiply vectors? If you know some physics, you may have heard of the dot product and the cross product. They are ways to "multiply" vectors. But in our multiplication, we want to multiply two pairs of numbers, and then get another pair of numbers out. Dot products give us one number back, not two, and cross products are just for three-dimensional vectors. None of our normal tricks work! Curses! But hey, as I showed you, we can write down operations all day! Suck it, mathematicians! We're going to define our complex multiplication right now! Let's do the obvious way. (a, x(c, d) = (ac, bd) It worked for addition, so hey, maybe we just multiply the components together here, too, and everything will work out! Spoiler alert: it doesn't. This multiplication will fail two very important properties of a field. Pasted from wikipedia, for a Field F: 1. There is an element, called the multiplicative identity element and denoted by 1, such that for all a in F, a · 1 = a. Okay, nothing special here. In the real numbers, if you multiply anything by 1, you get that same thing back. 1 doesn't do anything. It's the "identity" in that you get a's "identity" back. In this definition, "1" is not necessarily the number 1. We are denoting our multiplicative identity by 1. Our multiplicative identity is a member of our set, so that means our "identity" is another pair of numbers. For the way we defined our multiplication, it's pretty clear that (1, 1) is our identity. (a, x(1, 1) = (a*1, b*1) and since a and b are both real numbers, that means we get (a, back. Cool, we have our identity. However, there's another property we need for a field, and it deals with division. In the language of abstract algebra, this is the "multiplicative inverse"--it's called that because really, division isn't a new operation, it's the reverse of multiplication. (There's also the "additive inverse", which for the real numbers is just subtraction). Here's the definition: 2. For any a in F other than 0, there exists an element a^−1 in F, such that a · a^−1 = 1. Okay, what the heck does that mean? Read a^-1 as "a inverse". For the real numbers, the inverse element is just 1/a. Because you'll get a* (1/a) = 1, and 1 is our multiplicative identity, so we're done. At first, you think everything is fine with our multiplication, right? Because we could just pick our inverse element of (a, to be (1/a, 1/B). Then we'd just get: (a, x(1/a, 1/B) = (a*(1/a), b*(1/B) = (1, 1) Which is our identity! Sweet, we are done. But wait. Not so fast. This must work for any two pairs of numbers. We have some problems right at 0, right? Now, in "pairs of numbers" land, 0 is (0, 0). So let's try to find an inverse to this number: (0, a) Hmmm. My inverse needs to get me (1, 1) back. Is that even possible? (0, a)x(x, y) = (0, ay). Nope, I'm never going to get (1, 1), because that first component is always 0. Crap! That means division totally fails, because division needs to work for all pairs of numbers (except (0, 0)). This is not the multiplication we are looking for. So, what is a multiplication that gives us the property of a field? You're not going to guess it, so I'm just going to write the answer down: (a, x(c, d) = (ac - bd, ad + bc) This is why I went through all that effort earlier to convince you that operations could be pretty weird, because this one is certainly weird. No one remembers this formula. But, you can verify the properties for multiplication, which I can go through for you. In this case, we our multiplicative identity is (1, 0), and we don't have the same problem as before. For this multiplication and our earlier addition, we get a field out! That's cool. This is obviously a doctored-up example, because we would've never guessed that. And why does no one remember this formula? Call (a, = a + ib and (c, d) = c + id. (a, x(c, d) = (a + ib)(c + id) Well, we can FOIL just like we always do! (a + ib)(c + id) = ac + iad + ibc + i*i*bd I haven't told you what i is yet. But look if we define i = √(-1), so i2= -1. Look what happens! ac + iad + ibc + i*i*bd = ac + iad + ibc + (-1)bd. Let me group this in a more suggestive fashion: (ac - bd) + i(ad + bc). Hey, lookit that. Going back to our "ordered pair" notation (ac - bd) + i(ad + bc) = (ac - bd, ad + bc), which is exactly the multiplication we defined before. What does this mean? That means multiplying complex numbers, with i being the square root of -1, it actually corresponds to this "abstract" field of pairs of numbers with a weird addition and multiplication. We can go from to the other freely. i is our shorthand for that weird multiplication in the abstract sense. So, if it makes you feel better, complex arithmetic really is just arithmetic on pairs of numbers with addition and a weird multiplication. No square root of -1 needed for it to work! But come on, it's a lot easier to FOIL with (a+ib)(c+id). They are equivalent! So I might as well use i, because I'm lazy and life is short. Now, okay, we defined multiplication. It's a field. But really does that multiplication correspond to, geometrically? For this, polar form is preferred. Let's let the x-axis be the real axis, and the y-axis be the imaginary axis. For a general complex number z = x + iy, we can write it like this, in polar form. We've written that complex number in terms of R, the radius from (0,0)--in complex analysis we call this the modulus of a number, and theta, the angle from the x-axis. This angle is called the argument of a number. Why these words? I don't know. In complex land mathematicians give everything a new name. It's annoying. Multiplying two numbers actually does this: the two moduli multiply together--so the new "length" of the complex number is the two lengths multiplied together, which is kind of what you expect--but the arguments don't multiply, they add. Like, for i, which (0,1), it's radius from the origin is 1, and the angle is π/2 (in radians). Multiplying i * i, we'd sure like that to be -1. Well, i * i, we multiply our moduli (1*1 = 1), but we add our angle, π/2 + π/2 = π. If you know your unit circle, you know