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Can someone please also do this for: 1. Matrices, especially matrix multiplication. Unlike matrix addition, multiplication is defined in a very weird way. I

by alok-g 14y ago

Can someone please also do this for: 1. Matrices, especially matrix multiplication. Unlike matrix addition, multiplication is defined in a very weird way. I think I understand where it is coming from -- defining it that way allows representing and solving linear equations. More insights, however, would help. 2. Dot and cross products. E.g., the magnitude of dot product in 3D is a.b.cos(theta), while for cross product, it is a.b.sin(theta). I never got sure what happens to other combinations like a new "vector" product whose magnitude is a.b."cos"(theta)? 3. How is a set and "belongs-to" operator defined? Most books I have come across just assume these (and later define natural numbers and addition from them). 4. Why is 0.9999... considered to be "equal" to 1. I understand them to be equal "under the limit", but not without. This seems to be in my way of understanding Cantor's infinities.

nhaehnle 14y ago
I'll try to give some short pointers. 1. Matrices represent linear functions between finite-dimensional vector spaces. That is, if you have a function f from V to W that satisfies f(ax + by) = af(x) + bf(y), then there is a matrix A such that f(x) = Ax, and vice versa. Once you understand that, try to figure out what happens to those matrices when you compose functions. In other words, when you define h(x) = g(f(x)) (assuming that f and g are linear maps that can be composed in this way), then given the matrices for g and f, what will the matrix for h look like? You will end up with exactly the rules for matrix multiplication. The reason that you call this result "multiplication" is simply that it behaves very much like the multiplication that you are used to from the reals. In particular, you get a ring on square matrices, with (matrix) addition and multiplication that satisfy a distributive law. 2. I personally think that those angles are a bit of a red herring. In particular, the cross product generalizes in a somewhat more complicated way to higher dimensions, and then you have to talk about determinants instead of angles. That would take too much time and space to explain properly here. 3. Set and the "element-of" relation are not defined in the usual sense. They are indeed simply assumed, and you just postulate the properties that they need to satisfy, a.k.a. the axioms of set theory. It's a way of thinking that takes some getting used to, but as an analogy, try to work through Euclid. He doesn't define points or lines, either, but only postulates properties that they need to satisfy. 4. Because 0.9999... is usually interpreted as a real number, and not as an infinite sequence of characters. As a real number, 0.9999... has no meaning except as a limit, and hence they must be equal. As infinite sequences of characters, 0.9999... and 1 are of course different, but that's not how we usually interpret them. If you think that this is in your way when understanding Cantor's infinities, perhaps you should try to use the diagonalisation argument on infinite bit strings instead of on real numbers. That way, those kinds of subtleties simply do not arise.
- alok-g 14y ago
  Thanks!! >> then you have to talk about determinants instead of angles I certainly never heard this before. (I know how to calculate determinants, but never quite developed intuition around them.) Can you please say some more on this? :-)
  - ColinWright 14y ago
    Take a matrix M. This is a linear transformation. Look at what happens to the unit cube, and ask what volume the result has? The answer is the determinant of the matrix.
  - nhaehnle 14y ago
    Ah, what the heck. Here's how I think about this, which is heavily influenced by working on some problems related to lattices (from the geometry of numbers). How could you generalize the cross product to higher dimension? What the cross product does is that it takes two linearly independent 3-dim vectors and gives you some kind of canonical vector orthogonal to both of them. Can you find a reasonably canonical operation that, given two linearly independent d-dim vectors gives you some kind of canonical vector orthogonal to both of them? The answer is basically no. However, you can find a reasonable operation that, given d-1 linearly independent d-dim vectors a(1) ... a(d-1), gives you a canonical vector v orthogonal to all of them. This vector should be the unique solution of a system of linear equations aj * v = 0, z * v = 1. One way to resolve such a system is using Cramer's rule: each coordinate of v is a determinant of the matrix A of the system with one column replaced by the right-hand side, divided by the determinant of A. There are a number of reasons for not liking this division: (1) it is impractical if you prefer to work with coefficient in a ring (such as the integers) instead of a field; (2) the cross product is defined without any divisions; (3) the determinant of A depends on your choice of z, but the determinants in the numerators