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.
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.
Here is what I am confused about with regards to diagonalisation:
Start with binary non-negative integers:
000
001
010
011
100
101
... (goes to infinity)
This set now includes all possible bit strings of infinite length since the way these are iteratively generated includes all possibilities.
This is also an enumerable set by definition.
Let's now reverse the bits and put them after a decimal. These are just real numbers now going from zero to one. (This step is actually unnecessary I think.)
.000
.100
.010
.110
.001
.101
.011
...
This must be enumerable set too.
Using diagonalisation argument, .111111 is never to be found in this set. This is exactly where I am stuck. This number comes into the set from flipping of infinity in the original set, which includes all possible stings of infinite length.
I immediately read into your message on treating these as bit strings instead of real numbers. I still am stuck though. (Do you also see a connection to 0.99999... by the way?)
> Start with binary non-negative integers:
> 000 001 010 011 100 101 ... (goes to infinity)
> This set now includes all possible bit strings of
> infinite length
No, it only contains the strings of finite length. There are infinitely many of them, but each one stops after a while. In particular, then n^th one only has log2(n) places before it then becomes all 0s.
> This is also an enumerable set by definition.
Yes.
> Let's now reverse the bits and put them after a decimal.
> These are just real numbers now going from zero to one.
> ... .000 .100 .010 .110 .001 .101 .011 ...
But not all of them, since these are only those numbers that have a finite number of 1's in them.
> This must be enumerable set too.
Yes it is.
> Using diagonalisation argument, .111111 is never to be
> found in this set.
Irrelevant.
> This is exactly where I am stuck. This number comes
> into the set from flipping of infinity in the original set,
"Infinity" was never in your original set. And if it was, it wouldn't be produced by the diagonalisation argument.
> which includes all possible strings of infinite length.
The set is still enumerable since this is just binary encoding mapping to the set {0, 1, 2, 3, ...}
The question still is if it covers all possible bit strings of infinite length.
For units place, we covered both zero and one. For (n+1)th place, we cover both zero and one together with all combinations for the first (n) bits. As n -> infinite, all possibilities get covered.
The question is if 111111111... is also there in this set. But isn't it there too?
> For the first set, I meant to write:
> [Prepend each string with infinite zeroes]
> ...000
> ...001
> ...010
...
If there are infinitely many zeros on the front, you can't actually append anything. That doesn't end up being well-defined.
(Well, actually, there are transfinite ordinals, but that would confuse the issue. It's not what you mean, and it doesn't help)
> Now all bit strings here have infinite length.
If you want to talk about an infinite "decimal" string, you need to talk about the things that come after the decimal point, in order. As such, they come in order, and you can't have infinitely many zeros and then a finite string on the end.
> The string is still enumerable since this is just binary
> encoding mapping to the set {0, 1, 2, 3, ...}
You need to be more careful about how you actually define the strings. Strings have a start, then they go on one place by one place.
> The question still is if it covers all possible bit strings of
> infinite length.
Well, you haven't actually properly defined strings, but even so, no. Everything you have starts with a zero.
> For units place, we covered both zero and one.
No, you don't seem to have.
> For (n+1)th place, we cover both zero and one
> together with all combinations for the first (n) bits.
> As n -> infinite, all possibilities get covered.
No, because as soon as you have a one in your expansion the string is finite, so not all possibilities are covered.
> The question is if 111111111... is also there in this set.
> But isn't it there too?
You defined the set - tell me where it is. Even leaving alone the fact that these aren't proper strings, it doesn't appear to be there.
>>> [Prepend each string with infinite zeroes]
> ...000
> ...001
> ...010
>> If there are infinitely many zeros on the front, you can't actually append anything.
I am lost. In this first set, I do not have a decimal point anywhere. Why cannot I have an infinitely many zeros to the left of 1. It will still be just one when looked at as a number.
>> Strings have a start, then they go on one place by one place
As far as representing it as a string, I may still start from the right and work towards the left.
I understand your point for the decimal case, infinitely many zeroes on the right of decimal cannot be followed a finite string. My argument does not require this however. (I change ...00000011010 from the first set to .0101100000000... in the second set.)