that we're on the unit circle with angle pi, which is exactly the point (-1,0). Cool. In a more general sense, here's what 2+i and 3+i multiplied together looks like. I don't really like that picture, but you can kind of see that the radii are being multiplied, but the angles are adding. Now, I don't know how much calculus you have, so I can't prove you the first and most important consequence of complex multiplication, but let's say, for kicks and giggles, that we can express a complex number like this: z = r*eiθ Where r is that radius (the modulus), and θ is its angle. What happens when we multiply two z's together? z1*z2 = r1*eiθ1 * r2*eiθ2 = (r1*r2)*ei(θ1 + θ2) The radii multiplied, like we want, and the angles added. Huh. This might signal to you that the exponential is a good way write a complex number in polar form. But... that's kind of weird. The exponential function is increasing. And how would we write numbers in polar form normally? If you've never seen this transformation, don't worry, but it's pretty easy to show with trigonometry that if you have (x, y) and you want to write that ordered pair in terms in of r and θ, you do this: x = rcosθ y = rsinθ So, (x, y) = (rcosθ, rsinθ) That might look familiar to you. But here, Chaos is telling you that you that you can write a complex number like this: (x, y) = x + iy = reiθ. Whaaaaat?? Doesn't this mean... (x, y) = (rcosθ, rsinθ) = rcosθ + i*rsinθ = reiθ. That means that... eiθ = cosθ + isinθ Let's put a box around that. This is important. This is Euler's equation. So the exponential function--this nice easy increasing function which looks like this--if you have a complex exponential... that means it is a sum of sines and cosines? If you are staring at those graphs and wondering just what the hell happened that makes that even work at all, well. You aren't the only one. The answer is that we have this i in here--this weird multiplication--which is making things weird. Apparently, eiπ = -1. If you just said "This post has been reported for attempting to skirt the rules", well, I introduce you to this xkcd comic: http://xkcd.com/179/ It's straightforward to prove Euler's Equation. While I kind of derived it, I assumed that you could write z = r*eiθ to do it. I'd like to prove that I am actually allowed to do that in the first place. To do this, I need calculus 2. Specifically, I need power series. If you've had any of that before, well, then I'll give you the quick version: if you take the power series for ex and substitute x = iθ, that i will give you some negative terms in the power series. By separating the real and imaginary parts, you get the power series for cosine + i * the power series for sine. If that made no sense, that's okay. You kind of need some basic calculus for you to be even convinced that works. But there you have it, that's what i does. It's a weird multiplication which forms an algebraic field. When you look at that field, it turns out that some weird things happen, like exponentials being cosines and sines. If you have any calculus, then I can tell you a few more weird results. Differentiation in the complex sense is a much, much stronger condition in the complex numbers than in the real numbers. For one thing, a function must satisfy the Cauchy-Riemann Equations for them to be differentiable at all. We haven't gotten there yet in complex analysis class, but my professor keeps telling me the weirdest result: if you have one derivative defined on the complex numbers, then the function is also infinitely differentiable. One derivative gets you all of them. Apparently this has something to do with the fact you can build a power series out of any differentiable function, and we know that power series are infinitely differentiable. So... that's kind of weird, because in the real case, it's easy to write down a function that has one derivative, but you can't get a second derivative. In the complex case, if you're differentiable, you're infinitely differentiable. So in this sense, complex functions are much smoother than in the real case. There's a lot of awesome properties of the complex numbers. For one thing, it's a generalization of the real numbers, so it's a much bigger set than the reals. It's the largest algebraic field, in fact. You get sexy, sexy results like the Fundamental Theorem of Algebra. Is there anything that complex numbers don't give you? Well, a few things. Earlier in the thread, I explained how √(x*y) is not always √(x)√(y) in the complex sense, and since square roots are exponents, you might well suspect this is because of our weird multiplication. You need to be more precise in polar form for things to work out. Perhaps the easiest property in the complex numbers that isn't true for the reals is that the complex numbers aren't "ordered". In the real numbers, for two real a and b, there are three possibilities: a = b a < b a > b The real numbers are "ordered". Either you are bigger or you are smaller than a number. There's a strict ordering to this number system. That's why we can draw the real number "line". Lines have an order to them. This isn't true in the complex numbers, because how I can determine what a complex inequality is? It's not a line, but an entire plane. Sure, we can say that 1 + i > -1 - i, because both components in the first are bigger than the second. But let's compare 1 - i to -1 + i. Is one big than the other? Nope, not really. The complex numbers are not ordered. Though, actually, this is a consequence of them being very similar R2 rather than any weirdness with i. Any vector in Rn for any n obviously isn't ordered. So in the complex world, we get a bunch of cool properties, but we have to give up some nice things that we liked in the real numbers. But the things we lose have little practical significance. In the case of being ordered, while the complex numbers are not themselves ordered, we can always take a complex number's modulus (that radius from 0), and moduli are always positive real numbers. So, if we need to order things, we can take the modulus and compare things that way.

