Chaos he/him Posted March 28, 2011 Report Share Posted March 28, 2011 Do you have a question about math? Is there something you never quite understood about it? Well, prepare to be confused no more; Chaos is here to help! Post your questions here. I can't promise I'll get to them quickly, though. If you need them answered quickly, find me on AIM or Skype. Link to comment Share on other sites More sharing options...

Eerongal he/him Posted March 28, 2011 Report Share Posted March 28, 2011 Ok. Simplify (n-a)(n-(n-c)...(n-x)(n-y)(n-z) 1 Link to comment Share on other sites More sharing options...

Chaos he/him Posted March 28, 2011 Author Report Share Posted March 28, 2011 (edited) Ok. Simplify (n-a)(n-(n-c)...(n-x)(n-y)(n-z) Oh fun, we're making this a stump Chaos session EDIT: Ah, zero. Edited March 28, 2011 by Chaos 1 Link to comment Share on other sites More sharing options...

Eerongal he/him Posted March 28, 2011 Report Share Posted March 28, 2011 Oh fun, we're making this a stump Chaos session EDIT: Ah, zero. curses! you saw through my cunning ruse to make you go "uhhh.....screw 'dat!" with a simple problem! Link to comment Share on other sites More sharing options...

Chaos he/him Posted March 28, 2011 Author Report Share Posted March 28, 2011 Hehe. But seriously, I'd rather not turn this into Stump Chaos. I want to help people learn math. Link to comment Share on other sites More sharing options...

Eerongal he/him Posted March 28, 2011 Report Share Posted March 28, 2011 Hehe. But seriously, I'd rather not turn this into Stump Chaos. I want to help people learn math. awww, but where's the fun in that? Link to comment Share on other sites More sharing options...

Chaos he/him Posted March 28, 2011 Author Report Share Posted March 28, 2011 awww, but where's the fun in that? Math is its own fun Link to comment Share on other sites More sharing options...

leinton Posted March 28, 2011 Report Share Posted March 28, 2011 If I ever get out of the easy that is statistics and get back into real math, you'll probably see me posting here. This is really thoughtful, man. And I agree... math is its own fun. 1 Link to comment Share on other sites More sharing options...

Shivertongue he/him Posted March 28, 2011 Report Share Posted March 28, 2011 Math is fun. But you know what is more fun that doing math? NOT doing math! 4 Link to comment Share on other sites More sharing options...

Joe ST he/him Posted March 28, 2011 Report Share Posted March 28, 2011 I like seeing threads on maths, but even better, I like seeing threads on maths with 9 posts NOT asking questions Link to comment Share on other sites More sharing options...

Shivertongue he/him Posted March 28, 2011 Report Share Posted March 28, 2011 Okay, I have a math question for you... Solve 5x^-2y^10 over 2x^-1(-3x^-3y^-1)^-2 and then use it in an example that involves monkeys, cheese, and a Pewter Misting that's been spiked with a Bronze Hemalurgical spike granting your choice of Allomantic mental abilities. This isn't to stump you. It's honestly been driving me crazy for days. Link to comment Share on other sites More sharing options...

Joe ST he/him Posted March 28, 2011 Report Share Posted March 28, 2011 Okay, I have a math question for you... Solve 5x^-2y^10 over 2x^-1(-3x^-3y^-1)^-2 and then use it in an example that involves monkeys, cheese, and a Pewter Misting that's been spiked with a Bronze Hemalurgical spike granting your choice of Allomantic mental abilities. This isn't to stump you. It's honestly been driving me crazy for days. Use Wolfram|Alpha for symbolic manipulation http://www.wolframalpha.com/input/?i=5x^-2y^10+over+2x^-1%28-3x^-3y^-1%29^-2 (5/9)x^3y^7 Link to comment Share on other sites More sharing options...

Chaos he/him Posted March 28, 2011 Author Report Share Posted March 28, 2011 Of course, I can't "solve" it, because that's an expression, not an equation. Link to comment Share on other sites More sharing options...

Shivertongue he/him Posted March 28, 2011 Report Share Posted March 28, 2011 Use Wolfram|Alpha for symbolic manipulation http://www.wolframalpha.com/input/?i=5x^-2y^10+over+2x^-1%28-3x^-3y^-1%29^-2 (5/9)x^3y^7 Failed. You did not use it in an example with the required subjects. Link to comment Share on other sites More sharing options...

Puck he/him Posted March 28, 2011 Report Share Posted March 28, 2011 So, I remember spending some time in a metaphysics course or two about the reality of multiple infinities. I remember it was explained mostly by showing how not all series of numbers map to each other (but I could be remembering it wrong). Could you walk me through that reasoning? Link to comment Share on other sites More sharing options...

SOM1else he/him Posted April 4, 2011 Report Share Posted April 4, 2011 How do you find the the 20th precentile in a set of data with 42 data points? Would it be 42*.20=8.4, so then it would be the 8th data point because 8.4 rounds to 8? Next question. To find quartiles you find the median of a set of data and then find the median of each half of the data set, the median of the upper half is Q3 and the median of the lower half is Q1, is that right? Assuming I am right about how I should find the Quartiles does Q3-Q1=IQR/Inter Quartile Range? Not really a math question but if I am right about what Q1 and Q3 are what is Q2? is it just the median of the whole data set? Link to comment Share on other sites More sharing options...