do not depend on the choice of z. So we just forget about the division, which is the same as choosing z such that det(A) = 1. That gives you a higher-dimensional generalization of the cross product. Okay, so now how long is v going to be? Well, for that you have to understand that the volume of the parallelepiped (skewed box) spanned by linearly independent vectors is equal to the determinant of the matrix that contains those vectors as rows or columns (or the square-root of the matrix multiplied with its transpose if you don't have a full-dimensional set). There are a number of ways to see this; one of them is that the parallelepiped is obtained via a linear transformation of the unit cube by that matrix, and linear transformations scale volume by the (absolute value of) the determinant. So the determinant of A is the height of z over the hyperplane spanned by a(1) ... a(d-1) times the (d-1)-dimensional volume of the parallelepiped spanned by them. This means that the height of z is the inverse of that volume (becaues we chose z so that det(A) = 1). But then since z * v = 1, the length of v must be equal to that volume, so that the volume and its inverse cancel. Now go back to the case d=3 and compute the size of the parallelepiped (aka parallelogram) spanned by your two starting vectors. Voila, you get exactly the formula for the length of the cross product. Compute the square-root of the determinant that you get when you multiple the matrix containing the two vectors as rows with its transpose. Voila, you get exactly the same formula again :) (In fact, if you look at how the components of the cross product in d=3 are computed, you'll find the 2x2 subdeterminants of that 2x3 matrix - and that fits perfectly with what I wrote above about using Cramer's rule.)
gryphonic 14y ago
Im going to let someone give a better response for some of these, but i will touch on a few. 2) The dot product can be thought of the projection of one vector onto another, or: If i shine a light directly at one vector with the 2nd vector in between, how long will the shadow of the 2nd vector in the system be? Do an image search for dot product to see this, and it will become clear (if you know basic trig) why the equation is ABcos(theta). The cross product is defined as having a magnitude equal to the area of the parallelogram that the vectors create. So have both vectors start at the same point, and then mirror them to make a parallelogram (see wiki on parallelogram). 4) How are the different? If you take 1-0.9999.... = x, what is x exactly? You can't tell me the difference. This is essentially how limits work.
jacobolus 14y ago
For #2 and to some extent #1, see the earlier link I posted about Geometric Algebra. For #1, you want to find a good linear algebra book or resource. I never watched them, but I’ve heard that Gil Strang’s MIT 18.06 lectures are good. http://www.youtube.com/course?list=ECE7DDD91010BC51F8 http://www.youtube.com/course?list=ECE7DDD91010BC51F8 For #3, try to find Halmos’s book Naive Set Theory. It has a very nice and accessible explanations.
Jun8 14y ago
Here's a fantastic article explaining scalars, matrices and tensors: http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tens... Explanation of why dot and cross products are defined that way: http://physics.stackexchange.com/questions/14082/what-is-the-physical-significance-of-dot-cross-product-of-vectors-why-is-divi http://physics.stackexchange.com/questions/14082/what-is-the...
ColinWright 14y ago
I don't have time now to provide complete answers of the type you want, but I can give you a little insight. > Matrices, especially matrix multiplication. Consider a transformation of space, specifically a shear, or a rotation, or an expansion (or contraction). All of these leave the origin unmoved, and a line will always become a line. They are what we call linear transformations. So given a point (x,y,z) in 3D space, the new values of x, y, and z are obtained by equations like a * x + b * y + c * z. If you then chase through how to combine two of these transformations, one after the other, the definitions of matrix multiplication drop out. This, by the way, also explains why you can't always divide by a matrix. Sometimes the transformation collapses the space into a lower dimensional space, and that can't be undone. > Dot and cross products. For unit vectors u and v, the dot product tells you how much of u points in the direction of v. Then we want it to be true that doubling the vector length doubles the size of the product. Then the result you ask about - the magnitude of dot product in 3D is a.b.cos(theta) - automatically follows. With the cross product, that can be defined as the area of a parallelogram that has the vectors as the side. Again, thinking about that interpretation shows how we get the answer you ask about. > How is a set and "belongs-to" operator defined? Set theory is often regarded as axiomatic, and so you can go back to the axiomatic definitions. We have a universe of discourse that has a bunch of atomic things. A "set" is then a collection of things, and then each set becomes a new thing in our universe. Thus sets