>> > For units place, we covered both zero and one.
>> No, you don't seem to have.
The units place is the rightmost below. Both zero and one are covered.
...000
...001
Stating the above for the decimal case, let n=1 be the place right after the decimal, n=2 to the right of it, and so on. Now for place n=1, the both zero and one are covered (first two cases below). For n=2 place, again both zero and one are covered for all possible combinations above for n=1 place (first four cases below).
Using the mathematical induction argument, all combinations are covered. This must include 0.1111111111... It sits exactly where (simple) infinity sits in the enumerable set {0, 1, 2, 3, ... }
OK, so you're not talking about the usual diagonalisation. I'll try to follow what you've said and respond as I go.
> In this first set, I do not have a decimal point anywhere.
OK, fine. But then you talked about flipping them around to come after the decimal point. When you do that you have only those strings that only have a finite number of 1s in them.
> As far as representing it as a string, I may still start
> from the right and work towards the left.
Yes you can, but that's not what people do when talking about Cantor and diagonalisation, so it's now completely unclear what you're talking about.
However, you start with finite strings of 0s and 1s, basically the non-negative whole numbers, represented as binary strings. Note that these are all finite, and the non-zero parts are still finite, even if you prepend an infinite number of 0s.
> Stating the above for the decimal case, let n=1 be the
> place right after the decimal, n=2 to the right of it, and
> so on.
See, now you're talking about stuff after the decimal point. I'll continue ...
> Now for place n=1, the both zero and one are covered
> (first two cases below).
But for diagonalisation that's irrelevant. We only ask what is the first digit of the first number.
> For n=2 place, again both zero and one are covered for
> all possible combinations above for n=1 place
Again, irrelevant. We only ask what is the 2nd digit of the 2nd number.
So here if we construct the diagonal of this sequence as you've listed it we have 0.00000... Let me highlight the diagonal for you from this quoted section:
Edit: This comment should be read after my comment below this. It shows up first on HN.
If infinity was in the original set, what would diagonalisation produce? [Genuinely asking, I am unclear on this.] I am flipping all the bits along a diagonal and they are all zeros before flipping.
If 0.99999... = 1, then using my argument of flipping the bits around the decimal, wouldn't infinity be in the set?
What does this mean? Your complete imprecision is making it impossible to answer the questions, despite wanting to help, because they don't make sense.
What set? Define it clearly. Don't talk about infinite strings of zeros followed by stuff, because in the context of decimal or binary expansions that doesn't make sense. Strings have a start, then they go on one place at a time.,
> I am flipping all the bits along a diagonal and they
> are all zeros before flipping.
If they are all zero before flipping then 0.11111... isn't there. You've stated that the n^th number has 0 in the n^th place. That means 0.11111... is not the n^th number for any n.
If 0.99999... = 1, then using my argument of flipping the bits around the decimal, wouldn't infinity be in the set?
>> If they are all zero before flipping then 0.11111... isn't there.
This helps. Since the first set has infinitely many zeros on the left, flipping around the decimal means there would have to be infinitely many zeros somewhere on the right. 0.1111.... however has infinitely many ones before these supposed infinitely many zeroes, which cannot be for the same reason that a finite string with one cannot be after infinitely many zeroes at the right of a decimal.
This means that the first set does not have 111111...
So if I keep incrementing binary numbers, I'll never reach 11111... within the limits of enumerability.
I need to think and read more. The above seems to imply that the simple infinity is not in the enumerable set. But I may be confused again.
>> You've stated that the n^th number has 0 in the n^th place. That means 0.11111... is not the n^th number for any n.
I have been confused about this. Somehow 0.11111... is not in the set, in spite of the mathematical induction proof I supply. I need to think more. The proof must be incomplete (or imprecise as you say).
> This set now includes all possible bit strings of infinite length since the way these are iteratively generated includes all possibilities.
That's not correct -- in fact, it contains no infinite bit strings at all (to prove this to yourself, ask at what position the first infinite string appears).