There are tags in the interview database for "brandon on writing", so definitely keep them. Keep it all, haha.

Verbatim as possible, yes. I think you can remove superfluous conjunctions, but content, no. As for the questioner... hmmm. For the transcript of the interview as a whole, I think that every word that is said is important. Like, if you're transcribing this into a coppermind, that intro stuff is important. However, thinking in context of the interview database, that superfluous stuff might not be important for an individual question entry. But I don't know. This is something that requires external judgment, on a case-by-case basis. Zas, unless you have any opinion otherwise--because you are, of course, the boss on this matter--I suggest that everyone gets as perfect transcription of the event as possible. Ums and ahs and obvious artifacts of vocalizing the words (like superfluous conjunctions), but all content, even if possibly unuseful, should remain. Once Zas has the transcript, he can pare it down to the format required of the interview database. It's always better to have more, and maybe another editor can look at it and take out some of the artifacts. Does that make sense, and do you agree, Zas?

Actually, this is excellent point which helps the theory. Ruin never saw far enough into the future to understand Preservation's plan. Even with Ruin's meddling of the Prophecies, there are plenty of passages which still successfully point to Sazed being the Hero of Ages--"hold the fate of the world on his arms", for example. Ruin didn't get it. He didn't get who would be the Hero. If Ruin truly saw into the future, he could've killed Sazed, and very easily block Preservation's plan from every coming into fruition. Instead, Ruin acted more selfishly, focusing on himself rather than the future. "I'm going to manipulate the Prophecies for my own gain, rather than just trying to stopping Preservation." "Even after I'm freed, I'm going to continue using Vin as my puppet to find the atium, even though Preservation chose her to ascend to godhood and she'll totally kill me later. Nothing could go wrong with this plan." In fact, that latter one was something I just thought up now. If Ruin saw the future effectively, he should've been able to counter Preservation. If Ruin killed Vin directly upon being freed, it's smooth sailing from there. But no. As is made clear from annotations, Ruin always acted with lusts. This always bothered me until reading this theory, Zas. As is obvious from atium, the direct essence of a Shard lets you see into the future--there was also that quote about Wyrn's future sight where Brandon said seeing into the future required the direct essence of a Shard, or something along those lines. It's certainly possible for any Shard to see into the future, so why couldn't Ruin? It always confused me. It had to be something with a Shard's intent, since that's the predominant differentiator between Shards, and Zas perfectly explains it. You know, it just occurred to me that atium is the exactly the future sight that Ruin would subscribe to. Not something which would let you to make long-term plans, but something almost immediate, so you can dominate and control right now. What else can you do with short-term future sight? Always win in battle, and that's about it. This substance is perfectly of Ruin. So, Zas, thank you. This is brilliant. It elegantly and easily explains problem with Ruin's future sight. I simply can't give you enough upvotes for this. (Well, I guess in the Austre is Endowment topic, I already gave you two upvotes, and now another one here, but still ) And now, food for thought: what do you think Odium is? I could go either way, really. I'm hoping he's more like Ruin--with more focusing on hatred right now--than something who bides his time, seeing into the future and making the best plan to exert his hatred for the long time. Someone is impulsive is predictable, and Ruin was predictable in this way. If Odium can see into the future... wow, we are so screwed. Then again, Odium isn't like Ruin, and Odium is biding his time very well. Only two Shards killed and Splintered, over centuries or millennia. Many, many Desolations. Like the back of the hardcover of the Way of Kings says, "Perhaps they saw that the heat and the hammer only make for a better grade of sword. But ignore the steel long enough, and it will eventually rust away." That... seems like a lot of foresight on Odium's part. Crap. EDIT: Well, I suppose Ruin plans too, but with Odium it seems like planning on a much vaster scale. That scares me.