can contain other sets. So a set is defined by what things are in it, so the "belongs-to" concept is fundamental to how the set in question is specified. > Why is 0.9999... considered to be "equal" to 1. People trip up on this in part because they think somehow that something is moving, that 0.999... "approaches" something. But no, when we write "0.999..." we have a representation of a point on the real line. Deciding what it is that you are defining requires that you be clear and detailed about what you think 0.999... means. So what does it mean? We drop back to saying that it's the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ... so we need to ask what real number can possibly be that limit? Let's call the limit L, and ask what it might be. Well, clearly: L > 0.9 L > 0.99 L > 0.999 L > 0.9999 ... and so on. Now, for every number X that's less than 1, there'll be something in that sequence that will be bigger than X. That means that the limit can't be less than 1. But equally, none of those numbers is bigger than 1, so the limit can't be bigger than 1. In fact, the smallest upper bound of { 0.9, 0.99, 0.999, 0.9999, ... } is 1. And then, why should it not be 1? We have different representations of points elsewhere. 2/3 is also 4/6 and 6/9, and sqrt(8) is the same as 2 * sqrt(2). These are simply different ways of specifying the same place on the real line. Likewise, writing 0.9999... is just a different representation of the point more often written as 1. You said: > I understand them to be equal "under the limit", > but not without. You tell me what you mean by 0,9999... and then I'll answer your question in more detail. > This seems to be in my way of understanding Cantor's infinities. Well, there are two different types of infinities based on set theory, but you're probably asking about cardinal infinities, the infinity of counting stuff - "How many things are in this set." So here, I have a sack with a collection of cubes, and they're numbered from 1 onwards. Call that collection A. Now let's take a cube and divide one face into two sections. Flip a coin, and write the result in one of the sections. in the other section, divide into two, flip a coin, write the result in one section, and then lather, rinse, repeat. We can fit infinitely many coin flips onto the cube in this way. In sack B put a cube with every possible result of this process. I claim that the cubes from sack A and sack B cannot be paired off with nothing left over. I won't prove that here, I'm now out of time. So, ask questions.
shou4577 14y ago
I can try: 1. Matrix multiplication is indeed defined oddly. To understand why, it can help to think of matrices as "linear transformations". For example, the matrix | 0 -1 | | 1 0 | corresponds to counterclockwise rotation of vectors by 90 degrees (go ahead, try multiplying the vector <1,0> by this matrix - the answer should be <0,1>). Let's call this matrix A. Also, the matrix |-1 0 | | 0 1 | corresponds to reflection over the y-axis (call this matrix B). If we begin with a vector v = <1,1>, then rotate twice by 90 degrees, then reflect over the y-axis, we should get the vector <1,-1> (it helps to draw a picture, but I am limited in these comments). We can figure this out by repeatedly multiplying by matrices: Av = <-1,1>, A(Av) = <-1, -1>, and B(A(Av)) = <1,-1> (you may want to multiply these out to check, then check by evaluating the operations geometrically). But wait - there is also a matrix which represents this transformation (that is, the transformation which is rotation counterclockwise by 90 degrees twice then reflection over the y axis). It's the matrix | 1 0 | | 0 -1 | which we'll call C (go ahead and check that this matrix works the way that I just said). Since multiplying a vector by C is the same as multiplying the vector by A then A then B, we have mathematically that Cv = B(A(Av)) Now, it sure would be nice if there was a way to "multiply" matrices together so that the product B(AA) = C. In abstract terms, we want multiplication of matrices to correspond to composition of functions, since we can use matrices to represent functions. If we define multiplication of matrices in the usual "weird" way, then this property holds! This lets us examine composing functions by multiplying matrices, which is a very powerful tool indeed. Another interesting thing to notice is that this explains why multiplication of matrices is not commutative (that is, order matters). If we rotate counterclockwise by 90 degrees, then reflect over the y axis, that's different than reflecting over the y axis then rotating counterclockwise by 90 degrees. So BA is not equal to AB. 2. I'm not sure that I understand your question here. 3. This question is beyond my ability to answer, so I'll leave it to someone else. 4. On the one hand, there are tricks to see this (e.g. 0.333... = 1/3, so 0.9999... = 3 * 0.3333... = 3 * (1/3) = 1), but I expect that you have seen them and been unsatisfied by them (I don't believe that these are satisfying explanations, personally, because I have simply replaced the unknown fact 0.999... = 1 with the equally unfounded fact 0.333.... = (1/3)). In order to really understand what's going on, we have to understand what we really mean by 0.999... What is meant by 0.9? This is equal to 9/10. How about 0.99? This is 9/10 + 9/100. 