>> 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? :-)
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.)
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.
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.
Thanks for clarifying the strict superset part, I'm not actually a mathematician.
To clarify my convention comment; it is only convention that mathematicians have decided that any given set of properties are useful or interesting to be used pervasively and alternate number systems are not. To any person who is asking why .9 repeating is 1 those reasons are entirely outside the scope of their knowledge and so entirely unrelated to the question; is it not true that the answer is "based on the axioms of which this number line is created due to complex reasons that you can't possibly know at this stage, this is effectively decided to be true as an axiom".
If one of these alternate number lines was in common use and reals were only uncommonly used then I'm pretty sure there would be tons of people saying "why is .99.. NOT equal to 1 in ($reals-replacement)" and all the people who have been taught it would sigh condescendingly because they believe it is just an intuitively obvious feature of numbers and not simply an axiom of the number line they are choosing to use. I'm interested in whether you disagree with that sentiment.
Re: your proof, trivially could be refused based on the other number line not being closed over division, then there would simply be no such number Z=(X+Y)/2 for X=.999.. and and Y=1. Then the proof where be like those silly ones where someone uses a /0 and arrives at a contradiction. Actually it seems that reals themselves already are not closed over division since x/0 doesn't result in a real.
My point was that .999=1 is a direct result from the fact that the number line is continuas.
Imagine that you had a number system which does have holes in it (for example, the difference between .999... and 1). Now, take 2 object, one at mass .999..., and the other one of mass 1. In this hypothetical system, these objects would have a different mass. Now, take a third object, whose mass is between the 2 of them.
I consider it a critical property of our number system that we can be guaranteed to have a number representing the third mass.
As a historic note, the Greeks believed that any 2 numbers could be expressed as a multiple of some unit. For example 2 and .333... could both be expressed as multiples of 1/3. Eventually, they proved that this system could not accuratly moddel their world (the diagonal of a square for example).
I suspect that if we did have a number system such as you described, we would eventually discover that it did not work well enough for us, and switch to a continuas one.
> The hyper-reals of non-standard analysis are, in fact, a strict superset of the reals.
And .9999==1 in the hyper-reals
>>If it were, then any operation involving only real numbers
>>would behave identically to the real number system.
>Why is that a problem? Seems to me that that's desirable.
Consider the expression ".999... - 1" in the real number system. As has been established earlier, this simplifies to 0.
Consider the same expression ".999... - 1" in the hyper-real number system. As all the numbers are real, this expression simplifies just as it would in the real number system. Meaning that ".999... - 1 =0" in the hyper-real system. Adding 1 to both sides ".999... = 1" in the hyper-real system.
One could also argue that because the statement ".999... = 1" is true in the real number system, and the hyper-reals are a superset of the reals. Then ".999... = 1" is true in the hyper-reals.
Ah, I see your point. I was trying to address the more general "intuition" that people have about this, that 0.9999... falls short of 1.0, and that there's some sort of infinitesimal between them. You can use that to talk about the hyper-reals, but, as you say, it still doesn't "solve" the "problem" that people perceive.
So you're right, and I was answering a different (although related) question.
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.
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
For #3, try to find Halmos’s book Naive Set Theory. It has a very nice and accessible explanations.
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").
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.
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.
Thanks for a detailed explanation. The one on matrices helps too even though others also posted the same idea.
>> We have a universe of discourse that has a bunch of atomic things. A "set" is then a collection of things
This is where I get stuck. In set theory, they define numbers (and addition) starting from a set and element-of operator. I cannot imagine a bunch of atomic things without imagining numbers/counting first. (PS: "collection" would be another word for a set, so does not help to define it.) So it seems to me that unless set and element-of are left undefined (as others here have suggested), natural numbers are more fundamental than sets.
>> 0,9999... = 1
What I am getting is that the "limit" is implicit in the statement above even though often unstated. The ellipsis is what signifies the limit there, being otherwise not mathematically defined.
>> Call that collection A
I get the precise definition of sack B. I am missing it for sac A. I understand what happens on a given cube, and that cubes in sac A are numbered 1 though infinity. Is there anything to say about the numbers written on two different cubes in sac A?