Chaos he/him Posted April 4, 2011 Author Report Share Posted April 4, 2011 Puck, I'm getting to your question, I promise. I'm just being slow. 1. How do you find the the 20th precentile in a set of data with 42 data points? Would it be 42*.20=8.4, so then it would be the 8th data point because 8.4 rounds to 8? 2. Next question. To find quartiles you find the median of a set of data and then find the median of each half of the data set, the median of the upper half is Q3 and the median of the lower half is Q1, is that right? Assuming I am right about how I should find the Quartiles does Q3-Q1=IQR/Inter Quartile Range? 3. Not really a math question but if I am right about what Q1 and Q3 are what is Q2? is it just the median of the whole data set? 1. I'd say you are right, but I'm looking at Wikipedia for Percentiles, and the formula is apparently P/100*N + 1/2. This is the same as what you did, just with a +1/2. I'm not certain what the significance of that +1/2 is. If this is for classwork, I'd say look at your textbook and see what definition it uses. Maybe the +1/2 is there, maybe it isn't. 2. Yes, you understand it correctly. 3. Yeah, Q2 is always the median. How convenient! It's almost like statisticians defined it to be so Link to comment Share on other sites More sharing options...

SOM1else he/him Posted March 24, 2012 Report Share Posted March 24, 2012 Thread Necromancy FTW! So I have a new question about something that was on my math home work the other day, we were learning about how to simplify radical expressions in my math class and one of the problems that we were assigned was: [(-2)^(1/2)]*[(-2)^(3/2)] I rewrote this problem as: √(-2)*√(-2)^3 and then to: √(-2)*√(-8) I then saw that there isn't a square root of -2 or of -8 so there could be no possible solution to this problem. I wrote that down as my answer and went to the back of my text book to see if I was right, there it said that the answer was 4. I was confused about this for a while and then realized √(-2)*√(-8) = √(16) = 4 My question is, how is it that two numbers that don't exist can multiply together to equal something that does? I talked to my teacher about this and he told me that question shouldn't have been on the homework because it wasn't sometime we had covered yet. Link to comment Share on other sites More sharing options...

Joe ST he/him Posted March 24, 2012 Report Share Posted March 24, 2012 I rewrote this problem as: √(-2)*√(-2)^3 and then to: √(-2)*√(-8) You could also have rewritten it as √(-2)*√(-2)*√(-2)*√(-2) (which is what the ^3 means) and then combine the two pairs of (√(-2)*√(-2)) into √4 's and then again into 4 Link to comment Share on other sites More sharing options...

SOM1else he/him Posted March 24, 2012 Report Share Posted March 24, 2012 You could also have rewritten it as √(-2)*√(-2)*√(-2)*√(-2) (which is what the ^3 means) and then combine the two pairs of (√(-2)*√(-2)) into √4 's and then again into 4 That works to but unless I am not understanding something in your post it doesn't answer how numbers that don't exist can multiply each other into to numbers that do exist. From the problem above it shows that they obviously can do it, I just don't get why they can. The way I see it is (Mistwraith*Kollos=Human) Mistwraiths and Kolloses (Kollosi?) don't really exist but humans do. But if you can take a Mistwraith and Multiply it by a Kollos then you end up with a Human. Link to comment Share on other sites More sharing options...

Joe ST he/him Posted March 24, 2012 Report Share Posted March 24, 2012 That works to but unless I am not understanding something in your post it doesn't answer how numbers that don't exist can multiply each other into to numbers that do exist. From the problem above it shows that they obviously can do it, I just don't get why they can. The way I see it is (Mistwraith*Kollos=Human) Mistwraiths and Kolloses (Kollosi?) don't really exist but humans do. But if you can take a Mistwraith and Multiply it by a Kollos then you end up with a Human. well the problem is that square roots of negative numbers DO 'exist' (they dont actually exist in the world, but they appear and are applicable to many areas of science, math, and engineering). No doubt Eric will back me up when he finally finishes his thesis on the subject that he's writing to answer your question. They're called imaginary numbers (or at least, the squareroot of -1 is the imaginary number). Link to comment Share on other sites More sharing options...

Eerongal he/him Posted March 24, 2012 Report Share Posted March 24, 2012 Yeah, imaginary numbers are something you will learn later, but essentially they help you to deal with negative square roots and continue computing Also, when you multiply square roots, you can write them under the same square root bracket. It's hard to show on here, but suffice to say: __ __ _____ \/-2 *\/-2 = \/-2*-2 Edit: so that doesnt display well at all. Let's try code brackets. extra edit: a bit better. Link to comment Share on other sites More sharing options...

Chaos he/him Posted March 24, 2012 Author Report Share Posted March 24, 2012 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? 1 Link to comment Share on other sites More sharing options...

Chaos he/him Posted March 24, 2012 Author Report Share Posted March 24, 2012 (edited) 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. Yeah, imaginary numbers are something you will learn later, but essentially they help you to deal with negative square roots and continue computing Also, when you multiply square roots, you can write them under the same square root bracket. It's hard to show on here, but suffice to say: __ __ _____ \/-2 *\/-2 = \/-2*-2 Edit: so that doesnt display well at all. Let's try code brackets. extra edit: a bit better. 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. Edited March 24, 2012 by Chaos 1 Link to comment Share on other sites More sharing options...

Chaos he/him Posted March 24, 2012 Author Report Share Posted March 24, 2012 (edited) That works to but unless I am not understanding something in your post it doesn't answer how numbers that don't exist can multiply each other into to numbers that do exist. From the problem above it shows that they obviously can do it, I just don't get why they can. The way I see it is (Mistwraith*Kollos=Human) Mistwraiths and Kolloses (Kollosi?) don't really exist but humans do. But if you can take a Mistwraith and Multiply it by a Kollos then you end up with a Human. 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. Edited March 24, 2012 by Chaos Link to comment Share on other sites More sharing options...

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