0.999 = 9/10 + 9/100 + 9/1000. Following this reasoning, 0.999... is equal to the infinite sum 9/10 + 9/100 + 9/1000 + ... + 9/10^k + ... Since there are infinitely many 9s in the decimal, we must add an infinite number of fractions of the form 9/10^k. We've opened up a can of worms here - how do we add up infinitely many things? Also, if we add up infinitely many things, how can the result possibly be finite? A Calculus II course will answer these questions rigorously, but let me attempt to make an explanation here. Think of it as a game of wits. You start. Choose a number less than 1 - any number at all, just so long as it is less than 1. I'll try to find some number k such that 0.9999...999 (with k 9s in it) is closer to 1 than your number. Suppose that you pick 0.97. That's pretty close to 1, only 0.03 away. But 0.01 < 0.03, and 1 - 0.99 = 0.01, so 0.99 is closer to 1 than 0.97. I win. You try again, this time picking 0.999878787. That's much closer to 1, but still 0.000121213. Since 0.0001 < 0.000121213, 0.9999 is closer to 1 than 0.999878787. I win again. In fact, let's suppose that you pick the number x. Since x is less than 1, 1 - x is some positive (but very small) number. No matter how small it is, though, I can find some k such that 1/(10^k) is even smaller (this fact is called the Archimedean principle, and it is one of the fundamental theorems of the real numbers). Another way to think about it is that 1/(1-x) may be very big, but I can find some k such that 10^k is even bigger. But then 0.999...99 (with k 9s) is going to be closer to 1 than your number, and I'll win again. Now, what does this mean? Since 0.999... is bigger than 0.999...99 (with any finite number of 9s), it is bigger than every number which is less than 1 (by the game we just played). The smallest number which is bigger than every number less than 1 is just 1. Anything smaller just won't do (since there would be numbers between it and 1). This shows that 0.999... is at least 1. I hope that helps. It isn't as elegant as the featured answer, but maybe it can be a little bit helpful to you.
esrauch 14y ago
Regarding 4, don't believe anyone who claims it's an intuitive result; it is simply a properly of real numbers that you cannot have nonzero infinitesimals and that any two distinct numbers have a number between them that is not equal to either (infinitely many in fact). You can construct alternate number lines that do allow nonzero infinitesimals and then .9999... actually is not equal to 1 under that number line; this number line is a strict superset of Reals just as Reals are a strict superset of Rationals. Most of the practical properties of real numbers would still apply to this new number line; it is actually only a historical convention note that .999... is actually equal to 1.
- alok-g 14y ago
  Do you know of something that I can read further on this? I wonder about the impact this may have on mathematics. I did not know of this property of real numbers of nonzero infinitesimals. (Possibly there is where my confusion is starting from.)
  - Evbn 14y ago
    Look up nonstandard analysis. It is a formalism that can define calculus, but doesn't have any practical difference on physical computations on real world entities.
    - Someone 14y ago
      For a more playful and, some would say, practical view, read "Winning Ways for Your Mathematical Plays" (http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathematical_Plays http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathemati...)
- gizmo686 14y ago
  There do exist number systems where .999.../=1, however, they are not a strict superset of the reals. If it were, then any operation involving only real numbers would behave identicly to the real number system. Also, this property is not a mere convention, but rather a nessasary result of what we want the number line to be. For example, assume that X<Y. Consider Z=(X+Y)/2. Z=X/2+Y/2. X<Z<Y. I have just shown, using basic algebra, that for any 2 distinct numbers, their is a third number between them. If that were not the case, then at least one of my steps must have been invalid.
Natsu 14y ago
Why is 0.999... one? It's an ambiguous decimal expansion. 1 - 0.999... = 0.000... Some part of you might think that there "must" be a 1 at the end of all those zeros. The problem is that there is no end at which to put a 1. And you'd end up with tiny holes everywhere if they're not equal. 1 = 1/3 + 1/3 + 1/3 = 0.333... + 0.333... + 0.333... = 0.999... oops. Where did our missing one go this time? Surely it's clear from this that we have nothing but 9s in that expansion and that three 3s can never be larger than 9, even if you repeat them over and over forever? And there can't be any funny business going on at the end, because infinite lists do not have ends by definition. At some point you go through the rules and just accept that this is how they play out and that using other rules just leads to weirdness (AKA "nonstandard analysis").