Basically what you are saying is that sac A cannot have all possible combinations like sac B is defined to have. I am not sure why that is. I see that you just added a comment on another related comment from me. I'll read that first.
>> We have a universe of discourse that has a bunch
>> of atomic things. A "set" is then a collection of things
> In set theory, they define numbers (and addition)
> starting from a set and element-of operator. I cannot
> imagine a bunch of atomic things without imagining
> numbers/counting first.
The point is that you can start with nothing, and define the set that has nothing in it. That's {}. Now we have one thing. We can define a set containing that, so we have { {} }. Now we have two things, and we can definea set containing both of them, and so on.
This is what happens when people want to construct a model of numbers and arithmetic using set theory as the basis. It proves that we can useset theory as the foundation. It doesn't mean it's a sensible thing to do in real life- it's a lot like programming in machine code.
> So it seems to me that unless set and element-of are left
> undefined (as others here have suggested), natural numbers
> are more fundamental than sets.
What are numbers? What is your model for numbers? What is "723"?
>> 0,9999... = 1
> What I am getting is that the "limit" is implicit in the
> statement above even though often unstated. The
> ellipsis is what signifies the limit there, being otherwise
> not mathematically defined.
So what is your question? You seem to be saying that when you write down "0.99999..." that is intended to represent the limit of the sequence 0.9, 0.99, 0.999, ... Define what you mean by limit. Once you make a careful definition of "limit" wou find that the limit of the above sequence is 1.
>> Call that collection A
> I get the precise definition of sack B. I am missing it for sack A.
Sack A has a cube with the number 1 written on it. And it has a cube with the number 2 written on it. And it has a cube with the number 3 written on it. And it has a cube with the number 4 written on it. And it has a cube with the number 5 written on it. And so on.
> I understand what happens on a given cube,
I assume you mean a cube in sack B.
> ... and that cubes in sack A are numbered 1 though infinity.
You can't say that with precision, because it is infinity you are struggling with. What you can say is that for every number n there is a cube with n on it, and that every cube has exactly one number written on it.
This level of detail matters.
> Is there anything to say about the numbers written on
> two different cubes in sack A?
Yes - they're different.
> Basically what you are saying is that sac A cannot have
> all possible combinations like sack B is defined to have.
No. I'm saying I have two collections of objects. I have very carefully defined what these objects are. And I'm saying that you cannot pair them off, one-to-one, without having things from sack B left over.
> I am not sure why that is.
I haven't proved it yet, so I haven't explained why this is so. I have merely claimed it is true to see where your understanding fails so far.
>> You can't say that with precision, because it is infinity you are struggling with. What you can say is that for every number n there is a cube with n on it, and that every cube has exactly one number written on it.
This may be nailing it -- struggling with infinity.
When you say "every number n there is a cube with n on it", what does "every" mean. Does every number include infinity? Or should not not consider infinity to be a number? If the latter, this is probably where I went wrong.
> When you say "every number n there is a cube with n on it",
> what does "every" mean.
To be more precise, every finite number.
> Does every number include infinity?
No, in these sorts of discussions infinity is never considered to be a number. You need explicitly to be discussing transfinite arithmetic, and we're not.
> Or should not not consider infinity to be a number?
> If the latter, this is probably where I went wrong.
Absolutely you should not be thinking of infinity as a number.
>> The point is that you can start with nothing, and define the set that has nothing in it. That's {}. Now we have one thing. We can define a set containing that, so we have { {} }. Now we have two things, and we can definea set containing both of them, and so on.
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.
Indeed, I was aware of this. And as you said, it does not really answer since it just leads to the same question for 0.3333...
I understand the piece on calculus too. My maths background is mostly related to engineering mathematics, and I have generally excelled at it. I later discovered that I am not clear on some "pure" mathematics things. Hypothetically, if counting were taught to me in adulthood, I may have asked a lot of relevant questions that I did not ask in my pre-school and later lost them without knowing them. I am rediscovering some of these questions now.
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.