Yes, it only applies in the positive real case. It's certainly true with positive numbers, but I'm talking about the general complex case, where it is not always true. If it were, you could do this: 1 = √(-1*-1) = √(-1)√(-1) = i * i = -1. That equality, by the way, is why mathematicians way back when didn't like using i, because it gave those nonsense answers. You can resolve the ambiguity, but you have to resort to different methods. EDIT: In your third link, it does specify the requirement that they need to be positive. Obviously, for most people's purposes, you just use the real case, with no need for this complex arithmetic But in general, yeah, you need to go to the complex plane.

Well, unfortunately, imaginary numbers do exist, and you can easily get real numbers back out. If it makes you feel better, the word imaginary isn't a particularly good one, because it immediately suggests "Whoa! This doesn't exist!" Gauss--famous mathematician, who got people to call them "complex" numbers--tried to change "the imaginary component" to the "lateral component" of a complex number, but it never caught on. Sorry about the bad terminology; I blame Descartes. EDIT: One more thing, guys, in my previous proofs, I was using the principal square root function--as in, always taking the positive root. This is very typical in complex analysis.

I got ninja'd (my post kind of took a while ), so I'm just going to double post to respond to the things people have said since then. Actually, you aren't doing it right. As I show in my post, you cannot distribute square roots through multiplication in the general case. You must convert the numbers to polar form before applying the exponents. If you could do what you are suggesting, Eero, I could: √(-2)√(-2)=√(2*-1)*√(2*-1) = √2 * √2 * i * i = -2, which is not the same as 2 (which is what you'd get if you said √(-2)√(-2) = √(-2*-2) = √4 = 2). Rigorously, you can only distribute square roots in the very special case of when Arg(u*v) = Arg(u) + Arg(v), for complex u and v. This is not true when u=v=-1, because Arg(1) = 0, but the Arg(-1) + Arg(-1) = pi + pi = 2pi. But, the Arg function--the principal argument, not any argument--is only defined on -pi to pi, not 2pi. Thus, the statement is false, 0 is not 2pi. I am happy to give you a very rigorous analysis if that doesn't make sense But basically, guys, don't be flippant with your square roots, because you can get inconsistent results if you do.

Conveniently for you, I am in a 400 level Complex Analysis class, and can explain! The crux of the matter is if you name the square root of -1 to be i, you can actually get answers that you could never get otherwise. Now, the first thing you need to know about "imaginary" numbers (that is, numbers with the square root of -1) is that the name "imaginary" numbers was made by Descartes to make fun of ever even using the idea of the square root of negative numbers. There are plenty of problems that don't have real solutions. For example, x^2 + 1 = 0. If you graphed this, this is a parabola which never touches the x-axis--of course it doesn't have real solutions! But, x^2 = -1 has a solution in the complex plane, which includes the imaginary numbers. The complex plane is pretty easy to visualize. It's just like any other graph, only the x-direction is the "real" axis, and the y-direction is the "imaginary" axis. So the point (0,3) would correspond to 0 + 3i, where i is the imaginary unit. I just wanted to mention that description of complex numbers, because I'm not sure if you've seen it before. Basically, don't think of imaginary numbers as not existing. After all, draw a coordinate plane. The point (0,1) certainly exists, right? The complex numbers, at the core, are just addition, multiplication, etc, to pairs of numbers. Let's take a look at some complex numbers (complex numbers are a real number plus an imaginary number, by the way). 1 + 3i 2 + 7i 10 - 4i That's just a way of writing pairs of numbers. I could just as easily write them as (1,3) (2,7) (10,-4) Now our supposedly "complex" numbers that don't exist are described by two real numbers. Heck, we've been doing pairs of real numbers ever since we started graphing! The point is, we don't intrinsically need the "i" to describe imaginary numbers--the letter i actually represents a convenient mathematical shorthand. I'll happily tell you more about how to derive the start of complex numbers, but I just wanted to tell you a little of the derivation to try and persuade you that these "imaginary" numbers exist, and are perfectly properly defined. Don't be scared of them! However, mathematicians didn't like the idea of square roots of negative numbers (so if you're skeptical, don't worry, mathematicians totally were too). You can kind of ignore it in the case of x^2 + 1 = 0. Just throw out the imaginary solutions. They don't exist, so no problem. The problem was when they tried to derive solutions for cubic equations, instead