by alok-g 14y ago

nhaehnle 14y ago

I'll try to give some short pointers. 1. Matrices represent linear functions between finite-dimensional vector spaces. That is, if you have a function f from V to W that satisfies f(ax + by) = af(x) + bf(y), then there is a matrix A such that f(x) = Ax, and vice versa. Once you understand that, try to figure out what happens to those matrices when you compose functions. In other words, when you define h(x) = g(f(x)) (assuming that f and g are linear maps that can be composed in this way), then given the matrices for g and f, what will the matrix for h look like? You will end up with exactly the rules for matrix multiplication. The reason that you call this result "multiplication" is simply that it behaves very much like the multiplication that you are used to from the reals. In particular, you get a ring on square matrices, with (matrix) addition and multiplication that satisfy a distributive law. 2. I personally think that those angles are a bit of a red herring. In particular, the cross product generalizes in a somewhat more complicated way to higher dimensions, and then you have to talk about determinants instead of angles. That would take too much time and space to explain properly here. 3. Set and the "element-of" relation are not defined in the usual sense. They are indeed simply assumed, and you just postulate the properties that they need to satisfy, a.k.a. the axioms of set theory. It's a way of thinking that takes some getting used to, but as an analogy, try to work through Euclid. He doesn't define points or lines, either, but only postulates properties that they need to satisfy. 4. Because 0.9999... is usually interpreted as a real number, and not as an infinite sequence of characters. As a real number, 0.9999... has no meaning except as a limit, and hence they must be equal. As infinite sequences of characters, 0.9999... and 1 are of course different, but that's not how we usually interpret them. If you think that this is in your way when understanding Cantor's infinities, perhaps you should try to use the diagonalisation argument on infinite bit strings instead of on real numbers. That way, those kinds of subtleties simply do not arise.

alok-g 14y ago
Thanks!! >> then you have to talk about determinants instead of angles I certainly never heard this before. (I know how to calculate determinants, but never quite developed intuition around them.) Can you please say some more on this? :-)
- ColinWright 14y ago
  Take a matrix M. This is a linear transformation. Look at what happens to the unit cube, and ask what volume the result has? The answer is the determinant of the matrix.
- nhaehnle 14y ago
  Ah, what the heck. Here's how I think about this, which is heavily influenced by working on some problems related to lattices (from the geometry of numbers). How could you generalize the cross product to higher dimension? What the cross product does is that it takes two linearly independent 3-dim vectors and gives you some kind of canonical vector orthogonal to both of them. Can you find a reasonably canonical operation that, given two linearly independent d-dim vectors gives you some kind of canonical vector orthogonal to both of them? The answer is basically no. However, you can find a reasonable operation that, given d-1 linearly independent d-dim vectors a(1) ... a(d-1), gives you a canonical vector v orthogonal to all of them. This vector should be the unique solution of a system of linear equations aj * v = 0, z * v = 1. One way to resolve such a system is using Cramer's rule: each coordinate of v is a determinant of the matrix A of the system with one column replaced by the right-hand side, divided by the determinant of A. There are a number of reasons for not liking this division: (1) it is impractical if you prefer to work with coefficient in a ring (such as the integers) instead of a field; (2) the cross product is defined without any divisions; (3) the determinant of A depends on your choice of z, but the determinants in the numerators do not depend on the choice of z. So we just forget about the division, which is the same as choosing z such that det(A) = 1. That gives you a higher-dimensional generalization of the cross product. Okay, so now how long is v going to be? Well, for that you have to understand that the volume of the parallelepiped (skewed box) spanned by linearly independent vectors is equal to the determinant of the matrix that contains those vectors as rows or columns (or the square-root of the matrix multiplied with its transpose if you don't have a full-dimensional set). There are a number of ways to see this; one of them is that the parallelepiped is obtained via a linear transformation of the unit cube by that matrix, and linear transformations scale volume by the (absolute value of) the determinant. So the determinant of A is the height of z over the hyperplane spanned by a(1) ... a(d-1) times the (d-1)-dimensional volume of the parallelepiped spanned by them. This means that the height of z is the inverse of that volume (becaues we chose z so that det(A) = 1). But then since z * v = 1, the length of v must be equal to that volume, so that the volume and its inverse cancel. Now go back to the case d=3 and compute the size of the parallelepiped (aka parallelogram) spanned by your two starting vectors. Voila, you get exactly the formula for the length of the cross product. Compute the square-root of the determinant that you get when you multiple the matrix containing the two vectors as rows with its transpose. Voila, you get exactly the same formula again :) (In fact, if you look at how the components of the cross product in d=3 are computed, you'll find the 2x2 subdeterminants of that 2x3 matrix - and that fits perfectly with what I wrote above about using Cramer's rule.)

gryphonic 14y ago

Im going to let someone give a better response for some of these, but i will touch on a few. 2) The dot product can be thought of the projection of one vector onto another, or: If i shine a light directly at one vector with the 2nd vector in between, how long will the shadow of the 2nd vector in the system be? Do an image search for dot product to see this, and it will become clear (if you know basic trig) why the equation is ABcos(theta). The cross product is defined as having a magnitude equal to the area of the parallelogram that the vectors create. So have both vectors start at the same point, and then mirror them to make a parallelogram (see wiki on parallelogram). 4) How are the different? If you take 1-0.9999.... = x, what is x exactly? You can't tell me the difference. This is essentially how limits work.

jacobolus 14y ago

For #2 and to some extent #1, see the earlier link I posted about Geometric Algebra. For #1, you want to find a good linear algebra book or resource. I never watched them, but I’ve heard that Gil Strang’s MIT 18.06 lectures are good. http://www.youtube.com/course?list=ECE7DDD91010BC51F8 http://www.youtube.com/course?list=ECE7DDD91010BC51F8 For #3, try to find Halmos’s book Naive Set Theory. It has a very nice and accessible explanations.

Jun8 14y ago

Here's a fantastic article explaining scalars, matrices and tensors: http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tens... Explanation of why dot and cross products are defined that way: http://physics.stackexchange.com/questions/14082/what-is-the-physical-significance-of-dot-cross-product-of-vectors-why-is-divi http://physics.stackexchange.com/questions/14082/what-is-the...

ColinWright 14y ago

I don't have time now to provide complete answers of the type you want, but I can give you a little insight. > Matrices, especially matrix multiplication. Consider a transformation of space, specifically a shear, or a rotation, or an expansion (or contraction). All of these leave the origin unmoved, and a line will always become a line. They are what we call linear transformations. So given a point (x,y,z) in 3D space, the new values of x, y, and z are obtained by equations like a * x + b * y + c * z. If you then chase through how to combine two of these transformations, one after the other, the definitions of matrix multiplication drop out. This, by the way, also explains why you can't always divide by a matrix. Sometimes the transformation collapses the space into a lower dimensional space, and that can't be undone. > Dot and cross products. For unit vectors u and v, the dot product tells you how much of u points in the direction of v. Then we want it to be true that doubling the vector length doubles the size of the product. Then the result you ask about - the magnitude of dot product in 3D is a.b.cos(theta) - automatically follows. With the cross product, that can be defined as the area of a parallelogram that has the vectors as the side. Again, thinking about that interpretation shows how we get the answer you ask about. > How is a set and "belongs-to" operator defined? Set theory is often regarded as axiomatic, and so you can go back to the axiomatic definitions. We have a universe of discourse that has a bunch of atomic things. A "set" is then a collection of things, and then each set becomes a new thing in our universe. Thus sets can contain other sets. So a set is defined by what things are in it, so the "belongs-to" concept is fundamental to how the set in question is specified. > Why is 0.9999... considered to be "equal" to 1. People trip up on this in part because they think somehow that something is moving, that 0.999... "approaches" something. But no, when we write "0.999..." we have a representation of a point on the real line. Deciding what it is that you are defining requires that you be clear and detailed about what you think 0.999... means. So what does it mean? We drop back to saying that it's the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ... so we need to ask what real number can possibly be that limit? Let's call the limit L, and ask what it might be. Well, clearly: L > 0.9 L > 0.99 L > 0.999 L > 0.9999 ... and so on. Now, for every number X that's less than 1, there'll be something in that sequence that will be bigger than X. That means that the limit can't be less than 1. But equally, none of those numbers is bigger than 1, so the limit can't be bigger than 1. In fact, the smallest upper bound of { 0.9, 0.99, 0.999, 0.9999, ... } is 1. And then, why should it not be 1? We have different representations of points elsewhere. 2/3 is also 4/6 and 6/9, and sqrt(8) is the same as 2 * sqrt(2). These are simply different ways of specifying the same place on the real line. Likewise, writing 0.9999... is just a different representation of the point more often written as 1. You said: > I understand them to be equal "under the limit", > but not without. You tell me what you mean by 0,9999... and then I'll answer your question in more detail. > This seems to be in my way of understanding Cantor's infinities. Well, there are two different types of infinities based on set theory, but you're probably asking about cardinal infinities, the infinity of counting stuff - "How many things are in this set." So here, I have a sack with a collection of cubes, and they're numbered from 1 onwards. Call that collection A. Now let's take a cube and divide one face into two sections. Flip a coin, and write the result in one of the sections. in the other section, divide into two, flip a coin, write the result in one section, and then lather, rinse, repeat. We can fit infinitely many coin flips onto the cube in this way. In sack B put a cube with every possible result of this process. I claim that the cubes from sack A and sack B cannot be paired off with nothing left over. I won't prove that here, I'm now out of time. So, ask questions.