of just quadratic equations. You probably don't care about the gory details, it turns out that you can solve a cubic equation with imaginary numbers, and actually get real answers! The example in my complex analysis book is the equation x^3 - 15x - 4 = 0. It turns out a solution is x = 4. But, you might ask, how the crap did you pick x = 4? Well, apparently this happened in the 1500s (long before you could get your graphing calculator and see where the equation hit the x-axis), and the only way you can actually derive that x=4 is a solution is if you use imaginary numbers. Weird, huh? So basically, you know how there's a quadratic equation, which tells you all the roots of a quadratic? Well, there's a cubic equation, too, but you can't find all the (real) solutions to cubics without using imaginary numbers. With this discovery, mathematicians were like, "Okay well... dangit. I guess we have to freaking use imaginary numbers now, even though we really don't want to." Because obviously, in that time, being able to have a method to solve any polynomial was really appealing. (It turns out while you can solve cubic and quartic problems, you can't solve quintic or higher ones. Sucks for mathematicians) I'm just starting to get into interesting parts in my complex analysis class, but the short story is, there are problems that you can solve in the complex world that give you real answers, and you could never derive the answers by staying in the real world. A lot of trigonometric identities can be derived in this way, surprisingly enough--and even more surprisingly, it's really easy to do. It's hard to derive some of the really cool results of complex analysis for you without you knowing Calculus 2, but trust me, there's cool stuff going on. Now, to get back to your point. Hopefully I've sort of persuaded you that imaginary numbers at least are mathematically useful. So they exist. I suppose I've already answered your question: the kicker is that imaginary numbers can give you real answers! Like, check this out. Let's let x and y be real numbers, and i the square root of -1. If I do this, (x + iy)(x - iy) = x^2 + iyx - iyx - (i)(i)y^2 (this is just from FOILing) Then, the two middle terms cancel, and we know i^2 = -1, so we get (x + iy)(x - iy) = x^2 + y^2. Two complex numbers multiplied together gives you a perfectly real result. Now, to discuss the square root function more directly, I give you a word of caution. You know in the real numbers, √(x*y) = √(x)√(y) We can separate square roots across multiplication, right? Um... not in the complex world. You can't always say that for any complex numbers u and v, that √(u*v) = √(u)√(v). Now watch this. This is the equation that mathematicians said "in your face, imaginary numbers are nonsense!" 1 = √(-1*-1) = √(-1)√(-1) = i * i = -1. Obviously this is a bunch of crap, because 1 isn't -1. In fact, looking in my complex analysis book, you, um, can't actually solve your problem the way you are doing it for the same reason. 4 is the answer, but I need to use complex arithmetic to show it to you. √(-2)*√(-8) = √(16) = 4 = √(-2)*√(-8) = √(2)√(-1)√(8)√(-1) = √(16) * i * i = -4 Right? I mean, I'm just multiplying crap. So what's the answer, is it 4, or -4? In fact, I will not get the right answer if I try and do this on √(-2)*√(-8). The reason for this is that exponents are slightly odder in the complex world, because we multiply differently there (in fact, remember when I said "i is a shorthand?" It's a shorthand for complex multiplication, because it turns out there's only one way you can multiply pairs of numbers that give us a closed operation. But I digress) The answer to the problem is 4, but to do it properly, we can't use regular tricks. This next line will make no sense to you, but convince yourself that -2 = 2e^(i*π). I'm converting -2 its polar form. You can do this for any pair of real numbers, but the weirdness in the complex case is why the hell I'm using the exponential function for this. Anyway. Convince yourself that -2 = 2e^(i*π). So, we get (-2)^1/2 *(-2)^3/2 = (2e^(i*π))^1/2 * (2e^(i*π))^3/2 Then, by the laws of exponents, we can bring that 1/2 in if we multiply. So we get: 2^1/2 * e^(i*π/2) * 2^3/2 * e^(i*π*(3/2)) 2^1/2 * 8^1/2 * e^(i*π/2) * e^(i*π*(3/2) Adding exponents, we get 4*e^(i*2π) Basically, what the polar form represents is the angle around a circle. If you've had some trig, you know that in radians, the angle 0 and the angle 2π are the same. So we get 4*e^(i*2π) = 4*e^(i*0) = 4*e^(0) = 4. Phew! So yay, your book got the answer right. However, watch out! If I did this same thing on √(-2)*√(-8), I actually get -4. This is because (-2)^3/2 = -i*√(2), whereas (-8)^1/2 = i*√(2). There's a little sign change. This is easy to see in polar form, but is surely voodoo to you if you've never seen it before. Long story short, using the proper complex arithmetic, your book did indeed get the right answer of 4. Did your head explode yet?