shou4577 14y ago

I can try: 1. Matrix multiplication is indeed defined oddly. To understand why, it can help to think of matrices as "linear transformations". For example, the matrix | 0 -1 | | 1 0 | corresponds to counterclockwise rotation of vectors by 90 degrees (go ahead, try multiplying the vector <1,0> by this matrix - the answer should be <0,1>). Let's call this matrix A. Also, the matrix |-1 0 | | 0 1 | corresponds to reflection over the y-axis (call this matrix B). If we begin with a vector v = <1,1>, then rotate twice by 90 degrees, then reflect over the y-axis, we should get the vector <1,-1> (it helps to draw a picture, but I am limited in these comments). We can figure this out by repeatedly multiplying by matrices: Av = <-1,1>, A(Av) = <-1, -1>, and B(A(Av)) = <1,-1> (you may want to multiply these out to check, then check by evaluating the operations geometrically). But wait - there is also a matrix which represents this transformation (that is, the transformation which is rotation counterclockwise by 90 degrees twice then reflection over the y axis). It's the matrix | 1 0 | | 0 -1 | which we'll call C (go ahead and check that this matrix works the way that I just said). Since multiplying a vector by C is the same as multiplying the vector by A then A then B, we have mathematically that Cv = B(A(Av)) Now, it sure would be nice if there was a way to "multiply" matrices together so that the product B(AA) = C. In abstract terms, we want multiplication of matrices to correspond to composition of functions, since we can use matrices to represent functions. If we define multiplication of matrices in the usual "weird" way, then this property holds! This lets us examine composing functions by multiplying matrices, which is a very powerful tool indeed. Another interesting thing to notice is that this explains why multiplication of matrices is not commutative (that is, order matters). If we rotate counterclockwise by 90 degrees, then reflect over the y axis, that's different than reflecting over the y axis then rotating counterclockwise by 90 degrees. So BA is not equal to AB. 2. I'm not sure that I understand your question here. 3. This question is beyond my ability to answer, so I'll leave it to someone else. 4. On the one hand, there are tricks to see this (e.g. 0.333... = 1/3, so 0.9999... = 3 * 0.3333... = 3 * (1/3) = 1), but I expect that you have seen them and been unsatisfied by them (I don't believe that these are satisfying explanations, personally, because I have simply replaced the unknown fact 0.999... = 1 with the equally unfounded fact 0.333.... = (1/3)). In order to really understand what's going on, we have to understand what we really mean by 0.999... What is meant by 0.9? This is equal to 9/10. How about 0.99? This is 9/10 + 9/100. 0.999 = 9/10 + 9/100 + 9/1000. Following this reasoning, 0.999... is equal to the infinite sum 9/10 + 9/100 + 9/1000 + ... + 9/10^k + ... Since there are infinitely many 9s in the decimal, we must add an infinite number of fractions of the form 9/10^k. We've opened up a can of worms here - how do we add up infinitely many things? Also, if we add up infinitely many things, how can the result possibly be finite? A Calculus II course will answer these questions rigorously, but let me attempt to make an explanation here. Think of it as a game of wits. You start. Choose a number less than 1 - any number at all, just so long as it is less than 1. I'll try to find some number k such that 0.9999...999 (with k 9s in it) is closer to 1 than your number. Suppose that you pick 0.97. That's pretty close to 1, only 0.03 away. But 0.01 < 0.03, and 1 - 0.99 = 0.01, so 0.99 is closer to 1 than 0.97. I win. You try again, this time picking 0.999878787. That's much closer to 1, but still 0.000121213. Since 0.0001 < 0.000121213, 0.9999 is closer to 1 than 0.999878787. I win again. In fact, let's suppose that you pick the number x. Since x is less than 1, 1 - x is some positive (but very small) number. No matter how small it is, though, I can find some k such that 1/(10^k) is even smaller (this fact is called the Archimedean principle, and it is one of the fundamental theorems of the real numbers). Another way to think about it is that 1/(1-x) may be very big, but I can find some k such that 10^k is even bigger. But then 0.999...99 (with k 9s) is going to be closer to 1 than your number, and I'll win again. Now, what does this mean? Since 0.999... is bigger than 0.999...99 (with any finite number of 9s), it is bigger than every number which is less than 1 (by the game we just played). The smallest number which is bigger than every number less than 1 is just 1. Anything smaller just won't do (since there would be numbers between it and 1). This shows that 0.999... is at least 1. I hope that helps. It isn't as elegant as the featured answer, but maybe it can be a little bit helpful to you.

esrauch 14y ago

Regarding 4, don't believe anyone who claims it's an intuitive result; it is simply a properly of real numbers that you cannot have nonzero infinitesimals and that any two distinct numbers have a number between them that is not equal to either (infinitely many in fact). You can construct alternate number lines that do allow nonzero infinitesimals and then .9999... actually is not equal to 1 under that number line; this number line is a strict superset of Reals just as Reals are a strict superset of Rationals. Most of the practical properties of real numbers would still apply to this new number line; it is actually only a historical convention note that .999... is actually equal to 1.

alok-g 14y ago
Do you know of something that I can read further on this? I wonder about the impact this may have on mathematics. I did not know of this property of real numbers of nonzero infinitesimals. (Possibly there is where my confusion is starting from.)
- Evbn 14y ago
  Look up nonstandard analysis. It is a formalism that can define calculus, but doesn't have any practical difference on physical computations on real world entities.
  - Someone 14y ago
    For a more playful and, some would say, practical view, read "Winning Ways for Your Mathematical Plays" (http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathematical_Plays http://en.wikipedia.org/wiki/Winning_Ways_for_your_Mathemati...)
gizmo686 14y ago
There do exist number systems where .999.../=1, however, they are not a strict superset of the reals. If it were, then any operation involving only real numbers would behave identicly to the real number system. Also, this property is not a mere convention, but rather a nessasary result of what we want the number line to be. For example, assume that X<Y. Consider Z=(X+Y)/2. Z=X/2+Y/2. X<Z<Y. I have just shown, using basic algebra, that for any 2 distinct numbers, their is a third number between them. If that were not the case, then at least one of my steps must have been invalid.

Natsu 14y ago

Why is 0.999... one? It's an ambiguous decimal expansion. 1 - 0.999... = 0.000... Some part of you might think that there "must" be a 1 at the end of all those zeros. The problem is that there is no end at which to put a 1. And you'd end up with tiny holes everywhere if they're not equal. 1 = 1/3 + 1/3 + 1/3 = 0.333... + 0.333... + 0.333... = 0.999... oops. Where did our missing one go this time? Surely it's clear from this that we have nothing but 9s in that expansion and that three 3s can never be larger than 9, even if you repeat them over and over forever? And there can't be any funny business going on at the end, because infinite lists do not have ends by definition. At some point you go through the rules and just accept that this is how they play out and that using other rules just leads to weirdness (AKA "nonstandard analysis").